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Grade: 11
        A thin equilateral prism has refractive index of 1.5.find the angle of minimum deviation. Sir how to solve it?
11 months ago

Answers : (1)

Arun
15912 Points
							
Refractive index = 1.5
Hence
1.5 = [sin(A + \Delta)/2 ]/sin A/2
A = 60°
Now
1.5 sin60/2 = sin(60+\Delta)/2
1.5 * 1/2 = sin(60 + \Delta)/2
48.6 ° = (60 + \Delta)/2
97.2 -60 = \Delta
\Delta = 37.2°
 
Regards
Arun (askIITians forum expert)
11 months ago
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