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Question
Find the angle of deviation suffered by the light ray shown in figure. The refractive index μ = 1.5 for the prism material.
Solution
Given,
The refractive index of the prism material (μ) = 1.5
Angle of prism form the figure = 4˚
We know that,
\[\mu = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}}\]
\[ \mu = \frac{\left( \frac{A + \delta_m}{2} \right)}{\frac{A}{2}} \left( \text{ because for small angle } \sin \theta \approx \theta \right)\]
\[ = \frac{A + \delta_m}{2^\circ }\]
\[1 . 5 = \frac{4^\circ + \delta_m}{2^\circ}\]
\[ \delta_m = 2^\circ \]
Hence, the angle of deviation is 2˚
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