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  • (i) For a glass prism  the angle of minimum deviation is equal to the angle of prism. Calculate the angle of prism. <p

(i) For a glass prism (\mu = \sqrt{3}) the angle of minimum deviation is equal to the angle of prism. Calculate the angle of prism.

(ii) Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having a refractive index \mu = \sqrt{3}.

 

 

 

 
 
 
 
 

Answers (1)

(i) The relation between the refractive index of prism and angle of minimum deviation (D) and angle of the prism is (A).

    \mu = \frac{\sin \left ( \frac{A+D}{2} \right )}{\sin\left ( \frac{A}{2} \right )}

given A = D

\mu = \frac{\sin A}{\sin\left ( \frac{A}{2} \right )}=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}= 2\cos \frac{A}{2}

\therefore \frac{\sqrt{3}}{2}=\cos \frac{A}{2}

\frac{A}{2} = 30^{\circ}

A = 60^{\circ}

(ii)

\mu =\sqrt{3}=\frac{1}{\sin i_{c}}

\mu =\frac{1}{\sqrt{3}}=\sin i_{c}

\therefore i_{c} is between 30 and 45^{\circ}

TIR takes place.

Posted by

Safeer PP

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