Question

Find out the value of object distance $x$, such that image will form at the object position.
(Assume lens to be thin)

• A. $10\text{cm}$
• B. $15\text{cm}$
• C. $25\text{cm}$
• D. $30\text{cm}$

Answer 1

The correct option is A $10\text{cm}$

Here, the given lens is thin hence, object can be assumed to be at distance $x$ from common optical center.
The system is a combination of plane mirror and a convex lens.
$\Rightarrow \frac{1}{f_{eq}}=\frac{1}{f_m}-\frac{2}{f_L}$
For lens, focal length:
$\frac{1}{f_l}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow \frac{1}{f_l}=(1.5-1)(\frac{1}{+10}-\frac{1}{-10})$
$\Rightarrow \frac{1}{f_l}=0.5\times \left(\frac{1}{5}\right)\Rightarrow f_l=10\text{cm}$
For plane mirror,
$f_m=\infty$
$\Rightarrow \frac{1}{f_{eq}}=\frac{1}{\infty }-\frac{2}{10}$
$\Rightarrow f_{eq}=-5\text{cm}$

Thus, above system behaves as a concave mirror of focal length $5\text{cm}.$
$\Rightarrow$For image to coincide with object,
$|u|=|2f|$ (object should be at center of curvature of the concave mirror)
$\therefore |u|=x=2\times 5=10\text{cm}$

Why this question? Tip: Whenever a lens is silvered or placed in close contact of mirror, whole system behaves like a mirror of focal length $f_{eq}$.