The angle of minimum deviation produced by a | vedclass

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(C) For a thin prism, the angle of minimum deviation $\delta$ is given by the formula: $\delta = (n - 1)A$, where $n$ is the refractive index of the p

Vedclass · Accepted Answer

$\delta_2 = \frac{1}{4} \delta_1$

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(C) For a thin prism, the angle of minimum deviation $\delta$ is given by the formula: $\delta = (n - 1)A$, where $n$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.In air, $\delta_1 = (^a n_g - 1)A = (\frac{3}{2} - 1)A = \frac{1}{2}A$.When the prism is immersed in liquid, the refractive index of the prism relative to the liquid is $^l n_g = \frac{^a n_g}{^a n_l} = \frac{3/2}{4/3} = \frac{9}{8}$.The new angle of minimum deviation is $\delta_2 = (^l n_g - 1)A = (\frac{9}{8} - 1)A = \frac{1}{8}A$.Comparing $\delta_1$ and $\delta_2$: $\frac{\delta_2}{\delta_1} = \frac{1/8 A}{1/2 A} = \frac{2}{8} = \frac{1}{4}$.Therefore, $\delta_2 = \frac{1}{4} \delta_1$.

Similar Questions

How is the graph between the angle of deviation $(\delta)$ and the angle of incidence $(i)$ represented for a triangular prism?

$A$ certain prism is found to produce a minimum deviation of $38^\circ$. It produces a deviation of $44^\circ$ when the angle of incidence is either $42^\circ$ or $62^\circ$. What is the angle of incidence when it is undergoing minimum deviation (in $^\circ$)?

The refractive index of glass for red light is $1.520$ and for blue light is $1.525$. Let $D_1$ and $D_2$ be the angles of deviation for red and blue light,respectively,in a prism of this glass. Then:

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