Question

Find the angle of deviation suffered by the light ray shown in figure. The refractive index μ = 1.5 for the prism material.
Figure

Answer 1

Given,
The refractive index of the prism material (μ) = 1.5
Angle of prism form the figure = 4˚

We know that,
$$\begin{gathered} \mu =\frac{\sin \left({\displaystyle \frac{\mathrm{A}+{\delta }_{\mathrm{m}}}{2}}\right)}{\sin {\displaystyle \frac{A}{2}}} \\ \mu =\frac{\left({\displaystyle \frac{\mathrm{A}+{\delta }_{\mathrm{m}}}{2}}\right)}{{\displaystyle \frac{A}{2}}}\left(\because forsmallangle\sin \theta \approx \theta \right) \\ =\frac{\mathrm{A}+{\delta }_{\mathrm{m}}}{2°} \\ 1.5=\frac{4°+{\delta }_{\mathrm{m}}}{2°} \\ {\delta }_{\mathrm{m}}=2° \end{gathered}$$

Hence, the angle of deviation is 2˚