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Q.

A prism $(μ = 1.5)$ has the refracting angle of $30^{0} .$ The deviation of a monochromatic ray incident normally on its one surface in degrees will be $(\sin 49^{0} = 0.75)$

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answer is 19.

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Detailed Solution

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$\begin{array}{l} A = 30^{0} \\ Ray incident normally on 1^{st} surface \\ i = 0^{0} \Rightarrow r_{1} = 0^{0} \\ r_{1} + r_{2} = A \Rightarrow r_{2} = 30^{0} \\ Applying Snell ’ s law on 2^{nd} surface \\ {μsinr}_{2} = 1 \times sine \\ \frac{3}{2} \sin 30^{0} = sine \\ e = 49^{0} \\ δ = e - r_{2} = 49^{0} - 30^{0} = 19^{0} \end{array}$

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A prism (μ=1.5) has the refracting angle of 300.The deviation of a monochromatic ray incident normally on its one surface in degrees will be (sin 490=0.75)