Board Previous Year Questions

How to Use This Section

CBSE Board-pattern questions on Ray Optics by mark value, with examiner-rewarded model answers — sign convention declared, formula stated before substitution, ray diagrams described, and the derivations (mirror equation, lens maker's, prism formula, instrument magnifications) in the stepwise form mark schemes follow. Ray optics is the single largest Board contributor: expect a derivation, a numerical and a reasoning question every year.

1-Mark Questions (Definitions & Direct)

Q1. Write the relation between focal length and radius of curvature of a spherical mirror. Answer: $f = R/2$ (paraxial rays; signs by the Cartesian convention).

Q2. Define the critical angle. Answer: The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90 degrees; $\sin i_c = 1/n$ (n of denser w.r.t. rarer).

Q3. What is the SI unit of the power of a lens? Define it. Answer: The dioptre (D) = 1 m$^{-1}$: the power of a lens of focal length one metre.

Q4. A lens has power $-2.0$ D. What is its type and focal length? Answer: Concave (diverging); $f = 1/P = -0.5$ m $= -50$ cm.

Q5. At minimum deviation, how does the ray travel inside a prism? Answer: Parallel to the base, with $i = e$ and $r_1 = r_2 = A/2$.

Q6. Why does a diamond sparkle? Answer: Its very small critical angle (24.4 degrees, from n = 2.42) traps entering light in repeated total internal reflections, releasing it only through intended cut faces.

2-Mark Questions (Short Answer)

Q7. State the two conditions for total internal reflection. Answer: (i) Light must travel from an optically denser to a rarer medium; (ii) the angle of incidence in the denser medium must exceed the critical angle for the pair.

Q8. The bottom of a 4 m deep pond appears 3 m deep when viewed normally. Find the refractive index of water. Answer: $n = \dfrac{\text{real depth}}{\text{apparent depth}} = \dfrac{4}{3} \approx 1.33$.

Q9. A concave mirror and a convex lens are each held in water. How do their focal lengths change? Answer: The mirror's focal length is set purely by geometry ($f = R/2$) — unchanged in water. The lens's f depends on the relative index $(n_{21}-1)$, which falls in water — its focal length increases (the lens weakens).

Q10. Why are totally reflecting prisms preferred over plane mirrors in binoculars? Answer: TIR reflects 100 percent of the light (no metallic absorption/transmission loss, no coating to degrade), giving brighter, sharper images through multiple reflections; prisms also fold and invert the path compactly.

Q11. An object is at 2f of a convex lens. Find the image position and magnification. Answer: $u = -2f$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}$, so $v = +2f$, $m = v/u = -1$ — real, inverted, same size at 2f on the other side.

3-Mark Questions (Derivations & Numericals)

Q12. Derive the mirror equation for a concave mirror forming a real image. Answer: For paraxial rays, the right triangles formed by the object/image heights and the focus give similar-triangle relations; eliminating the heights yields $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, with magnification $m = -v/u$. With the Cartesian convention the result holds for all mirrors and images.

Q13. Derive the lens maker's formula for a thin lens. Answer: Apply $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}$ at the first surface (image $I_1$), then at the second surface with $I_1$ as object (indices reversed, radius $R_2$); adding the two equations (thin lens: same v for surface distances) gives $\frac{1}{f} = (n_{21}-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ on putting the object at infinity.

Q14. Derive the prism formula $n = \sin[(A+D_m)/2]/\sin(A/2)$. Answer: Geometry: $r_1 + r_2 = A$ and $d = i + e - A$. At minimum deviation $i = e$, $r_1 = r_2$, so $r = A/2$ and $i = (A+D_m)/2$; Snell's law at one face then gives the formula.

Q15. An object 10 cm tall stands 30 cm before a convex lens of focal length 10 cm. Find the image position, size and nature. Answer: $\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$: $v = +15$ cm; $m = \frac{15}{-30} = -\frac{1}{2}$; image height $= -5$ cm. Real, inverted, half-size, 15 cm beyond the lens.

5-Mark Questions (Long Answer)

Q16. (a) Draw the labelled ray diagram of a compound microscope (final image at infinity) and derive its magnifying power. (b) Why should both focal lengths be small? (c) For $f_o$ = 1 cm, $f_e$ = 2 cm, L = 20 cm, compute m. Answer:

1. (a) Objective forms a real, inverted, magnified image of the near object at the first focal plane of the eyepiece; the eyepiece (simple magnifier) then renders the final image at infinity. Objective's linear magnification $m_o = L/f_o$ (L = tube length); eyepiece's angular magnification $m_e = D/f_e$; total $m = \frac{L}{f_o}\cdot\frac{D}{f_e}$.
2. (b) Both $f_o$ and $f_e$ appear in denominators — small focal lengths (practically down to about 1 cm) maximise m.
3. (c) $m = (20/1)(25/2) = 250$.

Q17. (a) Describe the astronomical refracting telescope (normal adjustment) with a ray diagram and obtain $m = f_o/f_e$. (b) State the advantages of a reflecting (Cassegrain) telescope. (c) A telescope has $f_o$ = 144 cm, $f_e$ = 6 cm: find m and the tube length. Answer:

1. (a) Parallel rays from a distant object form a real inverted image at the objective's focus, coincident with the eyepiece's focus; emergent rays are parallel. Magnifying power = angle ratio $\beta/\alpha = f_o/f_e$; tube length $f_o + f_e$.
2. (b) Mirror objective: no chromatic aberration; parabolic figure removes spherical aberration; lighter and supportable over the entire back — so very large apertures (better light gathering and resolution) are feasible. The Cassegrain's convex secondary folds the beam through a hole in the primary, shortening the tube.
3. (c) $m = 144/6 = 24$; tube $= 150$ cm.