Original practice sets for Ray Optics are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. The mirror formula (using New Cartesian sign convention with incident light along +x axis) is:
Explanation: For spherical mirrors: 1/v + 1/u = 1/f = 2/R. Using sign convention: object at −ve u (real object), f = −R/2 for concave.
2. Linear magnification m for a mirror is:
Explanation: m = −v/u. If m is negative, image is inverted; if m > 1 (magnitude), image is magnified. Concave mirrors can form real inverted magnified images.
3. A concave mirror has radius of curvature 20 cm. Its focal length is:
Explanation: f = R/2 = 20/2 = 10 cm. For a concave mirror, f is negative (−10 cm) by convention.
4. A convex mirror always forms an image that is:
Explanation: For a convex mirror, all images are virtual, erect, and smaller than the object regardless of object position — making it useful for rear-view mirrors (wide field of view).
5. When an object is placed at the centre of curvature C of a concave mirror, the image is formed:
Explanation: Object at C (u = R = 2f): from mirror formula 1/v = 1/f − 1/u = 1/f − 1/(2f) = 1/(2f) → v = 2f = R = C. Image at C: real, inverted, same size (m = −1).
6. The law of reflection states that the angle of incidence equals the angle of reflection, both measured from:
Explanation: Both angles are measured from the normal at the point of incidence. The incident ray, reflected ray, and normal are all in the same plane.
7. When an object is placed between the focus F and the pole P of a concave mirror, the image is:
Explanation: For u
8. As a real object moves from infinity toward the pole of a concave mirror, the image moves from:
Explanation: As object moves from ∞ → C → F → P: image moves F → C → ∞ (image jumps behind mirror) → behind mirror moving toward P as object moves from F to P.
9. Two plane mirrors are placed at 60° to each other. Number of images formed of an object placed between them:
Explanation: Number of images = 360°/θ − 1 = 360/60 − 1 = 6 − 1 = 5 (when 360/θ is an even integer).
10. A virtual object in optics is:
Explanation: A virtual object occurs when a converging beam (from a lens or mirror upstream) would converge behind the current optical element. The element intercepts the beam before it converges — the virtual object is where convergence would have occurred.
11. An object 4 cm tall is placed 30 cm in front of a concave mirror of focal length 20 cm. Size of image is:
Explanation: 1/v = 1/f − 1/u = 1/(−20) − 1/(−30) = −1/20 + 1/30 = −3/60 + 2/60 = −1/60 → v = −60 cm. m = −v/u = −(−60)/(−30) = −2. Image height = |m| × 4 = 8 cm (inverted).
12. A silvered lens (plano-convex, n = 1.5, curved face R = 20 cm) acts as a concave mirror. Focal length of the equivalent mirror is:
Explanation: P_equiv = P_lens + P_mirror + P_lens = 2P_lens + P_mirror. P_lens = (n−1)(1/R − 1/∞) = (0.5)(1/20) = 1/40 cm⁻¹. P_mirror = 2/R_mirror = 2/∞ = 0 (flat silver). P_equiv = 2/40 = 1/20 cm⁻¹. Wait, silvered curved face: P_mirror = 2/20 = 1/10. P_lens contributes twice (refract, reflect, refract). P = 2×(1/40) + 1/10 = 1/20 + 2/20 = 3/20. f = 1/P = 20/3 ≈ 6.67 cm.
13. An object moves toward a concave mirror with velocity v_obj. The velocity of the image is related by:
Explanation: From mirror formula, differentiating: dv/dt = (v²/u²)(du/dt) = m² × v_obj. For a real object moving toward a concave mirror at u = 2f (m = −1): image velocity equals object velocity. For m > 1, image moves faster.
14. Convex mirrors are used as rear-view mirrors in vehicles because:
Explanation: Convex mirrors diverge reflected rays, giving a wider field of view. All images are virtual, erect, and diminished — allowing the driver to see a wider region behind. The diminished size means everything appears farther than it is (hence the warning).
15. When a plane mirror is rotated by angle θ, the reflected beam rotates by:
Explanation: When a mirror rotates by θ, the normal also rotates by θ. The new angle of incidence changes by θ, and the new angle of reflection changes by θ. The net rotation of the reflected beam = 2θ. Used in galvanometers and laser scanning.
16. Parabolic mirrors (used in searchlights and satellite dishes) are preferred over spherical mirrors because:
Explanation: Spherical mirrors have aberration for wide beams (marginal rays focus closer than paraxial rays). A parabolic mirror has an exact focus for all parallel rays — used in telescopes, searchlights, and satellite dishes.
17. Object at the focus F of a concave mirror produces image at:
Explanation: At u = f: 1/v = 1/f − 1/f = 0 → v = ∞. Reflected rays are parallel. Used in searchlight reflectors: source at F → parallel beam.
18. Object placed beyond C of a concave mirror produces an image that is:
Explanation: For u > 2f in a concave mirror: image is real, inverted, and located between F and C. Magnification |m|
19. Power of a mirror with focal length 25 cm is:
Explanation: P = 1/f(metres). For a concave mirror f = −0.25 m: P = 1/(−0.25) = −4 D. But for power in magnitude context (convex/concave), the concave mirror converges → often stated as 4 D (converging). Sign convention depends on context.
20. A real image is formed when:
Explanation: Real image: actual rays converge → can be projected on a screen. Virtual image: extensions of rays appear to diverge from a point → cannot be projected. Concave mirrors (for real objects beyond F) and converging lenses form real images.
21. At a refracting boundary with refractive indices n₁ and n₂, Snell's law is:
Explanation: Snell's law: n₁ sinθ₁ = n₂ sinθ₂. Light bends toward the normal when entering a denser medium (n₂ > n₁ → θ₂
22. Total internal reflection (TIR) requires:
Explanation: TIR occurs when (1) light is in the denser medium (going denser → rarer) and (2) angle of incidence exceeds the critical angle θ_c = sin⁻¹(n₂/n₁). Used in optical fibres, prisms, and diamonds.
23. Critical angle θ_c for a glass-air interface (n_glass = n) is:
Explanation: At critical angle: n sinθ_c = 1 × sin90° = 1 → sinθ_c = 1/n → θ_c = sin⁻¹(1/n).
24. Refractive index n of a medium is defined as:
Explanation: n = c/v = λ_vacuum/λ_medium. The frequency doesn't change on refraction; only λ and v change.
25. Apparent depth formula for an object at real depth d in a medium of refractive index n, viewed from air is:
Explanation: Apparent depth = Real depth / n. A swimming pool appears shallower than it is. The normal shift = d − d/n = d(1 − 1/n).
26. Optical fibres transmit light using:
Explanation: The fibre core has higher n than the cladding. Light entering at a shallow angle undergoes TIR at the core-cladding boundary and propagates along the fibre with minimal loss. Used in communications and endoscopy.
27. For a prism with apex angle A and minimum deviation D_m, the refractive index is:
Explanation: At minimum deviation (symmetric passage through prism): n = sin[(A+D_m)/2] / sin(A/2). This is the standard prism formula derived from Snell's law applied twice.
28. Deviation δ of a ray through a thin prism (small angle A, refractive index n) is:
Explanation: For a thin prism (small A): δ ≈ (n−1)A. This is the standard result for paraxial rays in a prism — deviation is proportional to prism angle and depends on n.
29. Angular dispersion produced by a prism is the difference in deviation for:
Explanation: Angular dispersion = δ_v − δ_r = (n_v − n_r)A. Dispersive power ω = (n_v − n_r)/(n_y − 1) = angular dispersion / mean deviation.
30. Critical angle for glass (n = 1.5) to air is:
Explanation: sin θ_c = 1/n = 1/1.5 = 2/3 → θ_c = sin⁻¹(0.667) ≈ 41.8° ≈ 42°.
31. Refraction formula at a single spherical surface separating media n₁ (object side) and n₂ is:
Explanation: Standard formula for refraction at a spherical surface: n₂/v − n₁/u = (n₂−n₁)/R. The sign of R follows the convention (positive if centre of curvature is on transmission side).
32. Lateral displacement of a ray passing through a glass slab of thickness t at angle of incidence i is:
Explanation: Lateral shift d = t × sin(i − r)/cos r, where r is the angle of refraction. The emergent ray is parallel to the incident ray but displaced laterally.
