A ray of light passes through an equilateral | vedclass

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(A) Given: Refractive index $\mu = 1.5$,Prism angle $A = 60^{\circ}$ (since it is an equilateral prism).The condition given is that the angle of incid

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$30$

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(A) Given: Refractive index $\mu = 1.5$,Prism angle $A = 60^{\circ}$ (since it is an equilateral prism).The condition given is that the angle of incidence $i$ is equal to the angle of emergence $e$,and $i = e = \frac{3}{4} A$.Calculating the angles:$i = e = \frac{3}{4} 	imes 60^{\circ} = 45^{\circ}$.The formula for the angle of deviation $\delta$ is:$\delta = i + e - A$.Substituting the values:$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ}$.$\delta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Therefore,the angle of deviation is $30^{\circ}$.

$A$ ray of light passes through an equilateral prism $(\mu = 1.5)$ such that the angle of incidence is equal to the angle of emergence,and the latter is equal to $3/4$ of the prism angle. The angle of deviation is....$^o$

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