Question

A thin equiconvex lens of glass $(\mu =\frac{3}{2})$ having a focal length of 30 cm in air is placed at a distance of 10 cm from a plane mirror, which, in turn is placed with its plane perpendicular to the optic axis of the lens. Water $(\mu =\frac{4}{3})$ fills the space between lens and mirror. A parallel beam of light is incident on the lens parallel to the principal axis. Choose the correct option(s):

• A. Final image is 18 cm left to the lens.
• B. Final image is 18 cm right to the lens.
• C. If mirror is rotated by $1^{\circ }$, as shown in figure, final image is displaced by \(\frac{\pi}{3}~cm\).
• D. If mirror is rotated by $1^{\circ }$, as shown in figure, final image is displaced by \(\frac{5 \pi}{9}~cm\).

Answer 1

Answer: (B), (C)

The correct options are
B

Final image is 18 cm right to the lens.

C

If mirror is rotated by $1^{\circ }$, as shown in figure, final image is displaced by \(\frac{\pi}{3}~cm\).

The focal length of the glass lens is 30 cm
$\frac{1}{f}(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{2(\mu -1)}{R}$
The radius of curvature, R = 30 cm.
The object distance, $u=\infty$.
Now we apply Gauss’s law at surface $S_1$ and $S_2$.
$\frac{\frac{3}{2}}{v_1}-\frac{1}{u}=\frac{\frac{3}{2}-1}{30}\dots \dots \left(1\right)\frac{\frac{4}{3}}{v}-\frac{\frac{3}{2}}{v_1}=\frac{-\frac{1}{6}}{-30}\dots \dots \left(2\right)\Rightarrow v=60cm$
After reflection from the mirror, the light rays appear to converge to a point 40 cm to the right of the convex lens. This serves as virtual object for the lens: u = +40cm
$\frac{\frac{3}{2}}{v_1}-\frac{\frac{4}{3}}{+40}=\frac{\frac{1}{6}}{30}\frac{1}{v}-\frac{\frac{3}{2}}{v_1}=\frac{-\frac{1}{2}}{-30}$
$\Rightarrow v=18cm$ to the right of convex lens.
If the mirror is rotated by $1^{\circ }$ the reflected ray rotates by $2^{\circ }$. The virtual object for the lens formed by the reflection from the mirror is displaced by:
$\Delta y_1=50\times \frac{2\pi }{180}cm$
The magnification due to the refraction at the two surfaces of the lens is
$m=m_1{m}_2=\left(\frac{\frac{v}{{\mu }_3}}{\frac{v_1}{{\mu }_2}}\right)\times \left(\frac{\frac{v_1}{{\mu }_2}}{\frac{u}{{\mu }_1}}\right)=\left(\frac{\frac{v}{{\mu }_3}}{\frac{u}{{\mu }_1}}\right)=\frac{\frac{18}{1}}{\frac{40}{\left(\frac{4}{3}\right)}}=\frac{18}{30}$
The displacement of the final image is
$\frac{18}{30}\times 50\times \frac{2\pi }{180}cm=\frac{\pi }{3}cm$