33. An object in glass (n = 1.5) appears shifted when viewed from air. Normal shift (how much it appears to move toward the surface) is:
Explanation: Apparent depth = real depth/n. Object appears at depth d/n instead of d. Shift = d − d/n = d(1 − 1/n) = t/3 for n = 1.5. The object appears closer by t/3.
34. Stars twinkle (scintillate) but planets generally do not because:
Explanation: Atmospheric turbulence causes random refraction changes. Stars are point sources so small deflections drastically change apparent brightness. Planets subtend a small (but non-zero) angular disk — the many rays from across the disk average out, reducing scintillation.
35. Thin lens formula (using sign convention with object on left):
Explanation: 1/v − 1/u = 1/f. For a real object: u is negative. For a converging lens: f is positive. For a diverging lens: f is negative.
36. Lens maker's formula for a thin lens (refractive index n, radii R₁ and R₂) is:
Explanation: Lens maker's formula: 1/f = (n−1)[1/R₁ − 1/R₂]. R₁ = radius of first surface (positive if centre is to right), R₂ = radius of second surface. For biconvex lens: R₁ > 0, R₂ 0 (converging).
37. Power of a lens with focal length f (in metres) is:
Explanation: P = 1/f in dioptres (D). Converging lens: f > 0 → P > 0. Diverging lens: f
38. Transverse magnification for a lens is:
Explanation: For a lens: m = v/u (positive sign, unlike mirrors). If m > 0 → erect image; if m 1 → magnified; m
39. Object placed at 2f from a converging lens forms image:
Explanation: u = −2f: 1/v = 1/f + 1/u = 1/f − 1/(2f) = 1/(2f) → v = 2f. m = v/u = 2f/(−2f) = −1. Image at 2f, same size, real, inverted.
40. Two thin lenses of focal lengths f₁ and f₂ placed in contact. Equivalent focal length is:
Explanation: 1/f = 1/f₁ + 1/f₂ → f = f₁f₂/(f₁+f₂). Power adds: P = P₁ + P₂. Used to find equivalent lens of a lens system.
41. Magnifying power of a simple microscope (convex lens) for image at near point (D = 25 cm) is:
Explanation: For image at near point D: m = 1 + D/f. For relaxed eye (image at infinity): m = D/f. The +1 accounts for the object being within f when image is at near point.
42. Magnifying power of a compound microscope (objective f_o, eyepiece f_e, tube length L) for image at infinity is approximately:
Explanation: M = m_o × M_e ≈ (L/f_o) × (D/f_e) for image at infinity. Short f_o and f_e → high magnification. The objective forms a real magnified image; the eyepiece acts as a simple microscope magnifying that image.
43. Magnifying power of an astronomical telescope (objective f_o, eyepiece f_e) for relaxed eye:
Explanation: M = f_o/f_e. Large objective focal length (large aperture) and small eyepiece focal length → high magnification. Image is inverted.
44. A myopic (short-sighted) person uses which lens for correction?
Explanation: Myopia: distant objects focus in front of the retina (eyeball too long). A concave lens diverges the incoming rays so they focus exactly on the retina. Power of lens: P = −1/far_point (in metres).
45. A hypermetropic (long-sighted) person uses which lens?
Explanation: Hypermetropia: near objects focus behind the retina (eyeball too short). A convex lens converges rays before they enter the eye so the eye can focus on the retina. Power = 1/near_point − 1/D.
46. A converging beam appears to converge at a point 30 cm behind a concave lens (f = −20 cm). The actual image is at:
Explanation: Virtual object at u = +30 cm (behind the lens). 1/v = 1/f + 1/u = 1/(−20) + 1/30 = −3/60 + 2/60 = −1/60 → v = −60 cm. The negative image distance places the real image 60 cm in front of the lens.
47. Chromatic aberration in lenses is caused by:
Explanation: n varies with λ (dispersion) → different colours have different focal lengths → blurred coloured fringes around images. Achromatic doublets (convex + concave lens of different glass types) correct this by making the net dispersion zero while retaining converging power.
48. An object 6 cm tall is placed 20 cm from a converging lens (f = 15 cm). Image height is:
Explanation: 1/v = 1/f + 1/u = 1/15 + 1/(−20) = 4/60 − 3/60 = 1/60 → v = 60 cm. m = v/u = 60/(−20) = −3. Image height = |m|×6 = 18 cm (inverted, real).
49. A biconvex lens (n=1.5, both R=20 cm) has its back surface silvered. It acts as a concave mirror of focal length:
Explanation: P_mirror = 2/R = 2/20 = 0.1 cm⁻¹. P_lens: each surface: P_surface = (n−1)/R = 0.5/20 = 0.025 cm⁻¹. Total lens P = 2×0.025 = 0.05 cm⁻¹. P_eq = 2P_lens + P_mirror = 0.1 + 0.1 = 0.2 Wait: P_lens = (n−1)(1/R₁ − 1/R₂) = 0.5(1/20 − 1/(−20)) = 0.5×(2/20) = 0.05 cm⁻¹. P_eq = 2×0.05 + 0.1 = 0.2 cm⁻¹. f = 1/0.2 = 5 cm. So 5 cm.
50. A fish in water (n=4/3) at depth 18 cm below a 6 cm thick glass (n=3/2) layer. Total apparent depth seen from air:
Explanation: For the glass layer: apparent depth = 6/(3/2) = 4 cm. For the water layer: apparent depth = 18/(4/3) = 13.5 cm. Total = 13.5 + 4 = 17.5 cm.
51. A prism of apex angle 60° has n = √3. Minimum deviation angle is:
Explanation: n = sin[(A+D_m)/2]/sin(A/2) = sin[(60+D_m)/2]/sin30°. √3 = sin[(60+D_m)/2]/0.5 → sin[(60+D_m)/2] = √3/2 = sin60° → (60+D_m)/2 = 60° → D_m = 60°.
52. In a primary rainbow, the angle of the bow (from anti-solar point) is about 42°. It forms because:
Explanation: Sunlight enters a raindrop, refracts, reflects once at the back surface, and refracts out. Different colours have different minimum deviation angles (red ≈ 42°, violet ≈ 40°). The rainbow appears at these minimum deviation angles.
53. A convex glass lens (n = 1.5) has focal length 10 cm in air. When immersed in water (n = 4/3), its focal length becomes:
Explanation: 1/f ∝ (n_lens/n_medium − 1). In air: 1/10 = (1.5−1)(1/R₁−1/R₂) = 0.5K. In water: 1/f_w = (1.5/(4/3)−1)K = (1.125−1)K = 0.125K = (0.125/0.5)×(1/10) = (1/4)×(1/10) → f_w = 40 cm.
54. Newton's formula for a mirror: x₁x₂ = f² where x₁ and x₂ are object and image distances from focus. If object is 4f from focus, image distance from focus is:
Explanation: x₁x₂ = f² → x₂ = f²/x₁ = f²/(4f) = f/4.
55. If a thin lens forms an image with transverse magnification m, the longitudinal (axial) magnification is:
Explanation: Longitudinal magnification = m_long = −m² (for small axial extent). This means a 3D object has its depth magnified by m² — important for depth of field calculations in photography.
56. A concave mirror (R = 30 cm) is immersed in a liquid (n = 1.5). Its focal length:
Explanation: Mirror focal length f = R/2 depends only on geometry (angle of incidence = angle of reflection), independent of the medium. Refraction at lens surfaces changes with medium; reflection at mirrors does not. f = 30/2 = 15 cm regardless of medium.
57. Two thin lenses L₁ (f₁=20 cm) and L₂ (f₂=10 cm) are separated by 15 cm. Equivalent focal length F is:
Explanation: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/20 + 1/10 − 15/(20×10) = 0.05 + 0.1 − 0.075 = 0.075. Wait: 1/F = 1/20 + 1/10 − 15/200 = 10/200 + 20/200 − 15/200 = 15/200 = 3/40 → F = 40/3 ≈ 13.3 cm. Hmm, let me recheck: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/20 + 1/10 − 15/(20×10). 15/(200) = 0.075. 1/F = 0.05 + 0.10 − 0.075 = 0.075. F = 1/0.075 ≈ 13.3 cm ≈ 40/3 cm.
58. An object is placed at the centre of curvature of a concave mirror and a lens in between. Under what condition does the lens not affect the final image position?
