Question

Find the radius of the sphere if the mark on the surface of the glass sphere having refractive index $1.5$ is viewed from a diametrically opposite position. It appears to be at distance $10cm$ from behind its actual position.
(A) $10cm$
(B) $8cm$
(C) $6cm$
(D) $5cm$

Answer 1

Hint The distance of the object will be the diameter of the glass sphere and the image distance will be the difference between $10cm$ and object distance. Then, use the lens formula for refraction between surfaces of two different refractive indices.

Complete step by step answer:
There are some basic laws of refraction:
The light bends towards the normal if the ray of light travels from rarer medium to denser medium.
All the rays including incident ray, refracted ray and the normal at the point of incidence, lie in the same plane.
The ray of light bends away from the normal if the ray of light travels from denser medium to rarer medium.
The incident ray and refracted ray all follow the Snell’ law of refraction which can be written as-
${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$
Now, according to the question, it is given that –
Let the refractive index of the glass sphere be ${\mu _1}$, distance of object be $u$ and the distance of image be $v$ and let the radius of sphere be $R$.
So,
$
  {\mu _1} = 1.5 \\
  u = 2R \\
 $
Distance from the actual position of the glass sphere is $10cm$.
Then, the image distance is –
$v = 10 - 2R$
Now, to calculate the radius of glass sphere we have to use the Lens formula of refraction between surfaces of two different refractive indices –
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R} \cdots \left( 1 \right)$
We know that, refractive index of air is $1$
$\therefore {\mu _2} = 1$
From sign convention –
$
  u = - u \\
  R = - R \\
 $
Putting the values of refractive indices, image distance and object distance in the equation $\left( 1 \right)$, we get –
\[
  \Rightarrow \dfrac{1}{{10 - 2R}} - \dfrac{{1.5}}{{ - 2R}} = \dfrac{{1 - 1.5}}{{ - R}} \\
  \Rightarrow \dfrac{1}{{10 - 2R}} = - \dfrac{{0.5}}{R} - \dfrac{{1.5}}{{2R}} \\
 \]
Multiplying by $2$ on numerator and denominator both on right hand side, we get –
\[
  \Rightarrow \dfrac{1}{{10 - 2R}} = - \dfrac{1}{{2R}} - \dfrac{3}{{4R}} \\
  \Rightarrow \dfrac{1}{{10 - 2R}} = - \dfrac{1}{{4R}} \\
  \Rightarrow 4R = 2R - 10 \\
  \Rightarrow R = - 5cm \\
 \]
Radius cannot be negative
So, \[R = 5cm\]
Hence, the radius of the glass sphere is $5cm$.

So, the correct option (D).

Note When using the lens formula points to be remembered are –
1. The sign convention should be used properly for the known values
2. Distances should be measured from the optical center.
3. Direction of incident rays is positive and the opposite direction is negative.