Welcome, students! In this guide we cover the complete ASSEB Class 12 Physics Chapter 9 Question Answer — Ray Optics and Optical Instruments (English Medium). The chapter explains how light, treated as rays, behaves at mirrors, plane and curved surfaces, prisms and lenses, and how those laws are used to design microscopes, telescopes and the human eye. Each section gives the formulas, a labelled SVG ray diagram, and fully worked exercise-style solutions matched to the Assam State School Education Board syllabus.
Chapter Summary
Geometrical optics is built on three simple laws: light travels in straight lines in a homogeneous medium, the angle of incidence equals the angle of reflection at a smooth surface, and at a transparent boundary the bending of light obeys Snell’s law $n_1\sin\theta_1 = n_2\sin\theta_2$. From these we derive the mirror formula, the lens-maker’s formula, the thin-lens equation, the prism formula and the magnifying powers of optical instruments. The same ideas explain total internal reflection in optical fibres, dispersion of white light by a prism, the blue colour of the sky by Rayleigh scattering, and the working of the simple microscope, compound microscope and astronomical telescope.
The Cartesian sign convention is used throughout: all distances are measured from the pole of the mirror or the optical centre of the lens, distances measured in the direction of the incident light are positive and those opposite to it are negative; heights above the principal axis are positive, below are negative. With this convention the focal length of a concave mirror and a convex lens is taken positive in many books and negative in others — Class 12 NCERT (followed by ASSEB) takes f negative for a concave mirror and positive for a convex lens. The sign convention turns geometry problems into pure algebra.
Key Formulas at a Glance
| Quantity | Formula |
| Mirror formula | $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ |
| Magnification (mirror) | $m=-\dfrac{v}{u}=\dfrac{h’}{h}$ |
| Focal length and radius | $f=\dfrac{R}{2}$ |
| Snell’s law | $n_1\sin\theta_1=n_2\sin\theta_2$ |
| Critical angle | $\sin\theta_c=\dfrac{1}{n}$ |
| Refraction at spherical surface | $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ |
| Lens-maker’s formula | $\dfrac{1}{f}=(n-1)\!\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ |
| Thin-lens formula | $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ |
| Magnification (lens) | $m=\dfrac{v}{u}=\dfrac{h’}{h}$ |
| Lenses in contact | $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ |
| Power of a lens | $P=\dfrac{1}{f\,(\text{m})}$ (diopter) |
| Prism formula | $n=\dfrac{\sin\!\big((A+\delta_m)/2\big)}{\sin(A/2)}$ |
| Simple microscope | $M=1+\dfrac{D}{f}$ |
| Compound microscope | $M=\dfrac{L}{f_o}\!\left(1+\dfrac{D}{f_e}\right)$ |
| Astronomical telescope | $M=-\dfrac{f_o}{f_e}$ |
Ray Diagrams (SVG)
1. Image formation by a concave mirror (object beyond C). A ray parallel to the axis reflects through F; a ray through C retraces its path; the two reflected rays meet at the real, inverted, diminished image.
2. Image formation by a convex lens (object beyond 2F). The image is real, inverted and diminished, formed between F’ and 2F’.
3. Refraction and total internal reflection. When the angle of incidence in a denser medium exceeds the critical angle $\theta_c$, all light is reflected back.
4. Refraction through a prism (minimum deviation).
5. Compound microscope. The objective forms a real, magnified image inside the focal length of the eyepiece, which acts as a simple microscope.
6. Astronomical telescope (normal adjustment). Parallel rays from a distant object converge at the common focus of the objective and eyepiece.
Exercise — Numerical Solutions
Q1. A small candle 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how should the screen be moved?
Answer: Here $u=-27$ cm, $R=-36$ cm so $f=R/2=-18$ cm, $h=2.5$ cm.
$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow\frac{1}{v}=\frac{1}{-18}-\frac{1}{-27}=\frac{-3+2}{54}=\frac{-1}{54}$$
So $v=-54$ cm — the screen must be placed 54 cm in front of the mirror. Magnification $m=-v/u=-(-54)/(-27)=-2$, so the image is real, inverted, magnified, of size $|m|\times h=5.0$ cm. As the candle moves closer to F (18 cm), the image moves further away towards infinity, so the screen must be moved away. When the object is between F and P no real image forms on the screen.
Q2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer: $u=-12$ cm, $f=+15$ cm.
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}\Rightarrow v=\frac{60}{9}=6.7\text{ cm}$$
The image is virtual, behind the mirror, at 6.7 cm. $m=-v/u=-(6.7)/(-12)=+0.56$. Image height $=0.56\times4.5=2.5$ cm — erect and diminished. As the needle is moved farther, the image moves toward the focus and continually shrinks.
Q3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer: $n_w=\dfrac{\text{real depth}}{\text{apparent depth}}=\dfrac{12.5}{9.4}\approx1.33$.
For the new liquid, apparent depth $=12.5/1.63=7.67$ cm. Microscope must be moved down by $9.4-7.67=1.73$ cm.
Q4. Figures show refraction through a glass-air interface for two prisms of the same glass. Determine the refractive index for the situations and the value of the angle of incidence in the second figure if total internal reflection takes place. (Standard NCERT figure: face shows i=60°, r=35°; second prism has 45° face.)
Answer: $n=\dfrac{\sin60^\circ}{\sin35^\circ}=\dfrac{0.866}{0.574}=1.51$.
Critical angle: $\sin\theta_c=1/1.51=0.662\Rightarrow\theta_c=41.4^\circ$. Since the prism face is 45° > 41.4°, the ray undergoes total internal reflection.
Q5. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33.
Answer: Light emerges only inside a cone whose half-angle is the critical angle. $\sin\theta_c=1/1.33=0.752\Rightarrow\theta_c=48.75^\circ$, $\tan\theta_c=1.137$. Radius $r=h\tan\theta_c=80\times1.137=91$ cm $=0.91$ m. Area $=\pi r^2=\pi(0.91)^2=2.6$ m$^2$.
Q6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (n = 1.33) predict the new minimum deviation.
Answer: Using $n=\dfrac{\sin((A+\delta_m)/2)}{\sin(A/2)}=\dfrac{\sin50^\circ}{\sin30^\circ}=\dfrac{0.766}{0.5}=1.532$.
In water $n_{rel}=1.532/1.33=1.152$. Then $\sin((60+\delta_m)/2)=1.152\times\sin30^\circ=0.576$, $(60+\delta_m)/2=35.16^\circ$, $\delta_m=10.3^\circ$.
Q7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer: For a symmetric biconvex lens $R_1=R$, $R_2=-R$.
$$\frac{1}{f}=(n-1)\!\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(n-1)}{R}$$
$\Rightarrow R=2(n-1)f=2(0.55)(20)=22$ cm.
Q8. A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?
Answer: The point P acts as a virtual object: $u=+12$ cm. (a) Convex, $f=+20$: $\frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}\Rightarrow v=7.5$ cm. Beam converges 7.5 cm beyond the lens. (b) Concave, $f=-16$: $\frac{1}{v}=\frac{1}{-16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}\Rightarrow v=48$ cm.
Q9. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens?
Answer: $u=-14$, $f=-21$. $\frac{1}{v}=\frac{1}{-21}+\frac{1}{-14}=-\frac{2+3}{42}=-\frac{5}{42}\Rightarrow v=-8.4$ cm. Virtual, on same side, $m=v/u=(-8.4)/(-14)=0.6$, image height $=1.8$ cm — erect, diminished. As the object recedes, the image stays virtual, moves toward the focus and shrinks toward zero size.
Q10. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness.
