Question

A clear transparent glass sphere $\left( {\mu = 1.5} \right)$ of radius $R$ is immersed in a liquid of refractive index $1.25$ . A parallel beam of light incident to it will cover it to a point. The distance of the point from the center will be
\left( A \right) - 3R \\\ \left( B \right) + 3R \\\ \left( C \right) - R \\\ \left( D \right) + R \\\

Answer 1

To solve this question, we are going to first consider the condition of the refraction for the first refracting surface, then, putting the values in it, we get the value of the distance at which the image is formed which serves as the virtual object, then, the final image distance is calculated.
The refraction condition is given as:
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Where, $R$ is the radius of the sphere, ${\mu _1}$ and ${\mu _2}$ the refractive indices of the two surfaces, $v$ and $u$ the image and the object distances.

Complete step by step answer:
Let us consider that the refractive index of the glass sphere be ${\mu _1}$ which is equal to $1.5$ and the refractive index of the liquid be ${\mu _2}$ which is equal to $1.25$ .
Now at the first surface the refraction condition is given as:
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Where, $R$ is the radius of the sphere
$u = \infty$
This implies that
$\dfrac{3}{{2v}} - 0 = \dfrac{1}{{4R}}$
Solving this equation further, we get
$v = 6R$
Now this image so formed serves as the virtual object for the second refracting surface.
The distance of this virtual object is:
$u = + 4R$
Now the refraction condition at this surface will be
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Putting the values,
\dfrac{5}{{4v}} - \dfrac{3}{{8R}} = \dfrac{1}{{4R}} \\\ \Rightarrow v = 2R \\\
So, the image is finally formed at $v = 2R$
Thus, the distance of the point of coverage from the center will be
$2R + R = 3R$
Hence, the option $\left( B \right) + 3R$ is the correct answer.

Note:
Whenever a sphere of radius $R$ is immersed in any other liquid whose refractive index is less than its own refractive index, then the final point of coverage is formed at the distance that is thrice than that of the radius. Thus, the object appears to be smaller than its actual size.