Q.

For a prism of prism angle  θ=600 , the refractive indices of the left half and right half are, respectively, n1  and  n2   (n2n1)    as shown in figure. The angle of incidence i is chosen such that the incident light rays will have minimum deviation if  n1=n2=n=1.5 .For the case of unequal refractive indices  n1=n  and  n2=n+Δn  ,(where is Δn  <<< n ) the angle of emergence is e=i+Δe  . Which of the following statement(s) is (are) correct? 

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a

  Δe lies between 1.0 and 1.6 miliradians, if Δn=2.8×103 

b

The value of  Δe is  (in radians) greater than that of   Δn

c

Δe  is proportional to  Δn

d

Δe lies between 2.0 and 3.0 miliradians, if  Δn=2.8×103

answer is B, C.

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Detailed Solution

Given angle of prism  A=600
   For minimum deviation.
 r1=r2=A/2=300
and from snell’s law,  n×sini=n1sinr1
1×sini=n1sinA2       sini=32×sin300=32×12sini=34

At another face of prism 
 n1sin300=1sin(e)
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On differentiating both sides  Δn  sin300 =Δe  cos(e)    

Δe=Δn2cos(e)   or   Δe=Δn21916=27Δn Δe=27ΔnΔe<Δn  and  ΔeαΔn Δn=2.8×103 Δe=2.8×103×27=2.11×103  rad=2.11mrad   

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