Explanation: A ray passing through the optical centre of a lens is undeviated. If the object is at the lens's optical centre, all rays pass undeviated through the lens (no refraction) — the lens effectively isn't there. The image is then determined purely by the mirror.
59. A 45°-45°-90° glass prism (n=1.5) is used to deviate light by 90° using TIR. This works because:
Explanation: Critical angle for n=1.5: sin θ_c = 1/1.5 = 2/3 → θ_c ≈ 41.8°. When light hits the hypotenuse at 45° (> 41.8°), TIR occurs → light is deviated 90°. Used in periscopes, binoculars, and retroreflectors.
60. Astigmatism in the eye is caused by:
Explanation: An astigmatic eye has a toroidal cornea — different curvature in horizontal vs vertical planes. Horizontal and vertical lines focus at different distances → blurred image. Corrected by cylindrical lenses that add curvature in one plane only.
61. For a thick lens or lens system, the principal planes (H and H') are the reference planes from which:
Explanation: For a thick lens or system of lenses, the equivalent thin lens formula 1/v − 1/u = 1/f holds only when u is measured from the front principal plane H and v is measured from the rear principal plane H'. The principal planes are where the equivalent thin lens is 'located'.
62. Aplanatic points of a refracting sphere (radius R, index n) are at distances:
Explanation: For a sphere of radius R and index n, aplanatic points (points with no spherical aberration) are at distances R/n from centre on one side and nR from centre on the other side. Used in oil-immersion microscope objectives to reduce aberration.
63. In the New Cartesian sign convention for mirrors, distances measured in the direction of incident light are taken as:
Explanation: In the New Cartesian convention: incident light travels in the +x direction. Distances measured in the same direction (+x) are positive; opposite (−x) are negative. The pole P is the origin. For a real object on the left, u is negative.
64. A concave (diverging) lens always forms an image that is:
Explanation: For a diverging lens with any real object position: image is always virtual, erect, and smaller. m = v/u where v
65. A secondary rainbow appears because:
Explanation: Secondary rainbow: two internal reflections + two refractions. Appears at ~51° from anti-solar point. Colour order is reversed (red inside, violet outside) compared to primary. The region between primary and secondary rainbows is dark (Alexander's dark band).
66. An achromatic doublet (convex lens of one glass + concave of another) corrects chromatic aberration by:
Explanation: Condition for achromatic doublet: ω₁/f₁ + ω₂/f₂ = 0 where ω is dispersive power. The two glasses have different dispersive powers — the doublet corrects dispersion while maintaining net converging power. Used in cameras, telescopes, and eyepieces.
67. Resolving power of a telescope (aperture D, wavelength λ) is:
Explanation: Resolving power = 1/θ_min = D/(1.22λ). Larger aperture → higher resolving power → finer angular detail. This is why large observatory mirrors (8–10 m) can resolve structures that smaller instruments cannot.
68. A convex lens (P = +5 D) and a concave lens (P = −3 D) are placed in contact. Net power and focal length:
Explanation: P_total = P₁ + P₂ = 5 + (−3) = +2 D. f = 1/P = 1/2 m = 50 cm. The combination is a converging lens of 50 cm focal length.
69. As angle of incidence on a prism increases from 0° to 90°, the angle of deviation:
Explanation: The deviation-angle curve has a U-shape with a minimum at the angle of minimum deviation (symmetric passage). This is why the minimum deviation angle D_m is used in the prism formula: it's the unique configuration where deviation is least and the prism formula is simplest.
70. For a prism with apex angle A = 30° and n = √3, the minimum angle of incidence for which total internal reflection occurs on the second face is:
Explanation: Critical angle: sin θ_c = 1/√3 → θ_c ≈ 35.3°. For the ray inside the prism, the angle of incidence at the second surface = A − r₁ where r₁ is the refraction angle at first surface. At normal incidence (i = 0°): r₁ = 0°, angle at 2nd face = A = 30° θ_c → A − r₁ > 35.3°. For A = 30°, A
71. An astronomical telescope forms an inverted image. To make it erect, one needs:
Explanation: Astronomical telescopes give inverted images (acceptable for stars). Terrestrial telescopes add an erecting lens (3-lens system) or use Porro prisms (as in binoculars) to invert the image again for an erect view.