For a thin prism,delta_1 is the angle of devi | vedclass

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(B) The angle of deviation $\delta$ for a thin prism is given by the formula $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism ma

Vedclass · Accepted Answer

$1: 4$

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(B) The angle of deviation $\delta$ for a thin prism is given by the formula $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.When the prism is in air,the refractive index is ${ }_{a} \mu_{g} = \frac{3}{2}$. Thus,$\delta_1 = (\frac{3}{2} - 1)A = \frac{1}{2}A$.When the prism is immersed in water,the refractive index of the prism relative to water is ${ }_{w} \mu_{g} = \frac{{ }_{a} \mu_{g}}{{ }_{a} \mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.Thus,$\delta_2 = (\frac{9}{8} - 1)A = \frac{1}{8}A$.The ratio $\delta_2 : \delta_1 = \frac{\frac{1}{8}A}{\frac{1}{2}A} = \frac{1}{8} 	imes 2 = \frac{1}{4}$.Therefore,the ratio is $1: 4$.

For a thin prism,$\delta_1$ is the angle of deviation produced when the prism is placed in air. When the prism is immersed in water,the angle of deviation produced is $\delta_2$. Given ${ }_{a} \mu_{g}=\frac{3}{2}$ and ${ }_{a} \mu_{w}=\frac{4}{3}$. The ratio $\delta_2: \delta_1$ is

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