PRAYOGX
ADV-2026-P2-PHY-Q05

Twin-Prism Mirror Bench — two prisms, one mirror: which statements hold?

JEE Advanced 2026 · Paper 2 · Physics · Q.5 · one or more correct options (+4 / partial / 0 / −1) · Ray Optics and Optical Instruments

01The question

Q.5 Consider two isosceles prisms 1 and 2 with prism angles A1 and A2 and refractive indices n1 and n2, respectively, as shown in the figure. The faces a1b1 and a2b2 are parallel to each other and perpendicular to the mirror M. If a ray of light is incident on the face a1c1 and emerges from the face a2c2, then the correct statement(s) is/are:
  1. (A)If both the prisms are at minimum deviation condition, then n2n1 = sin(A12) / sin(A22).
  2. (B)If prism 2 is at minimum deviation condition, then sin i1 = n2 sin(A22) is always true.
  3. (C)If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation δm1 and δm2, respectively, then θ = δm12(n1 − 1) + δm22(n2 − 1).
  4. (D)If prism 1 is at minimum deviation condition, then sin i2 = n1 sin(A12) is always true.

02The experiment

Target:the correct statement(s)

Ready.

40.0°
60.0°
1.500
50.0°
1.775

03Detailed solution

Answer:

  1. The mirror links the prisms. Faces a1b1 and a2b2 are both perpendicular to M, so their normals are parallel to the mirror. The ray leaves prism 1 at e1 to that normal, i.e. at e1 to the mirror's plane; reflection keeps that angle, so it reaches prism 2 at i2 = e1.
  2. Minimum deviation. A prism deviates least when the ray passes symmetrically: r = A/2 inside and equal angles outside, so sin i = sin e = n sin(A/2).
  3. (D) is true. With prism 1 at minimum deviation, sin e1 = n1 sin(A1/2), and i2 = e1, so sin i2 = n1 sin(A1/2) whatever prism 2 is.
  4. (A) is true. If prism 2 is also at minimum deviation, sin i2 = n2 sin(A2/2). Equating: n1 sin(A1/2) = n2 sin(A2/2), i.e. n2/n1 = sin(A1/2)/sin(A2/2).
  5. (B) is false. Prism 2 at minimum deviation gives sin i2 = sin e1 = n2 sin(A2/2). That equals sin i1 only if i1 = e1, i.e. only if prism 1 is also at minimum deviation. It is not "always" true: the experiment finds a counter-example.
  6. (C) is true. In the figure each dashed line through a1 (or a2) continues inside the prism and meets the base at a right angle: it is the bisector of the prism angle. With the faces a1b1, a2b2 parallel, the bisectors make A1/2 and A2/2 with them, so θ = (A1 + A2)/2. For thin prisms δm = (n − 1)A, so A = δm/(n − 1) and θ = δm1/(2(n1 − 1)) + δm2/(2(n2 − 1)).
  7. A trap. Reading θ as the angle between the extended faces a1c1 and a2c2 would give A1 + A2, twice the right value, and would wrongly reject (C). The right-angle marks on the bases show the dashed lines are bisectors.

Understand the concept

A prism's deviation depends on how the ray enters it: it is smallest when the path inside is symmetric, and then the angles of incidence and emergence are equal with sin i = n sin(A/2). Here a mirror joins two prisms whose facing faces are both perpendicular to it, so whatever angle the ray leaves prism 1 with is exactly the angle it meets prism 2 with. That single link decides every statement: a symmetric prism 1 fixes what prism 2 receives, and both can be symmetric together only when n1 sin(A1/2) = n2 sin(A2/2). For thin prisms the deviation is simply (n − 1)A, which turns the geometry of the bisectors into a statement about deviations.

Key takeaways

  • Minimum deviation: symmetric path, r = A/2, sin i = n sin(A/2), δm = 2i − A.
  • A plane mirror perpendicular to two faces carries the emergence angle of one into the incidence angle of the other.
  • "Always true" needs every case; one counter-example rejects it.
  • Read figures carefully: a right-angle mark on an isosceles prism's base marks the bisector of its apex angle.

Verification: an independent derivation (corrected after a first reading of θ as A₁ + A₂, by examining the right-angle marks in the figure); a vector ray trace with Snell's law at four faces and a mirror reflection, sharing no code with the page; every statement tested across many prism pairs; the thin-prism limit; and the page's own engine, run by tests/verify_p2phyq05.py. The MathonGo solution was read first only for the approach; it gives (A), (D), taking θ as the angle between the faces. The official final answer key (JEE Advanced 2026, Paper 2), read last, gives (A), (C), (D), in agreement with the corrected solve. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

How to use this simulation

  1. Press ▶ START EXPERIMENT. The laser beam is traced through prism 1, off the mirror and through prism 2, with every angle measured.
  2. The bench then tests the statements one by one: it scans i1 to find prism 1's minimum deviation (D), looks for a counter-example (B), tunes n2 until prism 2 is symmetric too (A), and switches to thin prisms to compare θ with the formula (C).
  3. The test board marks each statement true or false with the measured numbers; the target shows the correct set.
  4. Then move i1, A1, A2, n1 or n2, or use Set… to put prism 1, or both, at minimum deviation, or to switch to thin prisms.
  5. Replay runs it again. Reset clears it. Classroom fills the screen for a projector.