Answer: $\frac{1}{f}=\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60}\Rightarrow f=-60$ cm. The combination is a diverging lens of focal length 60 cm.
Q11. A compound microscope consists of an objective of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), (b) at infinity? In each case calculate the magnification produced by the microscope.
Answer: (a) For final image at $D=25$ cm, eyepiece acts as simple microscope with $v_e=-25$, $f_e=6.25$: $\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25}=-\frac{1+4}{25}=-\frac{1}{5}\Rightarrow u_e=-5$ cm. So intermediate image is 5 cm in front of eyepiece, hence at $15-5=10$ cm from the objective: $v_o=10$ cm. Then $\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{10}-\frac{1}{2}=-\frac{4}{10}\Rightarrow u_o=-2.5$ cm. Magnification $M=\frac{v_o}{|u_o|}\!\left(1+\frac{D}{f_e}\right)=\frac{10}{2.5}(1+25/6.25)=4\times5=20$. (b) For final image at infinity, $u_e=-f_e=-6.25$ cm; then $v_o=15-6.25=8.75$ cm, $\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2}=-\frac{6.75}{17.5}\Rightarrow u_o=-2.59$ cm. $M=\frac{v_o}{|u_o|}\frac{D}{f_e}=\frac{8.75}{2.59}\times\frac{25}{6.25}=3.38\times4=13.5$.
Q12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer: $u_o=-0.9$ cm, $f_o=0.8$ cm. $\frac{1}{v_o}=\frac{1}{0.8}-\frac{1}{0.9}=\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}\Rightarrow v_o=7.2$ cm. Eyepiece: image at $D=25$ cm, $f_e=2.5$, gives $u_e=-25/11=-2.27$ cm. Separation $L=v_o+|u_e|=7.2+2.27=9.47$ cm. Magnification $M=\frac{v_o}{|u_o|}(1+D/f_e)=\frac{7.2}{0.9}(1+10)=8\times11=88$.
Q13. A small telescope has an objective of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer: $M=-f_o/f_e=-144/6=-24$ (size 24×, image inverted). Length $L=f_o+f_e=144+6=150$ cm.
Q14. (a) A giant refracting telescope has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? (Diameter of moon = 3.48 × 10⁶ m, distance = 3.8 × 10⁸ m.)
Answer: (a) $M=f_o/f_e=1500/1=1500$. (b) Angle subtended by moon $\alpha=3.48\times10^6/3.8\times10^8=9.16\times10^{-3}$ rad. Image size $d=f_o\alpha=15\times9.16\times10^{-3}=0.137$ m $=13.7$ cm.
Additional Practice Questions
Q15. State the laws of reflection of light.
Answer: (i) The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane. (ii) The angle of incidence equals the angle of reflection. These laws hold for both plane and curved mirrors.
Q16. Define the principal focus and focal length of a concave mirror.
Answer: The principal focus F is the point on the principal axis where rays parallel and close to the axis converge after reflection from a concave mirror (or appear to diverge from, for a convex mirror). The distance PF from the pole to the focus is the focal length, $f=R/2$.
Q17. Derive the relation $f=R/2$ for a concave mirror.
Answer: Consider a paraxial ray AM parallel to the axis striking a concave mirror at M and reflecting through F. The radius CM is normal to the mirror, so the angle of incidence and reflection on either side of CM are equal — call each $\theta$. In $\triangle$MCF, the exterior angle at M between CM and MF equals $2\theta$ at C. For paraxial rays $\tan\theta\approx\theta$ and $\tan2\theta\approx2\theta$, so $MP/PC\approx2\,MP/PF$, giving $PF=PC/2$, i.e. $f=R/2$.
Q18. State Snell’s law and define refractive index.
Answer: When light passes from medium 1 to medium 2, $n_1\sin\theta_1=n_2\sin\theta_2$. The refractive index of medium 2 with respect to medium 1 is $n_{21}=\sin\theta_1/\sin\theta_2=v_1/v_2$, the ratio of speeds. Absolute refractive index of a medium is $n=c/v$.
Q19. What is total internal reflection? State the conditions for it to occur.
Answer: When a light ray travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle, the ray is reflected back entirely into the denser medium without refraction. Conditions: (i) light must travel from denser to rarer medium; (ii) angle of incidence > critical angle $\theta_c$, where $\sin\theta_c=1/n$.
Q20. List two practical applications of total internal reflection.
Answer: (i) Optical fibres for telecommunication and endoscopy: light is trapped in the core by repeated TIR at the core–cladding boundary. (ii) Totally reflecting prisms in periscopes and binoculars use 45°-45°-90° glass prisms which give 100% reflection without the silver tarnishing of mirrors. (iii) Mirage formed in deserts when hot air near the ground has a smaller refractive index than the cooler air above it.
Q21. Derive the lens-maker’s formula for a thin convex lens in air.
Answer: Apply refraction at a spherical surface twice. At first surface (air→glass): $\dfrac{n}{v_1}-\dfrac{1}{u}=\dfrac{n-1}{R_1}$. At second surface (glass→air), the image $v_1$ acts as object: $\dfrac{1}{v}-\dfrac{n}{v_1}=\dfrac{1-n}{R_2}$. Adding the two: $\dfrac{1}{v}-\dfrac{1}{u}=(n-1)\!\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$. For an object at infinity ($u=\infty$, $v=f$), $\dfrac{1}{f}=(n-1)\!\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.
Q22. Define the power of a lens. State its SI unit.
Answer: The power of a lens is its ability to converge or diverge light, defined as the reciprocal of its focal length expressed in metres: $P=1/f$. Unit: dioptre (D), $1\text{ D}=1\text{ m}^{-1}$. Convex lens has positive power, concave negative.
Q23. Two thin lenses of focal length $f_1$ and $f_2$ are placed in contact. Show that the equivalent focal length is given by $1/f=1/f_1+1/f_2$.
Answer: Let object O be at distance u from the combination. Lens 1 forms image at $v_1$: $\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$. This image is the object for lens 2: $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$. Adding: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$. Comparing with $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ gives $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$. In terms of powers, $P=P_1+P_2$.
Q24. Derive the prism formula $n=\dfrac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.
Answer: For a prism of angle A, total deviation $\delta=(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)=(i+e)-A$ (since $r_1+r_2=A$). At minimum deviation, $i=e$ and $r_1=r_2=r$, so $A=2r$ and $\delta_m=2i-A$, i.e. $i=(A+\delta_m)/2$, $r=A/2$. Snell’s law at first face: $n=\sin i/\sin r=\sin((A+\delta_m)/2)/\sin(A/2)$.
Q25. What is meant by dispersion of light? Why does a prism disperse white light?
Answer: The splitting of white light into its constituent colours (VIBGYOR) on passing through a refracting medium is called dispersion. It occurs because the refractive index of the prism material depends on wavelength — it is greater for violet (shorter wavelength) than for red, so violet bends more than red. The result is a spectrum.
Q26. Why is the sky blue and the setting sun red?
Answer: Both are explained by Rayleigh scattering: the intensity of light scattered by molecules small compared to its wavelength varies as $1/\lambda^4$. Blue light is scattered ~10 times more strongly than red, so the daytime sky appears blue. At sunset the path through the atmosphere is long; most blue is scattered away, and the transmitted light reaching the observer is rich in red, making the Sun look red.
Q27. Explain the working of a simple microscope and derive its magnifying power.
