A prism $(\mu=1.5)$ has the refracting angle of $30^{\circ}$. The deviation of a monochromatic ray incident normally on its one surface will be (given, $\sin 48^{\circ} 36^{\prime}=0.75$)

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Aicharya · Accepted Answer

The given situation is shown in the following figure.

For the surface $A B$, $\frac{1}{\mu}=\frac{\sin i}{\sin e} \Rightarrow \frac{1}{\mu}=\frac{\sin 30^{\circ}}{\sin e}$ $\begin{array}{rlrl} \Rightarrow & \sin e =\mu \sin 30^{\circ}=1.5 \times \frac{1}{2}=0.75 \\ & =\sin 48^{\circ} 36^{\prime} \quad\left[\therefore \sin 48^{\circ} 36^{\prime}=0.75\right] \\ \Rightarrow & e =48^{\circ} 36^{\prime} \end{array}$ From figure angle of deviation, $\delta=e-i=48^{\circ} 36^{\prime}-30^{\circ}=18^{\circ} 36^{\prime}$

A prism \((\mu=1.5)\) has the refracting angle of \(30^{\circ}\). The deviation of a monochromatic ray…

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