Q4. Assertion: apart from the minimum and grazing endpoints, every allowed transmitted deviation is produced by two angles of incidence. Reason: a ray’s path can be traced backwards, so a ray incident at e leaves at i, and δ = i + e − A is unchanged when i and e are swapped. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion (NCERT pages 239–240).
Q5. An equilateral prism has refractive index √2. Find its angle of minimum deviation and the angle of incidence at the minimum.
sin[(60° + Dm)/2] = √2 × sin 30° = 0.707, so (60° + Dm)/2 = 45°: i = 45° and Dm = 30° (NCERT Eqs. (9.37)–(9.38), page 240).
Q6. For a 60° prism with n = 1.5, find r1, r2 and the deviation when i = 40°.
sin r1 = sin 40°/1.5 ≈ 0.429, so r1 ≈ 25.4°; r2 = 60° − 25.4° = 34.6°; sin e = 1.5 sin 34.6° ≈ 0.852, so e ≈ 58.5°; δ = 40° + 58.5° − 60° ≈ 38.5°, a little more than Dm ≈ 37.2° (NCERT Eqs. (9.34)–(9.35), page 239).
Q7. For a 60° prism with n = 1.5, what is the smallest angle of incidence for which light still comes out of the second face?
The ray just emerges when r2 equals the critical angle, sin−1(1/1.5) ≈ 41.8°. Then r1 ≈ 18.2° and sin i = 1.5 sin 18.2° ≈ 0.468, so i ≈ 27.9°; for smaller i the light is totally reflected at the second face (reasoned from NCERT Eqs. (9.12) and (9.34)).
Q8. A prism of angle A has critical angle C. Show that no ray passes through it if A > 2C.
r1 is at most C (at i = 90°), so r2 = A − r1 ≥ A − C, which exceeds C if A > 2C: the ray is always totally reflected at the second face (reasoned from NCERT Eqs. (9.12) and (9.34)). For n = 1.5 the limit is about 83.6°.
Q9. A prism’s minimum deviation equals its angle A. Show that n = 2 cos(A/2).
NCERT Eq. (9.38) gives n = sin A/sin(A/2) = 2 cos(A/2); for A = 60°, n = √3 ≈ 1.73.