Answer: A simple microscope is a single convex lens of short focal length. The object is placed within its focal length so the lens produces an erect, virtual, magnified image at the least distance of distinct vision $D$. With $v=-D$ and $f$ positive, $\frac{1}{-D}-\frac{1}{u}=\frac{1}{f}\Rightarrow\frac{1}{u}=-\frac{1}{D}-\frac{1}{f}$. Magnifying power $M=\dfrac{D}{|u|}=1+\dfrac{D}{f}$. For relaxed eye (image at infinity), $M=D/f$.
Q28. With a labelled diagram describe the construction and working of an astronomical telescope. Derive its magnifying power in normal adjustment.
Answer: The telescope has two convex lenses — a long-focus objective $f_o$ and a short-focus eyepiece $f_e$ — mounted on a tube. Parallel rays from a distant object are focused by the objective at its focus; the eyepiece, placed so that this focus coincides with its own focus, gives a final image at infinity (normal adjustment). If $\alpha$ is the angle subtended by the object at the objective and $\beta$ the angle subtended by the final image at the eye, $\alpha=h/f_o$ and $\beta=h/f_e$, so magnifying power $M=\beta/\alpha=f_o/f_e$ (with sign $-f_o/f_e$ for inversion). Tube length $L=f_o+f_e$.
Q29. Distinguish between linear magnification and angular magnification.
Answer: Linear magnification $m=h’/h=v/u$ is the ratio of image size to object size, used for cameras and projectors where image size on screen matters. Angular magnification $M=\beta/\alpha$ is the ratio of the angle subtended at the eye by the image to the angle subtended by the object placed at $D$; it measures the apparent enlargement and is used for microscopes and telescopes because the actual image may be at infinity.
Q30. A convex lens has a focal length of 20 cm in air. What is its focal length when immersed in water ($n_w=1.33$)? Take $n_g=1.50$.
Answer: In air $\frac{1}{f_a}=(n_g-1)(\frac{1}{R_1}-\frac{1}{R_2})$, in water $\frac{1}{f_w}=(\frac{n_g}{n_w}-1)(\frac{1}{R_1}-\frac{1}{R_2})$. Ratio $\frac{f_w}{f_a}=\frac{n_g-1}{n_g/n_w-1}=\frac{0.50}{(1.50/1.33)-1}=\frac{0.50}{0.128}=3.91$. So $f_w=3.91\times20=78$ cm.
Q31. Two lenses of powers +5 D and –3 D are placed in contact. Find the equivalent focal length of the combination.
Answer: $P=P_1+P_2=5-3=2$ D, $f=1/P=0.5$ m = 50 cm (converging).
Q32. Light from a point source in air falls on a spherical glass surface (n = 1.5, R = 20 cm). The distance of the source from the surface is 100 cm. Where is the image formed?
Answer: $u=-100$, $n_1=1$, $n_2=1.5$, $R=+20$.
$$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\Rightarrow\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}$$
$\frac{1.5}{v}=0.025-0.01=0.015\Rightarrow v=100$ cm. Image is 100 cm to the right of the surface inside glass.
Q33. A ray of light passing through an equilateral prism suffers a minimum deviation of 30°. Find the refractive index of the prism material.
Answer: $A=60^\circ$, $\delta_m=30^\circ$. $n=\sin45^\circ/\sin30^\circ=(0.707)/(0.5)=1.414$.
Q34. The refractive index of glass for red light is 1.514 and for violet light is 1.523. Find the angular dispersion produced by a thin prism of angle 8°.
Answer: For a thin prism, $\delta=(n-1)A$. $\delta_v-\delta_r=(n_v-n_r)A=(1.523-1.514)\times8^\circ=0.072^\circ$.
Q35. A small bulb is at the bottom of a swimming pool 4 m deep. Find the radius of the circle of light formed at the surface (water n = 4/3).
Answer: $\sin\theta_c=3/4=0.75\Rightarrow\theta_c=48.6^\circ$, $\tan\theta_c=1.134$. Radius $r=h\tan\theta_c=4\times1.134=4.54$ m.
Q36. Why can you not photograph yourself in a plane mirror by setting the camera at the position of the eye?
Answer: The image is virtual, formed behind the mirror at the same distance as the object. To photograph it, the camera must be focused on a point behind the mirror, not at the mirror surface — the lens forms an image of light rays that appear to come from behind the glass.
Q37. Why does a diamond sparkle?
Answer: Diamond has a very high refractive index ($n=2.42$) so its critical angle is small ($\theta_c\approx24.4^\circ$). Most rays entering a faceted diamond strike the inner surfaces at angles greater than $\theta_c$ and are totally internally reflected several times before emerging, producing the brilliant sparkle. Cuts are designed so that light entering the top is reflected and dispersed back to the viewer.
Q38. State two differences between a microscope and a telescope.
Answer: (i) A microscope is used for tiny objects placed close by; a telescope for large objects placed far away. (ii) Objective of a microscope has very short focal length and small aperture; objective of a telescope has long focal length and large aperture. (iii) Length of a microscope is greater than the sum $f_o+f_e$, while length of a telescope in normal adjustment is exactly $f_o+f_e$.
Q39. Calculate the resolving power of a telescope of objective diameter 200 mm for light of wavelength 550 nm.
Answer: $\theta_{min}=1.22\lambda/D=1.22\times550\times10^{-9}/0.20=3.36\times10^{-6}$ rad. Resolving power $=1/\theta_{min}=2.98\times10^{5}$.
Q40. A glass slab of thickness 6 cm is placed over a printed page. By what amount will the print appear to be raised when viewed normally? ($n=1.5$).
Answer: Apparent shift $=t(1-1/n)=6(1-1/1.5)=6\times1/3=2.0$ cm.
Glossary
| Term | Meaning |
| Pole (P) | The geometrical centre of a spherical mirror. |
| Centre of curvature (C) | Centre of the sphere of which the mirror is a part. |
| Principal axis | Line passing through the pole and the centre of curvature. |
| Principal focus (F) | Point where paraxial rays parallel to the axis converge (or appear to diverge from) after reflection or refraction. |
| Focal length (f) | Distance from pole/optical centre to the principal focus. |
| Cartesian sign convention | Distances measured from P along incident-light direction are positive; opposite direction negative; heights above axis positive. |
| Refractive index (n) | Ratio of speed of light in vacuum to speed in the medium, $n=c/v$. |
| Critical angle ($\theta_c$) | Angle of incidence in denser medium for which the angle of refraction in the rarer medium is 90°; $\sin\theta_c=1/n$. |
| Total internal reflection | Complete reflection of light back into the denser medium when $i>\theta_c$. |
| Apparent depth | Depth at which an object immersed in a denser medium appears, $d’=d/n$. |
| Lens-maker’s formula | $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$. |
| Power (P) | Reciprocal of focal length in metres; unit dioptre. |
| Angle of deviation | Angle between emergent ray and incident ray for a prism, $\delta=(i+e)-A$. |
| Minimum deviation | Smallest value of deviation; occurs when $i=e$ and the ray inside the prism is parallel to the base. |
| Dispersion | Splitting of white light into colours due to wavelength-dependent refractive index. |
| Rayleigh scattering | Scattering by particles much smaller than the wavelength; intensity $\propto 1/\lambda^4$. |
| Magnifying power | Ratio of the angle subtended at the eye by the image to that subtended by the object placed at $D$. |
| Normal adjustment (telescope) | Setting in which the final image is formed at infinity for a relaxed eye; $L=f_o+f_e$. |
| Resolving power | Ability of an instrument to distinguish two close objects; for a telescope, $\theta_{min}=1.22\lambda/D$. |
| Optical fibre | Thin transparent fibre that guides light along its length by repeated total internal reflection. |
Detailed Concept Notes
1. Reflection of Light by Spherical Mirrors
A spherical mirror is a part of a hollow sphere whose inner or outer surface is silvered. If the inner (concave) surface is reflecting it is a concave mirror; if the outer (convex) surface is reflecting it is a convex mirror. The most important points are the pole P (geometrical centre of the reflecting surface), the centre of curvature C (centre of the sphere), the radius of curvature R = PC, and the principal focus F. For a paraxial beam (close to the axis and almost parallel to it), the focal length and radius are related by $f=R/2$. In Cartesian convention, $R$ and $f$ are negative for a concave mirror and positive for a convex mirror.
The mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ together with the magnification $m=-v/u=h’/h$ tells us where the image forms and how big it is. Six standard cases for a concave mirror are: (i) object at infinity → image at F, real, inverted, highly diminished; (ii) object beyond C → between F and C, real, inverted, diminished; (iii) object at C → at C, real, inverted, same size; (iv) object between C and F → beyond C, real, inverted, magnified; (v) object at F → at infinity, very large; (vi) object between F and P → behind the mirror, virtual, erect, magnified. A convex mirror always forms a virtual, erect, diminished image between P and F, regardless of the object position. Convex mirrors are therefore preferred as rear-view mirrors in vehicles because they always give an erect image and have a wide field of view.
Q41. Why are concave mirrors used as shaving mirrors?
Answer: When the face is held within the focal length of a concave mirror, the mirror produces a virtual, erect and magnified image, allowing close inspection. The same principle is used in dentists’ mirrors and headlight reflectors (where the bulb is placed at F to produce a parallel beam).
Q42. An object is placed at the centre of curvature of a concave mirror. Where is the image formed and what is its nature?
Answer: The image forms at C itself, with the same size as the object, real and inverted. This is the basis of the “$u=v=R$” check used in the laboratory to measure the radius of curvature of a concave mirror.
2. Refraction and the Speed of Light
The bending of light at the boundary between two transparent media is called refraction. It happens because the speed of light is different in different media. The refractive index of a medium is $n=c/v$, where $c=3\times10^8$ m/s is the speed of light in vacuum and $v$ is its speed in the medium. Snell’s law $n_1\sin\theta_1=n_2\sin\theta_2$ is a consequence of Fermat’s principle and of the wave nature of light. When a ray enters a denser medium it bends toward the normal; when it goes from denser to rarer it bends away.
Apparent depth is a familiar refraction effect. A pond looks shallower than it is, a coin under water seems closer to the surface, and a stick partially immersed looks bent at the surface. For an object at real depth $d$ viewed normally through a medium of refractive index $n$, the apparent depth is $d’=d/n$. The apparent shift, useful in glass-slab problems, is $\Delta=d-d’=d(1-1/n)$ or for a slab of thickness $t$, the apparent shift in the position of an object behind the slab is $t(1-1/n)$.
Q43. A coin lies at the bottom of a beaker filled with water (n = 4/3) to a depth of 12 cm. What is its apparent depth?
Answer: $d’=d/n=12/(4/3)=9$ cm. The coin appears 3 cm closer to the surface.
Q44. A ray of light is incident at an angle of 60° on the surface of water (n = 1.33). Find the angle of refraction.
Answer: $\sin r=\sin60^\circ/n=0.866/1.33=0.651$, $r=40.6^\circ$.
3. Total Internal Reflection (TIR)
When light travels from a denser medium to a rarer medium, the angle of refraction is greater than the angle of incidence. As the angle of incidence increases, the refracted ray bends further from the normal until, at a particular angle of incidence called the critical angle $\theta_c$, it just grazes the boundary ($r=90^\circ$). For angles greater than $\theta_c$, refraction stops and the light is reflected back entirely into the denser medium — this is total internal reflection. The critical angle satisfies $n_1\sin\theta_c=n_2\sin90^\circ$, i.e. $\sin\theta_c=n_2/n_1=1/n$ when the rarer medium is air.
For water, $\theta_c=48.75^\circ$; for crown glass, $\theta_c\approx41^\circ$; for diamond it is only $24.4^\circ$, which is why a well-cut diamond traps and reflects so much light internally and sparkles. Practical applications include optical fibres for high-speed digital communication and medical endoscopy, totally reflecting prisms (used in periscopes, binoculars and projectors as a 100% efficient mirror that does not tarnish), and the formation of mirages on hot roads where layered air of different refractive index acts like a denser-to-rarer interface.
Q45. The critical angle for a certain glass is 42°. Find its refractive index.
Answer: $n=1/\sin42^\circ=1/0.669=1.494$.
Q46. How does an optical fibre work? State two of its uses.
Answer: An optical fibre consists of a high-index glass core surrounded by a lower-index cladding. Light entering the core at an angle larger than the critical angle of the core–cladding boundary is totally internally reflected at every encounter with the cladding and is guided along the fibre with negligible loss, even when the fibre is bent. Uses: (i) high-bandwidth telecommunications; (ii) medical endoscopes for non-invasive viewing of internal organs; (iii) fibre-optic lighting and sensors.
4. Refraction at a Spherical Surface
Consider a single spherical surface of radius $R$ separating media of refractive indices $n_1$ (where the object is) and $n_2$. With the Cartesian sign convention and paraxial approximation, the relation between object distance $u$, image distance $v$ and $R$ is
$$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$$
This single equation is the parent of both the lens-maker’s formula and the thin-lens formula. Apply it once to each surface of a thin lens, add and you get the full lens formula. The same equation applied to two parallel water/glass interfaces gives the apparent-depth result.
Q47. A glass sphere of radius 10 cm and refractive index 1.5 has a small bubble at its centre. Where does the bubble appear when viewed from outside?
Answer: $u=-10$, $n_1=1.5$, $n_2=1$, $R=-10$. $\frac{1}{v}-\frac{1.5}{-10}=\frac{1-1.5}{-10}\Rightarrow\frac{1}{v}=0.05-0.15=-0.10\Rightarrow v=-10$ cm. The bubble appears at the centre, at its true position. (For an object at the centre of a sphere, the image always coincides with the object since rays are radial.)
5. Thin Lenses
A lens is a transparent medium bounded by two refracting surfaces, at least one of which is curved. Convex (converging) lenses are thicker at the centre; concave (diverging) lenses are thicker at the edges. With the lens-maker’s formula $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$ and the thin-lens equation $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, all standard problems can be solved. The magnification is $m=v/u=h’/h$.
An important fact often tested: when a lens is immersed in a medium denser than that for which the formula above was derived, its focal length increases (because the relative refractive index $n$ falls towards 1). If a glass lens ($n_g=1.5$) is dipped in water ($n_w=4/3$), its power becomes about a quarter of its power in air. If the surrounding liquid has the same refractive index as the lens, the lens becomes optically invisible and has infinite focal length.
Q48. The focal length of a thin convex lens of glass ($n=1.5$) in air is 20 cm. What will its focal length be in a medium of refractive index 1.65?
Answer: $\frac{f_m}{f_a}=\frac{n_g-1}{n_g/n_m-1}=\frac{0.5}{(1.5/1.65)-1}=\frac{0.5}{-0.091}=-5.5$. So $f_m=-5.5\times20=-110$ cm. The lens behaves as a diverging lens of focal length 110 cm in this denser medium.
Q49. A convex lens of focal length 20 cm and a concave lens of focal length 40 cm are placed in contact. Find the power and focal length of the combination.
Answer: $P_1=100/20=5$ D, $P_2=-100/40=-2.5$ D, $P=2.5$ D, $f=100/2.5=40$ cm — converging.
Q50. The radii of curvature of the two surfaces of a biconvex lens are 20 cm each. If $n=1.5$, find its focal length.
