The minimum deviation produced by a hollow pr | vedclass

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Question

(A) For a prism at the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) For a prism at the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2 = r$.Given that the angle of refraction $r = 30^o$,we know that for a prism,the angle of the prism $A = r_1 + r_2 = 2r$.Therefore,$A = 2 	imes 30^o = 60^o$.The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$,where $\delta_m$ is the angle of minimum deviation.Given $\delta_m = 30^o$ and $A = 60^o$,we have:$\mu = \frac{\sin(\frac{60^o + 30^o}{2})}{\sin(\frac{60^o}{2})} = \frac{\sin(45^o)}{\sin(30^o)}$.Substituting the values,$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $30^o$. The angle of refraction is also found to be $30^o$. The refractive index of the liquid is

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