Ray Optics and Optical Instruments — Questions & Answers

Sample questions from Ray Optics and Optical Instruments

1. Q1. For a convex lens of focal length f, the least possible distance between a real object and its real image is:

   • A.f
   • B.2f
   • C.4f✓
   • D.There is no minimum
   Solution

   Let object-image separation be D = |u| + v. From the lens formula, |u| + v is minimum when |u| = v = 2f, giving D_min = 4f. Equivalently, for a given D the equation x(D - x) = fD has real roots only if D^2 ≥ 4fD, i.e. D ≥ 4f.

2. Q2. A biconvex lens (n = 1.5) has radii of curvature 10 cm and 15 cm. Its focal length is:

   • A.12 cm✓
   • B.6 cm
   • C.25 cm
   • D.30 cm
   Solution

   1/f = (n - 1)(1/R1 - 1/R2) with R1 = +10 cm, R2 = -15 cm: 1/f = 0.5 × (1/10 + 1/15) = 0.5 × (5/30) = 1/12, so f = 12 cm.

3. Q3. Snell's law of refraction, relating the angle of incidence (theta1) and angle of refraction (theta2) at a boundary between two media with refractive indices n1 and n2, is stated as:

   • A.n1 cos(theta1) = n2 cos(theta2)
   • B.n1 sin(theta1) = n2 sin(theta2)✓
   • C.n1/theta1 = n2/theta2
   • D.n1 x theta1 = n2 x theta2 (a simple linear relation)
   Solution

   Snell's law states n1 sin(theta1) = n2 sin(theta2), relating the refractive indices of the two media to the SINES of the angles of incidence and refraction (measured from the normal to the interface) -- this fundamental relation governs how light bends when crossing a boundary between media of different optical density (refractive index), with light bending TOWARD the normal when entering a denser medium (n2>n1) and AWAY from the normal when entering a less dense medium (n2<n1). A student who uses the angles directly (rather than their sines) picks a wrong option.

4. Q4. A concave mirror is used to form a real image of an object on a screen. If the mirror is progressively covered around its edges (leaving only a small central region uncovered, near the pole), the resulting image on the screen becomes:

   • A.Disappears completely as soon as any part of the mirror is covered
   • B.Progressively smaller in overall size as more of the mirror's edge is covered
   • C.Shows only a small, correspondingly cropped portion of the original scene
   • D.Dimmer but remains the SAME complete image (same size, same features), consistent with the general principle that each part of a mirror/lens contributes to forming the WHOLE image✓
   Solution

   Just as with a lens (a related, previously-discussed principle), covering part of a MIRROR'S reflecting surface does NOT remove any PART of the resulting image -- every remaining uncovered portion of the mirror's surface still receives and reflects light from the ENTIRE object, contributing to forming the COMPLETE image -- covering the mirror's edges (leaving only a small central region) simply reduces the total light-gathering area (and hence the image's BRIGHTNESS), without cropping or reducing the SIZE or completeness of the image itself, consistent with the same general optical principle established for lenses. A student who assumes covering part of the mirror crops or shrinks the resulting image picks B or C, missing this general whole-image-formation principle.

5. Q5. An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.44. The critical angle at the core-cladding interface is closest to:

   • A.16.3°
   • B.41.8°
   • C.60.0°
   • D.73.7°✓
   Solution

   sin C = n_cladding/n_core = 1.44/1.50 = 0.96, so C = sin^-1(0.96) ≈ 73.7°. Rays striking the interface at angles above 73.7° are totally internally reflected and guided along the core.

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