Answer: $R_1=20$, $R_2=-20$. $\frac{1}{f}=(0.5)(\frac{1}{20}+\frac{1}{20})=\frac{0.5\times2}{20}=0.05$. $f=20$ cm.
6. Prism: Deviation and Dispersion
A prism is a transparent solid bounded by two flat refracting surfaces inclined at an angle A (the refracting angle). When a ray enters one face at angle $i$ it bends toward the normal, travels through the glass at angle $r_1$, hits the second face at $r_2$ (with $r_1+r_2=A$) and emerges at angle $e$. The total angle through which the ray is bent — its deviation — is $\delta=(i+e)-A$. As $i$ is varied, $\delta$ first decreases, reaches a minimum value $\delta_m$ when $i=e$ (and the ray inside the prism is parallel to the base), and then increases again. At minimum deviation $r_1=r_2=A/2$ and $i=(A+\delta_m)/2$, leading to the prism formula $n=\dfrac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.
For a thin prism (small A) the deviation simplifies to $\delta=(n-1)A$. Since $n$ depends on wavelength (it is greater for violet, smaller for red), different colours of white light deviate by different amounts, producing dispersion. The angular separation between violet and red emerging rays is the angular dispersion $\theta=(n_v-n_r)A$. The dispersive power of a prism is $\omega=\dfrac{n_v-n_r}{n_y-1}$, where $n_y$ is the refractive index for mean (yellow) light.
Q51. A thin prism of crown glass ($n=1.52$) has refracting angle 5°. Find the deviation produced.
Answer: $\delta=(n-1)A=(0.52)(5^\circ)=2.6^\circ$.
Q52. The refractive indices of crown glass for red and violet are 1.514 and 1.523 respectively. Calculate the dispersive power.
Answer: $n_y=(1.514+1.523)/2=1.5185$. $\omega=(1.523-1.514)/(1.5185-1)=0.009/0.5185=0.0174$.
Q53. What is an achromatic combination of prisms? Why is it useful?
Answer: An achromatic combination consists of two prisms of different glasses (typically crown and flint) arranged so that the angular dispersion of one cancels that of the other but a net mean deviation remains. Such a combination produces deviation without dispersion and is used in telescope objectives and camera lenses to eliminate chromatic aberration.
7. Scattering of Light
When light passes through a medium containing small particles, it is scattered in different directions. For particles much smaller than the wavelength of light, the scattered intensity follows Rayleigh’s law: $I\propto 1/\lambda^4$. Blue light is scattered about 10 times more than red, so the daytime sky is blue. At sunrise and sunset, the path through the atmosphere is long; most blue and violet are scattered out of the line of sight, leaving the transmitted light enriched in the longer-wavelength reds and oranges. Larger particles (water droplets in clouds, dust) scatter all wavelengths roughly equally, which is why clouds and fog look white. The same principle, applied to molecules of paint or biological tissue, accounts for the blueness of tobacco smoke against a dark background and the reddish appearance of blood-perfused skin in transmitted light.
Q54. Why does the Sun appear reddish at sunrise and sunset?
Answer: At sunrise and sunset, light from the Sun travels the longest path through the dense lower atmosphere. Rayleigh scattering removes most of the blue and violet light from the direct beam, and only the longer-wavelength red and orange components reach the observer’s eye, making the Sun look reddish.
Q55. Why is it dangerous to drive in fog with high-beam (white) headlights?
Answer: Fog droplets are larger than the wavelength of visible light and scatter all wavelengths nearly equally. With white headlights, much of the light is scattered back into the driver’s eyes, producing glare and reducing visibility. Yellow fog-lights, in contrast, contain less of the strongly-scattered blue component and penetrate fog better.
8. The Human Eye and Defects of Vision
The human eye is a remarkable optical instrument. The cornea (the transparent front bulge) does about two-thirds of the focusing; the eye-lens, supported by the ciliary muscles, fine-tunes the focal length so that images of objects at varying distances form sharply on the retina. The ability of the lens to change its focal length is called accommodation. The far point of a normal eye is at infinity; the near point (least distance of distinct vision) is at $D=25$ cm in young adults.
Common defects:
(i) Myopia (short-sight): far point comes nearer; corrected with a concave (diverging) lens of focal length equal to the distance of the far point.
(ii) Hypermetropia (long-sight): near point recedes beyond 25 cm; corrected with a convex (converging) lens.
(iii) Presbyopia: age-related loss of accommodation; corrected with bifocal lenses.
(iv) Astigmatism: cornea/lens has different curvatures in different planes; corrected with cylindrical lenses.
Q56. A person can see clearly only up to 1.5 m. What lens should he wear to see distant objects?
Answer: He is myopic with far point 1.5 m. He needs a concave lens of focal length $-1.5$ m and power $P=-1/1.5=-0.67$ D.
Q57. A hypermetropic person has near point at 75 cm. What power of lens does he need to read at the normal near point of 25 cm?
Answer: The lens must form a virtual image of an object at 25 cm at the patient’s actual near point, 75 cm. So $u=-25$, $v=-75$. $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=-\frac{1}{75}+\frac{1}{25}=\frac{-1+3}{75}=\frac{2}{75}$. $f=37.5$ cm $=0.375$ m. $P=1/0.375=+2.67$ D — a convex lens.
9. Microscopes — Simple and Compound
A simple microscope (magnifying glass) is a single converging lens of short focal length. With the object placed inside the focal length, it forms an erect, virtual, magnified image. Maximum angular magnification is obtained when this image is at the near point: $M=1+D/f$. For relaxed-eye viewing (image at infinity), $M=D/f$.
A compound microscope uses two convex lenses. The objective, of very short focal length $f_o$, produces a real, inverted and magnified image of an object placed just beyond its focus. This intermediate image lies inside the focal length of the eyepiece, which then magnifies it further as a simple microscope. The total angular magnification is $M=m_o\times M_e=\frac{L}{f_o}\!\left(1+\frac{D}{f_e}\right)$ when the final image is at the near point, or $\frac{L}{f_o}\frac{D}{f_e}$ when the image is at infinity. Here $L$ is the tube length (distance between the back focus of the objective and the front focus of the eyepiece). Magnifications of several hundred to two thousand are routinely achieved.
Q58. Why is the focal length of the objective of a compound microscope kept very small?
Answer: The linear magnification produced by the objective is $m_o=L/f_o$. To get a large $m_o$ for a given tube length, $f_o$ must be small. A short-focus objective also allows the object to be placed very close to it, which increases the angle subtended.
Q59. The objective and eyepiece of a compound microscope have focal lengths 1 cm and 5 cm respectively. The tube length is 20 cm and the final image is formed at infinity. Calculate the magnifying power.
Answer: $M=\frac{L}{f_o}\frac{D}{f_e}=\frac{20}{1}\times\frac{25}{5}=20\times5=100$.
10. Telescopes — Astronomical and Reflecting
A telescope is used to view distant objects which subtend very small angles at the unaided eye. The astronomical refracting telescope uses two convex lenses: a long-focus objective forming a real, inverted, diminished image of the distant object at its focus, and a short-focus eyepiece magnifying that image. In normal adjustment (final image at infinity), the focal points of the two lenses coincide, the tube length is $f_o+f_e$, and the angular magnification is $M=-f_o/f_e$.
For high magnification, $f_o$ should be large; for high light-gathering power and resolution, the diameter of the objective $D$ should be large (resolving power $\propto D/\lambda$). Practical refracting telescopes are limited by chromatic aberration and the difficulty of making large achromatic lenses. Reflecting telescopes (e.g. the Newtonian, Cassegrain) avoid these problems by using a concave parabolic mirror as the objective; chromatic aberration is absent because reflection does not depend on wavelength, and very large mirrors can be made comparatively inexpensively. The Hubble Space Telescope and most modern observatories use reflecting designs.
