A parallel beam of monochromatic light is inc | vedclass

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Question

(A) For a prism,the relation between the angle of incidence $(i_1)$,angle of emergence $(i_2)$,prism angle $(A)$,and angle of deviation $(\delta)$ is

Vedclass · Accepted Answer

Less than $41^o$

Vedclass · Answer

(A) For a prism,the relation between the angle of incidence $(i_1)$,angle of emergence $(i_2)$,prism angle $(A)$,and angle of deviation $(\delta)$ is given by: $i_1 + i_2 = A + \delta$ Given: $i_1 = 55^o$,$i_2 = 46^o$,and for an equilateral prism,$A = 60^o$. Substituting these values: $55^o + 46^o = 60^o + \delta$ $101^o = 60^o + \delta$ $\delta = 41^o$ We know that the angle of minimum deviation $(\delta_m)$ occurs when the angle of incidence equals the angle of emergence $(i_1 = i_2)$. Since $i_1 
eq i_2$ $(55^o 
eq 46^o)$,the deviation $\delta = 41^o$ is not the minimum deviation. Since the deviation curve is a parabola with the minimum at $i_1 = i_2$,any deviation measured at $i_1 
eq i_2$ must be greater than the minimum deviation. Therefore,$\delta_m < \delta$,which implies $\delta_m < 41^o$.

$A$ parallel beam of monochromatic light is incident at one surface of an equilateral prism. The angle of incidence is $55^o$ and the angle of emergence is $46^o$. The angle of minimum deviation will be:

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