A light ray enters through a right angled prism at point $P$ with the angle of incidence $3 0^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$ . The refractive index of the prism is:

Option 1 -

$\frac{\sqrt[]{5}}{4}$

Option 2 -

$\frac{\sqrt[]{5}}{2}$

Option 3 -

$\frac{\sqrt[]{3}}{4}$

Option 4 -

$\frac{\sqrt[]{3}}{2}$

6 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$A = 9 0^{?}$

In prism, $r_{1} + c = A$

$r_{1} = 9 0^{?} - c$ …(i)

$sin c = \frac{1}{μ} \Rightarrow cos c = \frac{\sqrt[]{μ^{2} - 1}}{μ}$

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

$1 \cdot sin 3 0^{?} = μ sin (r_{1}) \Rightarrow 1 \times \frac{1}{2} = μ \times sin (9 0^{?} - c)$

$\frac{1}{2} = μ \times \frac{\sqrt[]{μ^{2} - 1}}{μ}$

On squaring $\frac{1}{4} = μ^{2} - 1$

$\Rightarrow μ^{2} = \frac{5}{4} \Rightarrow μ = \frac{\sqrt[]{5}}{2}$

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A light ray enters through a right angled prism at point $P$ with the angle of incidence $3 0^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$ . The refractive index of the prism is:

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A light ray enters through a right angled prism at point P with the angle of incidence 3 0 ∘ as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC . The refractive index of the prism is:

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A light ray enters through a right angled prism at point $P$ with the angle of incidence $3 0^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$ . The refractive index of the prism is: