Q158 / 180 · KCET 2020 · Physics

The refracting angle of prism is $A$ and refractive index of material of prism is $\cot \frac{A}{2}$. The angle of minimum deviation is $A$

  • A
    $180^{\circ}-3 A$
  • B
    $180^{\circ}+2 A$
  • C
    $90^{\circ}-A$
  • Correct answer
    D
    $180^{\circ}-2 A$

Solution

Given, angle of prism $=A$ Refractive index of prism, $\mu=\cot \frac{A}{2}$ As, we know, $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ where, $\delta_{m}=$ angle of minimum deviation. $\Rightarrow \quad \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ $\Rightarrow \quad \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ $\Rightarrow \sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_{m}}{2}\right)$ $\Rightarrow \quad \frac{180^{\circ}-A}{2}=\frac{A+\delta_{m}}{2}$ $\Rightarrow \quad \delta_{m}=180^{\circ}-2 A$

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

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