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  • An equilateral glass prism has a refractive index  in air. calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index 

An equilateral glass prism has a refractive index 1.6 in air. calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index 4\sqrt{2}/5. 

 

 

 

 
 
 
 
 

Answers (1)

 

n=\frac{\sin \frac{\delta m +A}{2}}{\sin \frac{A}{2}}

Where, '\delta m' is the angle of minimum deviation,

  'n'  is the refractive index of the prism with respect to the medium 

 'A' is the angle of the prism 

Hence, 

The refractive index of the prism with respect to the medium will be - n=\frac{1.6}{\frac{4\sqrt{2}}{5}}=\frac{1.6}{4\sqrt{2}}\times \frac{5}{1}=\frac{8}{4\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}

 Now, substituting the above values, in the formula used:-

We have,

 \sqrt{2}=\frac{\sin (\frac{\delta m}{2}+\frac{60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}=\frac{\sin \left ( \frac{\delta m}{2}+30^{\circ} \right )}{\sin 30^{\circ}} \frac{\sqrt{2}}{2}=\sin \left ( \frac{\delta m}{2}+30^{\circ} \right )            \left [ \because \sin 30^{\circ} =\frac{1}{2}\right ]

\frac{\delta m}{2}+30^{\circ}=sin^{-1}(\frac{1}{\sqrt{2}})

  \\\Rightarrow \frac{\delta m}{2}+30^{\circ}=45^{\circ}

  \delta m=30^{\circ}

Posted by

Safeer PP

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