Q167 / 180 · KCET 2016 · Physics

A ray of light suffers a minimum deviation when incident on an equilateral prism of refractive
index \( \sqrt{2} \). The angle of incidence is

  • A
    \( 30^{\circ} \)
  • Correct answer
    B
    \( 45^{\circ} \)
  • C
    \( 60^{\circ} \)
  • D
    \( 50^{\circ} \)

Solution

Angle of deviation $\sigma=i_{1}+i_{2}-A$ where $\mathrm{A}$ is angle of prism $A=r_{1}+r_{2}$ $\Rightarrow \delta=i_{1}+i_{2}-\left(r_{1}+r_{2}\right)$ At minimum deviation $r_{1}=r_{2}$ and $i_{1}=i_{2}$ $\Rightarrow \delta=2 i_{1}-2 r_{1}$ and $A=2 r_{1}$ For equilateral prism, $\mathrm{A}=60^{\circ}$ $\Rightarrow 2 r_{1}=60^{\circ} \Rightarrow r_{1}=30^{\circ}$ Now we know $\sin i_{1}=\mu \sin r_{1} \Rightarrow i_{1}=\sin ^{-1}\left(\mu \sin r_{1}\right)$ $i_{1}=\sin ^{-1}\left(\sqrt{2} \sin 30^{\circ}\right)=\sin ^{-1}\left(\sqrt{2} \times \frac{1}{2}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ $\Rightarrow i_{1}=45^{\circ}$

Hence, the correct answer is option: 'B'.

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