Question

An object lying at the distance of $5\text{cm}$ from a spherical interface as shown in the figure, is moving with the velocity of $5\text{cm/s}$. If the interface is moving with a velocity of $3\text{cm/s}$, then calculate the velocity of the image formed by the spherical surface.
(Assume that spherical interface always seperates air and medium of refractive index $1.5$)

• A. $20\text{cm/s}$
• B. $50\text{cm/s}$
• C. $45\text{cm/s}$
• D. $15\text{cm/s}$

Answer 1

The correct option is C $45\text{cm/s}$

Given :
Velocity of object $\left(V_O\right)=5\text{cm/s}$
Velocity of spherical interface $\left(V_S\right)=-3\text{cm/s}$
Distance of object from the pole of the interface $\left(u\right)=-5\text{cm}$
Radius of the spherical interface $\left(R\right)=5\text{cm}$
Referactive index of the medium $\left({\mu }_2\right)=1.5$

Let the velocity of the image formed is $V_I$

We know that for spherical interface, position of the image is given by :

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{1.5}{v}-\frac{1}{-5}=\frac{1.5-1}{5}$

$\Rightarrow v=-15\text{cm}$

We know that the velocity of the image is given by :

$V_{IS}=\frac{{\mu }_1}{{\mu }_2}.\frac{v^2}{u^2}.V_{OS}$

$V_I-V_S=\frac{{\mu }_1}{{\mu }_2}.\frac{v^2}{u^2}.(V_O-V_S)$

$\Rightarrow V_I-(-3)=\frac{1}{1.5}\times \frac{(-15{)}^2}{(-5{)}^2}\times [5-(-3\left)\right]$

$\Rightarrow V_I+3=\frac{225}{1.5\times 25}\times 8$

$\Rightarrow V_I+3=48$

$\Rightarrow V_I=45\text{cm/s}$