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Question: Find the distance of the object placed in the slab of refractive index \(\mu \) from point \(P\) of ...

Find the distance of the object placed in the slab of refractive index μ\mu from point PP of the curved surface with radius RR so that the image is formed at infinity.

A) (μ1)Rμ\dfrac{{(\mu - 1)R}}{\mu }
B) μR(μ1)\dfrac{{\mu R}}{{(\mu - 1)}}
C) Rμ1\dfrac{R}{{\mu - 1}}
D) (μ1)R2μ\dfrac{{(\mu - 1)R}}{{2\mu }}

Explanation

Solution

In this question, use the concept of the refraction of the light due to the special surface of the object which depends on the refractive index of the material of the object and the medium. In this question, first discuss how the refraction phenomenon is happening in the medium. Then write down the formula of refraction in spherical surfaces and discuss various terms of the formula. Finally solve this by applying this formula.

Complete step by step solution:
As we know that when the light changes its direction while travelling from a medium to another medium this is called refraction. The main cause of refraction of light is the change in refractive index of medium.
As we know the formula of the refraction in spherical surfaces is,
μ2vμ1u=μ2μ1R\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}
Where, vv is the distance of the image from the spherical surface and it is negative by sign convention, uu is the distance of the object from the spherical surface μ1{\mu _1} is the refractive index of a medium from which rays are incident, and μ2{\mu _2} is the refractive index of another medium.
Here it is said that the image formation will be at infinity so v=v = \infty , also μ1=1{\mu _1} = 1 and μ1=μ{\mu _1} = \mu
Now we substitute the values in the above equation as,
1μu=μ1R\dfrac{1}{\infty } - \dfrac{\mu }{u} = \dfrac{{\mu - 1}}{R}
0μu=μ1R\Rightarrow 0 - \dfrac{\mu }{u} = \dfrac{{\mu - 1}}{R}
On simplification we get,
u=μRμ1\Rightarrow u = - \dfrac{{\mu R}}{{\mu - 1}}
Here the negative sign has come out due to sign convention.
So, the distance of the object placed in the slab will be μRμ1\dfrac{{\mu R}}{{\mu - 1}}.

Therefore, the correct option is B.

Note: While solving this problem we have to be very much aware about the sign convention of the terms in the formula. The refractive index of a medium is understood by the speed of the light in the first medium through which the light is travelling to the speed of the light in the second medium where the light gets refracted. The refractive index of the air is 11.