Q60. State two advantages of a reflecting telescope over a refracting telescope.
Answer: (i) Mirrors are free from chromatic aberration, lenses are not. (ii) Spherical aberration in mirrors can be eliminated by using a parabolic shape; for large lenses it is hard to correct. (iii) Large-diameter mirrors are far easier and cheaper to manufacture than large-diameter, defect-free lenses; mirrors can be supported from behind whereas lenses must be supported only at the rim, causing them to sag.
Q61. The objective of an astronomical telescope has focal length 100 cm and the eyepiece 5 cm. Find its magnifying power and the length of the telescope in normal adjustment.
Answer: $M=-f_o/f_e=-100/5=-20$. Length $L=f_o+f_e=105$ cm.
Q62. What is the angular magnification of the telescope in question 61 if the final image is at the near point?
Answer: $M=-\frac{f_o}{f_e}\!\left(1+\frac{f_e}{D}\right)=-\frac{100}{5}\!\left(1+\frac{5}{25}\right)=-20\times1.2=-24$.
Higher-Order Thinking and Conceptual Questions
Q63. A fish in a tank looks at the world above. Through what range of angles can the fish see external objects?
Answer: Light from above can reach the fish only by entering the water within Snell’s window — a cone of half-angle equal to the critical angle ($\approx 49^\circ$ for water). Any external scene therefore appears compressed into this cone above the fish’s head, while objects beyond the cone appear by reflection from the underside of the water surface.
Q64. Why does an empty test tube placed in a beaker of water appear to shine like silver?
Answer: Light entering the water hits the inner glass-air boundary of the tube at an angle greater than the critical angle and undergoes total internal reflection. The tube acts like a perfect mirror and looks silvery from outside.
Q65. A glass paper-weight on a printed page makes the print appear nearer. Explain.
Answer: Each printed character emits light upward through the glass into the air. Refraction at the upper glass-air surface bends the rays away from the normal, so when traced back the rays appear to come from a point higher up than the real character — i.e. the print appears raised by $t(1-1/n)$.
Q66. Why must the focal length of the eyepiece of a compound microscope be small? Should it be smaller than that of the objective?
Answer: The angular magnification of the eyepiece is approximately $D/f_e$, so a small $f_e$ gives a large eyepiece magnification. Although it is short, $f_e$ is usually larger than $f_o$, because the objective must produce a magnified intermediate image and that requires $f_o\ll L$.
Q67. State the conditions for the deviation produced by a prism to be minimum.
Answer: (i) The angle of incidence equals the angle of emergence ($i=e$). (ii) The refracted ray inside the prism is parallel to the base of the prism. (iii) The angles of refraction at the two faces are equal ($r_1=r_2=A/2$). At minimum deviation the path of the ray is symmetric.
Q68. The image formed by a convex lens of a real object is real, inverted and of the same size as the object. Where is the object placed?
Answer: Such an image is formed when the object is at $2f$. The image is also at $2f$ on the other side, with $|m|=1$. This configuration is used in optical-bench experiments to determine the focal length by the displacement method.
Q69. Two convex lenses of focal lengths 30 cm and 20 cm are kept coaxially 10 cm apart with their principal axes coincident. Find the equivalent focal length.
Answer: For lenses separated by $d$, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}=\frac{1}{30}+\frac{1}{20}-\frac{10}{600}=\frac{2+3-1}{60}=\frac{4}{60}=\frac{1}{15}$. $F=15$ cm.
Q70. A convex lens of focal length 20 cm is placed in front of a plane mirror at a distance of 5 cm. An object is placed at a certain distance from the lens such that its image coincides with the object. Find the position of the object.
Answer: For the image to coincide with the object, light after passing through the lens must strike the mirror normally, i.e. parallel to the axis after refraction. The object must therefore be at the focus of the lens-mirror system. For this combination (lens + plane mirror in contact pattern) the effective focal length equals that of the lens, so the object is placed 20 cm from the lens.
Q71. The radius of curvature of the convex face of a plano-convex lens is 12 cm and its refractive index is 1.5. (a) Find its focal length in air. (b) If the plane surface is silvered, find the equivalent focal length of the system.
Answer: (a) $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})=(0.5)(\frac{1}{12}-\frac{1}{\infty})=\frac{1}{24}$, $f=24$ cm. (b) Power of the silvered system $P_{eq}=2P_l+P_m$. Here $P_m=0$ for a plane mirror, so $P_{eq}=2/24=1/12$, giving an equivalent focal length of 12 cm — the system behaves as a concave mirror of focal length 12 cm.
Q72. The far point of a myopic eye is at 80 cm and the near point at 15 cm. What lens does the patient need (a) to see distant objects, (b) to read at the normal near point of 25 cm?
Answer: (a) Distance lens: image of object at infinity should be at far point, so $f=-80$ cm $=-0.8$ m, $P=-1.25$ D — concave. (b) Reading lens: image of an object at 25 cm should be at the actual near point of 15 cm: $\frac{1}{f}=\frac{1}{-15}-\frac{1}{-25}=\frac{-5+3}{75}=-\frac{2}{75}$, $f=-37.5$ cm. $P=-1/0.375=-2.67$ D. (For most myopia, distance correction alone is sufficient; bifocals are needed only when presbyopia is also present.)
More Worked Examples and Board-Style Questions
Q73. A point object is placed at a distance of 60 cm from a convex lens of focal length 20 cm. Find the position, nature and magnification of the image.
Answer: $u=-60$, $f=+20$. $\frac{1}{v}=\frac{1}{20}-\frac{1}{60}=\frac{3-1}{60}=\frac{1}{30}$, $v=+30$ cm. Real, inverted image 30 cm beyond the lens. $m=v/u=30/(-60)=-0.5$. Image is half the size of the object.
Q74. A convex lens forms a real image of a point source on a screen. Half the lens is now covered with an opaque sheet. What happens to the image?
Answer: The image still forms at the same position and same size — every part of the lens contributes to forming the image, so blocking half merely reduces the intensity of the image (by half), not its position or size.
Q75. A concave mirror produces an image five times the size of the object on a screen. If the object is at 12 cm from the mirror, what is the radius of curvature of the mirror?
Answer: Real magnified image, so $m=-5$, $u=-12$. $m=-v/u\Rightarrow v=-mu=-(-5)(-12)=-60$ cm. $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{-60}+\frac{1}{-12}=\frac{-1-5}{60}=-\frac{6}{60}=-\frac{1}{10}$, $f=-10$ cm, $R=20$ cm.
Q76. A ray of light passes from glass ($n=1.5$) to water ($n=1.33$). What is the critical angle for this interface?
Answer: $\sin\theta_c=n_{rarer}/n_{denser}=1.33/1.5=0.887$, $\theta_c=62.5^\circ$.
Q77. A double-convex lens made of glass of refractive index 1.55 has both radii equal to 30 cm. Calculate (a) its focal length in air, (b) its power.
Answer: $\frac{1}{f}=(0.55)(\frac{1}{30}+\frac{1}{30})=\frac{0.55\times2}{30}=\frac{1.10}{30}=0.0367$. $f=27.3$ cm. $P=1/0.273=+3.67$ D.
Q78. A convex lens of focal length 25 cm is placed in contact with a concave lens of focal length 50 cm. Find the power of the combination and state whether it is converging or diverging.
Answer: $P_1=100/25=4$ D, $P_2=-100/50=-2$ D. $P=4-2=2$ D, converging. $f=50$ cm.
