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A ray of light passing through a prism having mu =sqrt2 suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. What is the angle of prism. |
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Answer» SOLUTION :As the prism is in the position of MINIMUM deviation.`delta_m =(2i - A)` with r=A/2 According to given problem, `i = 2R = A (as r = A/2]` `delta_m = 2A - A = A` and hence from `mu = ( sin[ A+ delta_m)//2])/(sin (A//2))i.e., sqrt(2)= ( sin A)/(sin A//2)` (or) `sqrt(2)sin (A )/( 2 )= 2 sin(A )/(2)cos (A )/(2) i.e.,cos ""(A)/(2) =(1)/( sqrt(2)) `(or ) `A/2 = cos^(-1) [(1)/(sqrt2)]=45^@ ,i.e.,,A=90^@` |
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