(a) Draw the ray diagram showing refraction of ray of light through a glass prism. Derive the expression for the refractive index \mu of the material of prism in terms of the angle of prism A and angle of minimum deviation \delta _{m} .

(b) A ray of light PQ enters an isosceles right angled prism ABC of refractive index 1.5 as shown in figure.

     (i) Trace the path of the ray through the prism.

     (ii) What will be the effect on the path of the ray if the refractive index of the prism is 1.4 ?

 

 
 
 
 

Answers (1)
S safeer

i) 

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

b) 

i)

 

ii) If \mu=1.4 Total Internal Reflection will not occur

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