A prism of refractive index n and angle a

Asked by admin @ 16/08/2021 in Physics viewed by 344 People

A prism of refractive index n and angle A is placed in minimum deviation position. If the angle of minimum deviation is equal to the angle A then the value of A is

Answered by admin @ 16/08/2021

Given:

A prism of refractive index n and angle A is placed in minimum deviation position. The angle of minimum deviation is equal to the angle A.

To find:

Value of A ?

Calculation:

Expression for minimum deviation:

$\therefore \: \mu = \dfrac{ \sin( \frac{A + \delta}{2} ) }{ \sin( \frac{A}{2} ) }$

Since $A = \delta$ :

$\implies \: \mu = \dfrac{ \sin( \frac{A + A }{2} ) }{ \sin( \frac{A}{2} ) }$

$\implies \: \mu = \dfrac{ \sin( \frac{2A}{2} ) }{ \sin( \frac{A}{2} ) }$

$\implies \: \mu = \dfrac{ \sin(A) }{ \sin( \frac{A}{2} ) }$

$\implies \: \mu = \dfrac{ 2\sin( \frac{A}{2} ) \cos( \frac{A}{2} ) }{ \sin( \frac{A}{2} ) }$

$\implies \: \mu = 2 \cos( \frac{A}{2} )$

$\implies \: \cos( \frac{A}{2} ) = \dfrac{ \mu}{2}$

$\implies \: \sqrt{1 - { \sin}^{2} ( \frac{A}{2}) } = \dfrac{ \mu}{2}$

$\implies \: 1 - { \sin}^{2} ( \frac{A}{2}) = \dfrac{ { \mu}^{2} }{4}$

$\implies \: { \sin}^{2} ( \frac{A}{2}) = 1 - \dfrac{ { \mu}^{2} }{4}$

$\implies \: { \sin}^{2} ( \frac{A}{2}) = \dfrac{ 4 - { \mu}^{2} }{4}$

$\implies \: \sin( \frac{A}{2}) = \sqrt{ \dfrac{ 4 - { \mu}^{2} }{4} }$

$\implies \: \dfrac{A}{2} = { \sin}^{ - 1} \bigg \{ \sqrt{ \dfrac{ 4 - { \mu}^{2} }{4} } \bigg \}$

$\implies \: A = 2{ \sin}^{ - 1} \bigg \{ \sqrt{ \dfrac{ 4 - { \mu}^{2} }{4} } \bigg \}$

Since $\mu= n$ in this question:

$\implies \: A = 2{ \sin}^{ - 1} \bigg \{ \sqrt{ \dfrac{ 4 - {n}^{2} }{4} } \bigg \}$

So, final answer is:

$\boxed{ \bf \: A = 2{ \sin}^{ - 1} \bigg \{ \sqrt{ \dfrac{ 4 - {n}^{2} }{4} } \bigg \}}$

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