JEE Advanced 2026 · Paper 2 · Physics Q5

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PhysicsMultiple correct+4 / -1Question 23 of 54

Consider two isosceles prisms 1 and 2 with prism angles $A_1$ and $A_2$ and refractive indices $n_1$ and $n_2$ , respectively, as shown in the figure. The faces $a_1b_1$ and $a_2b_2$ are parallel to each other and perpendicular to the mirror $M$ . If a ray of light is incident on the face $a_1c_1$ and emerges from the face $a_2c_2$ , then the correct statement(s) is/are:

Two triangular prisms standing on either side of a horizontal mirror M. Prism 1 (left, apex a₁ at top, vertices c₁ and b₁; apex angle A₁, index n₁): a ray enters face a₁c₁ at incidence i₁ and leaves the vertical face a₁b₁ at angle e₁, going down to M. The reflected ray rises to prism 2 (apex a₂, vertices b₂ and c₂; apex angle A₂, index n₂), enters the vertical face a₂b₂ at angle i₂ and leaves face a₂c₂ at angle e₂. In each prism a dashed line runs from the apex perpendicular to the base (c₁b₁, c₂b₂; right-angle marks), i.e. along its symmetry axis; these two axes, extended upward beyond a₁ and a₂, meet above at angle θ.

If both the prisms are at minimum deviation condition, then

\dfrac{n_2}{n_1}=\sin\left(\dfrac{A_1}2\right)\Big/\sin\left(\dfrac{A_2}2\right)

If prism 2 is at minimum deviation condition, then

\sin i_1=n_2\sin\left(\dfrac{A_2}2\right)

is always true.

If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation

\delta_{m1}

and

\delta_{m2}

, respectively, then

\theta=\dfrac{\delta_{m1}}{2(n_1-1)}+\dfrac{\delta_{m2}}{2(n_2-1)}

If prism 1 is at minimum deviation condition, then

\sin i_2=n_1\sin\left(\dfrac{A_1}2\right)

is always true.

Answer and worked solution

Official answer: A, C, D

Key fact. The faces $a_1b_1$ and $a_2b_2$ are vertical and the mirror is horizontal. Their normals are horizontal, and reflection at $M$ only reverses the vertical component of the ray. So the ray leaving $a_1b_1$ at angle $e_1$ meets $a_2b_2$ at the same angle: $i_2=e_1$ .

At minimum deviation in an isosceles prism, the incidence and emergence angles are equal and $\sin i=n\sin\frac A2$ .

(D) Prism 1 at minimum deviation: $\sin e_1=n_1\sin\frac{A_1}2$ , and $i_2=e_1$ . True (no condition on prism 2 is needed).
(A) Both at minimum deviation: $n_1\sin\frac{A_1}2=\sin e_1=\sin i_2=n_2\sin\frac{A_2}2$ , so $\dfrac{n_2}{n_1}=\dfrac{\sin(A_1/2)}{\sin(A_2/2)}$ . True.
(B) Prism 2 at minimum deviation gives $\sin i_2=n_2\sin\frac{A_2}2$ , i.e. a condition on $e_1$ , not on $i_1$ . It holds for $i_1$ only if prism 1 is also symmetric. Not always true.
(C) The dashed lines are the prisms' symmetry axes. Each makes angle $\frac{A}2$ with the vertical face, so $\theta=\frac{A_1}2+\frac{A_2}2$ . For thin prisms $\delta_m=(n-1)A$ , so
$\theta=\frac{\delta_{m1}}{2(n_1-1)}+\frac{\delta_{m2}}{2(n_2-1)}.$ True.

Topics:Prism and dispersion Reflection and plane mirrors

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JEE Advanced 2026 · Paper 2, Physics question 5