Q79. At what angle should a ray of light be incident on the face of an equilateral prism so that it just suffers total internal reflection at the other face? ($n=1.5$).
Answer: Critical angle $\sin\theta_c=1/1.5=0.667$, $\theta_c=41.81^\circ$. For just-TIR at the second face, $r_2=\theta_c=41.81^\circ$. Then $r_1=A-r_2=60-41.81=18.19^\circ$. Snell’s law: $\sin i=1.5\sin18.19^\circ=1.5\times0.312=0.468$, $i=27.91^\circ$.
Q80. The focal length of a concave mirror is 20 cm. Where should an object be placed in front of the mirror so that its image is three times its size and (a) real, (b) virtual?
Answer: $f=-20$. (a) Real, $m=-3$, so $v=-3u\cdot(-1)$. Use $m=-v/u=-3\Rightarrow v=3u$ (note: with $u$ negative and $v$ negative, $v=3u$). $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\Rightarrow\frac{1}{-20}=\frac{1}{3u}+\frac{1}{u}=\frac{4}{3u}\Rightarrow u=-\frac{4\times20}{3}=-26.7$ cm. (b) Virtual, $m=+3$: $v=-3u$ but with opposite signs ($v$ positive, $u$ negative), so $\frac{1}{-20}=\frac{1}{-3u}+\frac{1}{u}=\frac{-1+3}{3u}=\frac{2}{3u}\Rightarrow u=-\frac{2\times20}{3}=-13.3$ cm.
Q81. White light passes through a prism. Which colour is deviated the most and which the least?
Answer: Violet (shortest wavelength, highest refractive index) is deviated the most; red (longest wavelength, lowest refractive index) is deviated the least.
Q82. State Cauchy’s formula for the variation of refractive index with wavelength.
Answer: $n(\lambda)=A+\dfrac{B}{\lambda^2}+\dfrac{C}{\lambda^4}+\ldots$, where $A,B,C$ are material constants. The first two terms suffice for visible light. Cauchy’s relation shows quantitatively why short wavelengths bend more than long ones.
Q83. A tank holds water to a height of 1 m. A light source is placed 30 cm below the surface. Find the radius of the bright spot seen on the surface from above.
Answer: Critical angle for water $\theta_c=48.75^\circ$, $\tan\theta_c=1.137$. Radius $r=h\tan\theta_c=30\times1.137=34.1$ cm.
Q84. A converging beam of light is incident on a concave lens of focal length 16 cm so that the beam, in the absence of the lens, would converge at a point 12 cm behind the lens. Find the actual point of convergence.
Answer: Virtual object at $u=+12$, $f=-16$. $\frac{1}{v}=\frac{1}{-16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}$, $v=+48$ cm. The beam converges 48 cm beyond the lens.
Q85. A glass prism has a refracting angle of 4° and a refractive index of 1.5. Calculate the deviation produced by it.
Answer: Thin prism: $\delta=(n-1)A=(0.5)(4^\circ)=2^\circ$.
Q86. Light of wavelength 600 nm passes from air into a medium of refractive index 1.5. Find (a) the wavelength inside the medium, (b) the speed of light in the medium.
Answer: (a) $\lambda_m=\lambda_0/n=600/1.5=400$ nm. (b) $v=c/n=3\times10^8/1.5=2\times10^8$ m/s. Frequency is unchanged across the boundary.
Q87. State the laws of refraction.
Answer: (i) The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane. (ii) Snell’s law: $n_1\sin\theta_1=n_2\sin\theta_2$, equivalent to saying the ratio $\sin\theta_1/\sin\theta_2$ is constant for any two media.
Q88. A ray of light strikes a glass slab of thickness 4 cm and refractive index 1.5 at an angle of 60°. Calculate the lateral shift of the emergent ray.
Answer: $\sin r=\sin60^\circ/1.5=0.866/1.5=0.577$, $r=35.26^\circ$. Lateral shift $s=t\sin(i-r)/\cos r=4\sin(24.74^\circ)/\cos35.26^\circ=4\times0.418/0.817=2.05$ cm.
Q89. Why does a convex lens of glass appear to disappear when immersed in a liquid of the same refractive index?
Answer: If $n_l=n_g$ then $n-1=0$ in the lens-maker’s formula, so $1/f=0$, i.e. $f=\infty$. The lens does not refract light at all; rays pass through undeviated, making the lens invisible.
Q90. A small object is placed 50 cm from a thin convex lens of focal length 25 cm. Where would the image be formed if a plane mirror were placed 10 cm beyond the lens?
Answer: First lens image: $u=-50$, $f=25$, $\frac{1}{v}=\frac{1}{25}-\frac{1}{50}=\frac{1}{50}$, $v=50$ cm — would form 50 cm right of lens, i.e. 40 cm right of mirror. The mirror reflects to form an object 40 cm left of it (= 30 cm right of lens). Now lens reverses with $u=+30$ (object beyond lens on right): $\frac{1}{v}-\frac{1}{30}=\frac{1}{25}$, $\frac{1}{v}=\frac{1}{25}+\frac{1}{30}=\frac{6+5}{150}=\frac{11}{150}$, $v=13.6$ cm to the left of the lens — final image is 13.6 cm to the left of the lens.
Q91. The objective and eyepiece of an astronomical telescope have focal lengths 200 cm and 4 cm. Calculate (i) magnifying power for normal adjustment, (ii) length of the telescope, (iii) magnification when image is at the near point.
Answer: (i) $M=-200/4=-50$. (ii) $L=200+4=204$ cm. (iii) $M=-\frac{f_o}{f_e}\!\left(1+\frac{f_e}{D}\right)=-50(1+4/25)=-50\times1.16=-58$.
Q92. A telescope has an objective of diameter 100 mm. Calculate its limit of resolution for light of wavelength 5500 Å.
Answer: $\theta_{min}=1.22\lambda/D=1.22\times5500\times10^{-10}/0.10=6.71\times10^{-6}$ rad $=1.4”$ of arc.
Q93. A magnifying glass produces an angular magnification of 5 when the image is formed at the near point. Find the focal length of the lens.
Answer: $M=1+D/f=5\Rightarrow D/f=4\Rightarrow f=D/4=25/4=6.25$ cm.
Q94. A real image of an object is formed at a distance of 24 cm from a convex lens. The size of the image is half that of the object. Find (a) the focal length, (b) the position of the object.
Answer: $|m|=v/|u|=1/2$, so $|u|=2v=48$ cm, $u=-48$ cm. $\frac{1}{24}+\frac{1}{48}=\frac{1}{f}\Rightarrow\frac{2+1}{48}=\frac{1}{f}\Rightarrow f=16$ cm. Object 48 cm to the left of the lens.
Q95. Distinguish between real and virtual images.
Answer: A real image is formed by the actual intersection of refracted/reflected rays; it can be obtained on a screen and is always inverted with respect to the object (in single-stage imaging). A virtual image is formed by the apparent intersection of rays after extending them backward; it cannot be projected on a screen and is always erect.
Q96. The radius of curvature of the surfaces of a concavo-convex lens are 10 cm and 20 cm. The refractive index is 1.5. Find its focal length.
Answer: Concavo-convex with convex face encountered first by light: $R_1=+10$, $R_2=+20$. $\frac{1}{f}=(0.5)(\frac{1}{10}-\frac{1}{20})=(0.5)(\frac{1}{20})=\frac{1}{40}$, $f=40$ cm.
Q97. A 1.5 cm tall object is placed at 30 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.
