A prism with refractive index mu = 1.5 has a | vedclass

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(A) The prism angle is $A = 30^\circ$. The ray is incident normally on one surface,so the angle of incidence at the first surface is $i_1 = 0^\circ$,w

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$18^\circ 36'$

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(A) The prism angle is $A = 30^\circ$. The ray is incident normally on one surface,so the angle of incidence at the first surface is $i_1 = 0^\circ$,which implies the angle of refraction at the first surface is $r_1 = 0^\circ$.Since $A = r_1 + r_2$,we have $r_2 = A - r_1 = 30^\circ - 0^\circ = 30^\circ$.At the second surface,the angle of incidence is $r_2 = 30^\circ$. Let the angle of emergence be $e$. Using Snell's Law: $\mu \sin(r_2) = 1 \cdot \sin(e)$.$1.5 \cdot \sin(30^\circ) = \sin(e) \Rightarrow 1.5 \cdot 0.5 = \sin(e) \Rightarrow \sin(e) = 0.75$.Given $\sin(48^\circ 36') = 0.75$,so $e = 48^\circ 36'$.The deviation $\delta$ produced by the prism is given by $\delta = e - A$ (since $i_1 = 0$ and $r_1 = 0$).$\delta = 48^\circ 36' - 30^\circ = 18^\circ 36'$.

$A$ prism with refractive index $\mu = 1.5$ has a prism angle of $30^\circ$. If a ray is incident normally on one surface,calculate its deviation. (Given: $\sin(48^\circ 36') = 0.75$)

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