JEE Main 2026 Jan 23 Shift 2, Physics Q15: Critical Angle and Total Internal Reflection

JEE Main2026Jan 23, Shift 2Physics

A prism of angle $7 5^{\circ}$ and refractive index $3$ is coated with thin film of refractive index $1.5$ only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must not be ______.

( $sin 1 5^{\circ} = 0.25$ and $sin 2 5^{\circ} = 0.43$ )

A
between $1 5^{\circ}$ and $2 0^{\circ}$
B
$1 5^{\circ}$
C
$3 0^{\circ}$
D
$< 1 5^{\circ}$

Solution

At the prism's back surface, the light goes from medium with $μ = 3$ into the film with $μ = 1.5$ . For TIR at this surface:

$3 sin r_{2} \geq 1.5 sin 9 0^{\circ} \Rightarrow sin r_{2} \geq \frac{1.5}{3} = \frac{3}{2}$ , so $r_{2} \geq 6 0^{\circ}$ .

Inside the prism, $r_{1} + r_{2} = 7 5^{\circ}$ (prism angle). So $r_{1} \leq 1 5^{\circ}$ .

At the front face (air to prism, $μ = 3$ ): $sin i = 3 sin r_{1} \leq 3 sin 1 5^{\circ} = 3 \cdot 0.25 \approx 0.433$ .

So $i \leq sin^{- 1} (0.433) \approx 2 5^{\circ}$ , i.e. the incident angle must be no larger than about $2 5^{\circ}$ .

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JEE Main 2026 Jan 23 Shift 2, Physics Q15: Critical Angle and Total Internal Reflection

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