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NEET 2024
Exam
NEET
Year
2024
Subject
Physics
Chapter
Ray Optics

NEET 2024 · Physics · Ray Optics

A light ray enters through a right angled prism at point P with the...

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

Options

  1. A52\frac{\sqrt{5}}{2}
  2. B34\frac{\sqrt{3}}{4}
  3. C32\frac{\sqrt{3}}{2}
  4. D54\frac{\sqrt{5}}{4}

Correct answer

A. 52\frac{\sqrt{5}}{2}

Step-by-step solution

A light ray enters through a right angled prism at point P with the angle of incidence 30^ as shown in figure. It
In prism, r1+c=Ar1=90∘−c…(1)sin⁡c=1μ⇒cos⁡c=μ2−1μ⇒r_1+c=A \begin{gathered} r_1=90^{\circ}-c \ldots(1)\\ \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{gathered} \Rightarrow Apply Snell's law, on incidence surface 1⋅sin⁡30∘=μsin⁡(r1)⇒1×12=μ×sin⁡(90∘−c)12=μ×μ2−1μ\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 14=μ2−1⇒μ2=54⇒μ=52\frac{1}{4}=\mu^2-1 \Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}
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