Answer: $u=-30$, $f=+15$. $\frac{1}{v}=\frac{1}{15}+\frac{1}{30}=\frac{2+1}{30}=\frac{1}{10}$, $v=+10$ cm. Virtual, behind the mirror. $m=-v/u=-(10)/(-30)=+0.333$. Height $=0.333\times1.5=0.5$ cm — erect, diminished.
Q98. Light from a galaxy of wavelength 6000 Å reaches us through space (n=1) and then enters Earth’s atmosphere (n≈1.0003). What change does it suffer?
Answer: Frequency is unchanged. Speed reduces slightly to $c/n\approx2.999\times10^8$ m/s and wavelength reduces to $\lambda_0/n=5998.2$ Å. The change is small but measurable in precise astrometry.
Q99. The refractive index of water for sound is greater than 1, but for light it is greater than 1 also — yet sound speeds up in water while light slows down. Why?
Answer: The optical refractive index $n=c/v$ of water is greater than 1 because the speed of light in water (2.25×10⁸ m/s) is less than in vacuum. For sound, the conventional “refractive index” is defined as $n=v_1/v_2$ — the ratio of speed in the first medium to speed in the second; for sound from air to water the ratio is large (sound moves faster in water). The two indices follow opposite conventions because Snell’s law is expressed in terms of speeds.
Q100. Why are the eyepieces of telescopes and microscopes always made achromatic doublets?
Answer: A simple eyepiece would form coloured fringes around the image because of chromatic aberration. An achromatic doublet — a pair of crown and flint lenses chosen so that the dispersions cancel while the deviations add — produces a sharp, colour-free image.
Q101. State and prove that a convex lens placed between an object and its image, on a fixed bench, has two positions for which a sharp image of the object is obtained on a screen.
Answer: Let the object-screen distance be $D$ and the lens position be $x$ from the object. Then $u=-x$, $v=D-x$, and the lens formula $\frac{1}{D-x}+\frac{1}{x}=\frac{1}{f}$ gives a quadratic $x^2-Dx+Df=0$ with two solutions $x=\frac{D\pm\sqrt{D^2-4Df}}{2}$, provided $D\geq 4f$. The two positions are symmetric about the midpoint and the displacement between them is $d=\sqrt{D^2-4Df}$, leading to the displacement-method formula $f=(D^2-d^2)/(4D)$.
Q102. State and explain the principle of reversibility of light.
Answer: If a light ray, after suffering any number of reflections and refractions, has its final path reversed, it retraces the same path back to the source. The principle follows from the symmetry of Snell’s law and the laws of reflection and is widely used in proving relations between refractive indices, e.g. ${}_1n_2\times{}_2n_1=1$.
Q103. Why does a convex lens behave as a converging lens when surrounded by air but as a diverging lens when surrounded by carbon disulphide ($n=1.63$)?
Answer: The factor $(n_g/n_m-1)$ in the lens-maker’s formula determines the sign of the focal length. With $n_g=1.5$ and $n_m=1.63$ this factor is negative, so $f$ becomes negative for what was a converging lens — it diverges light. The shape of the lens is no longer enough to specify its action; the surrounding medium also matters.
Q104. What is meant by spherical aberration in a lens? How can it be minimised?
Answer: Spherical aberration is the failure of a spherical lens or mirror to bring all parallel rays to a single point — paraxial rays focus farther away than marginal rays. It produces a blurred image. It can be minimised by (i) using lenses of large focal length and small aperture, (ii) using stops to block marginal rays, (iii) combining a convex and a concave lens (one over-corrects, the other under-corrects), or (iv) using parabolic mirrors instead of spherical ones in telescopes.
Q105. What is chromatic aberration and how is it corrected?
Answer: Chromatic aberration is the inability of a single lens to focus light of different colours at the same point because $f$ depends on $\lambda$ through $n$. It is corrected by using an achromatic combination — typically a convex crown-glass lens cemented to a concave flint-glass lens — chosen so that the dispersive powers cancel: $\omega_1/f_1+\omega_2/f_2=0$, while the net focal length is finite.
Q106. Explain why a single optical fibre can transmit signals from one end to the other even when the fibre is bent.
Answer: Light entering the core at a sufficiently steep angle hits the core-cladding boundary at greater than the critical angle and is totally internally reflected. Each successive bounce keeps the light trapped inside the core. Provided bends are gentle enough that the angle of incidence at the boundary stays above $\theta_c$, the light follows the fibre, even round corners.
Q107. State the order of magnitude of the radius of curvature, focal length and aperture diameter of the cornea-lens system in a normal eye.
Answer: The eye-lens has a focal length of roughly 2.3 cm in air-equivalent terms (the actual lens sits in a medium of refractive index ~1.34 and the optics is dominated by the cornea). The radius of curvature of the cornea is about 8 mm and that of the lens about 10 mm. The pupil aperture varies from 2 mm in bright light to 8 mm in the dark.
Q108. The objective of a microscope has a numerical aperture of 0.9 and is used with light of wavelength 500 nm. Calculate its limit of resolution.
Answer: $d_{min}=\dfrac{1.22\lambda}{2\,N\!A}=\dfrac{1.22\times500\times10^{-9}}{2\times0.9}=3.39\times10^{-7}$ m $\approx340$ nm.
Q109. Why does a swimmer under water see external objects only within a circular patch directly above?
Answer: External light can reach the swimmer’s eye only by entering the water at angles less than 90° to the surface, i.e. entering Snell’s window — a cone of half-angle equal to the critical angle for water (~49°). All external scenery is therefore compressed inside this cone; rays from beyond it cannot enter the water by refraction.
Q110. Show that for two thin lenses of focal lengths $f_1$ and $f_2$ separated by distance $d$, the equivalent focal length is given by $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}$.
Answer: Take an object at infinity. Lens 1 forms its image at its focus $f_1$. This image is the object for lens 2 at a distance $u_2=-(f_1-d)$. Then $\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}\Rightarrow\frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{u_2}=\frac{1}{f_2}-\frac{1}{f_1-d}$. The equivalent system focal length $F$ is given by the back focal distance from the second lens to the final image plus a correction; algebraic simplification (see textbook derivation) yields $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}$.
Quick-Revision Mind Map
| Topic | Key formula / fact |
| Mirror | $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, $f=R/2$, $m=-v/u$. |
| Refraction | $n_1\sin\theta_1=n_2\sin\theta_2$, $n=c/v$. |
| Apparent depth | $d’=d/n$, shift $=t(1-1/n)$. |
| TIR | $\sin\theta_c=1/n$. Used in fibres, prisms, mirage. |
| Spherical surface | $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$. |
| Lens-maker’s | $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$. |
| Thin lens | $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, $m=v/u$. |
| Lenses in contact | $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$ (powers add). |
| Lenses separated | $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}$. |
| Power | $P=1/f$ (m); unit dioptre. |
| Prism | $\delta=(i+e)-A$; $n=\frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$. |
| Thin prism | $\delta=(n-1)A$. |
| Dispersive power | $\omega=(n_v-n_r)/(n_y-1)$. |
| Rayleigh | $I\propto 1/\lambda^4$. |
| Simple microscope | $M=1+D/f$ (image at $D$); $D/f$ (relaxed). |
| Compound microscope | $M=\frac{L}{f_o}(1+D/f_e)$. |
| Telescope | $M=-f_o/f_e$, $L=f_o+f_e$. |
| Resolving power | Telescope: $1.22\lambda/D$. Microscope: $2\mu\sin\beta/\lambda$. |
Master the sign convention, draw a clear ray diagram for every problem, and substitute carefully — most “tough” optics questions become routine once these habits are formed. Good luck with your ASSEB Class 12 Physics examination!