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(a) Deduce the expression, by drawing a suitable ray diagram, for the refractive
index of a triangular glass prism in terms of the angle of minimum deviation (D)
and the angle of prism (A).
Draw a plot showing the variation of the angle of deviation with the angle of
incidence.
(b) Calculate the value of the angle of incidence when a ray of light incident on one
face of an equilateral glass prism produces the emergent ray, which just grazes
along the adjacent face. Refractive index of the prism is \sqrt{2}.

 

 

 
 
 
 
 

Answers (1)

a )
       
at a minimum angle of deviation
i= e\, ,r_{1}= r_{2}
\therefore D_{m}= i+e-A
             = 2i-A
i= \frac{A+D_{m}}{2}
By snell's law
n_{1}\sin i= n_{2}\sin r
n_{1}\sin\left ( \frac{A+D_{m}}{2} \right )= n_{2}\sin \left ( \frac{A}{2} \right )
\frac{n_{2}}{n_{1}}= \frac{\sin \left ( \frac{A+D_{m}}{2} \right )}{\sin \left ( \frac{A}{2} \right )}

b)
   \frac{\sin r_{2}}{\sin 90}= \frac{1}{\sqrt{2}}
r_{2}= 45^{\circ}
r_{1}+r_{2}= A
r_{1}+45= 60^{\circ}
r_{1}= 15^{\circ}
\frac{\sin i}{\sin r_{1}}= \sqrt{2}
\sin i= \sqrt{2}\sin 15^{\circ}
i= \sin^{-1}\left ( \sqrt{2}\sin 15 \right )= 21\cdot 47^{\circ}

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Safeer PP

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