JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 21)

In the given figure, the face $$A C$$ of the equilateral prism is immersed in a liquid of refractive index '$$n$$'. For incident angle $$60^{\circ}$$ at the side $$A C$$, the refractive light beam just grazes along face $$A C$$. The refractive index of the liquid $$n=\frac{\sqrt{x}}{4}$$. The value of $$x$$ is ____________.

(Given refractive index of glass $$=1.5$$ )

JEE Main 2022 (Online) 26th July Evening Shift Physics - Geometrical Optics Question 86 English

Answer
27

Explanation

Given prism is equilateral so, angel of prism, A $=60^{\circ}$

On first surface light is passing straight without any deviation, hence angle of refraction $\mathrm{r}_1=0^{\circ}$.

$$ \begin{aligned} & r_1+i_2=\mathrm{A} \\\\ & 0+i_2=60^{\circ} \\\\ & \Rightarrow i_2=60^{\circ} \end{aligned} $$

On $2^{\text {nd }}$ surface ray, refracting ray is becoming parallel to the junction of surfaces, so $i_2$ will act as critical angle.

From snell's law $\mu_1 \sin \theta_1=\mu_2 \sin \theta_2$

$$ \begin{aligned} & \Rightarrow \frac{3}{2} \sin i_2=\mu \sin \theta_2 \\\\ & \Rightarrow \mu=\frac{3}{2} \times \frac{\sin 60^{\circ}}{\sin 90^{\circ}}=\frac{3}{2} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} \text { or } \frac{\sqrt{27}}{4} \end{aligned} $$

$\Rightarrow$ Comparing above equation with

given $n=\frac{\sqrt{x}}{4}$, we get $x=27$

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