NEET 2024 — Physics solution

NEET2024Physics

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 BC and emerges along the face AC. The refractive index of the prism is:

A
$\frac{3}{4}$
B
$\frac{3}{2}$
C
$\frac{5}{4}$
D
$\frac{5}{2}$

Solution

Let the refraction angle inside the prism at face AB be $r$ . Since the ray inside runs parallel to base BC, geometry of the right-angled prism gives the angle at face AC such that the ray emerges grazing along AC; this means the refracted ray inside meets AC at the critical angle $θ_{c}$ , where $sin θ_{c} = 1/ μ$ .

Applying Snell's law at face AB: $sin 3 0^{\circ} = μ sin r$ , with the geometry constraint $r + θ_{c} = 9 0^{\circ}$ giving $cos r = sin θ_{c} = 1/ μ$ , so $sin r = 1 - 1/ μ^{2}$ .

Substituting: $\frac{1}{2} = μ 1 - \frac{1}{μ ^{2}} = μ^{2} - 1$ .

$\Rightarrow μ^{2} - 1 = \frac{1}{4} \Rightarrow μ^{2} = \frac{5}{4} \Rightarrow μ = \frac{5}{2}$ .

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