Question

A sphere of radius 1m and refractive index 1.5 is silvered at its back. A point object is kept at a distance of 1 m from the front face as shown in figure. What is the position of final image w.r.t. P?

• A. $\frac{13}{7}m$ ;right side
• B. $\frac{13}{7}m$ ; left side
• C. $\frac{7}{13}m$ ; right side
• D. $\frac{7}{13}m$ ; left side

Answer 1

The correct option is A. $\frac{13}{7}m$ ;right side.

For the first event $n_1=1;n_2=1.5;u=–1m;R=+1m$. Therefore we get
$\frac{1.5}{v_1}-\frac{1}{-1}=\frac{1.5-1}{1}$
$v_1=–3m$.

This image will act as an object for the rear silvered surface. However, for taking the object distance, we have to measure it from the pole of the mirror. As we have discussed before.
$u_2=v_1–x$
$u_2=–3–2=-5m$.
Applying mirror formula, we get (radius of curvature is –1m, so focal length of the mirror is $-\frac{1}{2}m$)

$\frac{1}{v_2}+\frac{1}{-5}=\frac{1}{-0.5}$

$v_2=-\frac{5}{9}m$.

These reflected rays will again be refracted from the spherical surface. And we need to take the distance of the image from point P, which presently is from the pole of the mirror. So take the difference.

$u_3=v_2–x$

$u_3=\left(\frac{-5}{9}\right)+2=\frac{+13}{9}m$.

Now, $\frac{1}{v_3}-\frac{1.5}{\frac{13}{9}}=\frac{1-1.5}{1}=-0.5$

$v_3=\frac{13}{7}m$.

Thus, the position of the final image w.r.t. P is $\frac{13}{7}m$ towards the right side.