Question

A prism has minimum angle of deviation at ${40}^{\circ }.$ Refracting angle of prism is ${60}^{\circ }.$ If the prism is placed in water of refractive index is $1.33,$ what will be new minimum angle of deviation?

Answer 1

Solution

we know , $\int =A(\mu -1)$

Initially,

${40}^{\circ }=\left({60}^{\circ }\right)(\frac{\mu prism}{\mu air}-1)$

$\Rightarrow \mu$ prism = $\frac{40}{60}+1=1.66$

After requiring it in water,

$f^{\prime }=\left({60}^{\circ }\right)(\frac{\mu prism}{\mu water}-1)=60[\frac{1.66}{1.33}-1]$

$f^{\prime }-{14.89}^{\circ }$