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Geometrical Optics question

2024 · Q170
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Geometrical Optics question

2024 · Q170

NEETPhysicsGeometrical OpticsMCQ+4 / −1

A light ray enters through a right angled prism at point PPP with the angle of incidence 30∘30^{\circ}30∘ as shown in figure. It travels through the prism parallel to its base BCB CBC and emerges along the face ACA CAC. The refractive index of the prism is:

NEET 2024 Physics - Geometrical Optics Question 4 English

  1. A
    54\frac{\sqrt{5}}{4}45​​
  2. B
    52\frac{\sqrt{5}}{2}25​​
  3. C
    34\frac{\sqrt{3}}{4}43​​
  4. D
    32\frac{\sqrt{3}}{2}23​​
View written solutionFree

Correct answer: B

NEET 2024 Physics - Geometrical Optics Question 4 English Explanation

In prism, $r_1+c=A$

$$\begin{aligned} \quad \& r_1=90^{\circ}-c \quad \text{.... (1)}\\ \& \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$$

$\Rightarrow$ Apply Snell's law, on incidence surface

$$\begin{aligned} 1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right) \\ \frac{1}{2}=\mu \times \frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$$

On squaring $\frac{1}{4}=\mu^2-1$

$$\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$$

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