If the polarising angle of a piece of glass f | vedclass

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(A) According to Brewster's law,the refractive index $\mu$ is given by $\mu = 	an 	heta_p$,where $	heta_p$ is the polarising angle.Given $	heta_p

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) According to Brewster's law,the refractive index $\mu$ is given by $\mu = 	an 	heta_p$,where $	heta_p$ is the polarising angle.Given $	heta_p = 54.74^o$,we have $\mu = 	an 54.74^o = 1.414 = \sqrt{2}$.For an equilateral prism,the angle of the prism $A = 60^o$.The formula for the refractive index of a prism is $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$,where $\delta_m$ is the angle of minimum deviation.Substituting the values: $\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(60^o/2)}$.$\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(30^o)}$.Since $\sin(30^o) = 0.5 = 1/2$,we get $\sqrt{2} = 2 \sin((60^o + \delta_m)/2)$.$\sin((60^o + \delta_m)/2) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.This implies $(60^o + \delta_m)/2 = 45^o$.$60^o + \delta_m = 90^o$.$\delta_m = 30^o$.Therefore,option $A$ is correct.

If the polarising angle of a piece of glass for green light is $54.74^o$,then the angle of minimum deviation for an equilateral prism made of the same glass is......$^o$ (Given $\tan 54.74^o = 1.414$)

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