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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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The correct answer is:
\(18^{\circ} 36^{\prime}\)
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,
\(\delta=e-i=48^{\circ} 36^{\prime}-30^{\circ}=18^{\circ} 36^{\prime}\)
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,
\(\delta=e-i=48^{\circ} 36^{\prime}-30^{\circ}=18^{\circ} 36^{\prime}\)
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