Question

A clear transparent glass sphere $(\mu =1.5)$ of radius $R$ is immersed in a liquid of refractive index $1.25$. A parallel beam of light incident on it will converge to a point. The position of this point measured from the centre of the sphere will be

• A. $-3R$
• B. $+3R$
• C. $-R$
• D. $+R$

Answer 1

The correct option is B $+3R$

For refraction at first spherical surface,
The rays are coming from infinity, $u=-\infty$
The radius of curvature is $R=+R$
So,
$\Rightarrow \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
or, $\frac{1.5}{v}-\frac{1.25}{-\infty }=\frac{1.5-1.25}{+R}$
or, $\frac{1.5}{v}=\frac{0.25}{R}$
$\therefore v=+6R$


Again for refraction at the second spherical surface, the object distance is:
$u=+(6R-2R)=+4R$
$R(R.O.C)=-R$
Thus, $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{1.25}{v}-\frac{1.5}{(+4R)}=\frac{1.25-1.5}{-R}$
or, $\frac{1.25}{v}=\frac{0.25}{R}+\frac{1.5}{4R}$
or, $\frac{1.25}{v}=\frac{1}{4R}+\frac{1.5}{4R}=\frac{2.5}{4R}$
$\Rightarrow v=\frac{4R}{2}=2R$
Thus, image (1) is formed at distance $2R$ from the pole of the spherical surface.
$\therefore$ Distance from centre $=2R+R=3R$
or, position of image $\Rightarrow +3R$

Why this question? Tip: Virtual image formed due to refraction at first spherical surface will serve as virtual object for refraction at second spherical surface.