Physics | VITEEE 2015 | Discussion Page

The refractive index for a prism is given as $\mu=\cot\frac{A}{2}$ . Then , angle of minimum deviation in terms of angle of prism is

$90^{0}-A$

B) 2A

$180^{0}-A$

$180^{0}-2A$

Option D

Using prism formula

$\mu=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin(\frac{A}{2})}$ .....(i)

where , A= angle of prism

$\delta_{m}$ = angle of minimum deviation

Given, $\mu=\cot(\frac{A}{2})=\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

So, from eq.(i)

$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$\Rightarrow \sin (\frac{\pi}{2}-\frac{A}{2})=\sin\left(\frac{A}{2}+\frac{\delta_{m}}{2}\right)$

$\Rightarrow \delta_{m}=\pi-2A=180^{0}-2A$

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