Angle of the prism is $A$ and one of its refracting surfaces is silvered. If light rays falling at an angle of incidence $2 A$ on the first surface returns back through the same path after suffering reflection at the silvered surface, the refractive index $μ$ , of the prism is:

2 sin A

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2 cos A

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\frac{1}{2} cos A

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tan A

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Solution

The correct option is B $2 cos A$

To retrace its path, light rays should fall normally on the reflecting surface.
So,
$\Rightarrow r_{2} = 0, \Rightarrow r_{1} = A - r_{2} \Rightarrow r_{1} = A$

Now applying snell's rule between incident ray and refracted ray.
$\Rightarrow s i n i = μ s i n r_{1} \Rightarrow s i n 2 A = μ s i n A \Rightarrow 2 s i n A c o s A = μ s i n A$
$\Rightarrow μ = 2 c o s A$

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Angle of prism is ‘A’ and its one surface is silvered. Light rays falling at an angle of 2A and travels through the same path after suffering reflection at second silvered surface. The refractive index of the material of the prism is

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Angle of the prism is A and one of its refracting surfaces is silvered. If light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface, the refractive index μ, of the prism is:

The correct option is B 2cosA To retrace its path, light rays should fall normally on the reflecting surface. So, ⇒r2=0,⇒r1=A−r2⇒r1=A Now applying snell's rule between incident ray and refracted ray. ⇒sin i=μ sin r1⇒sin 2A=μ sin A⇒2 sin A cos A=μ sin A ⇒μ=2 cos A

Angle of the prism is $A$ and one of its refracting surfaces is silvered. If light rays falling at an angle of incidence $2 A$ on the first surface returns back through the same path after suffering reflection at the silvered surface, the refractive index $μ$ , of the prism is: