Question

A convex lens of focal length, $f=1.5\text{m}$ is placed in a system of coordinate axes such that its optical centre is at origin $\text{O}$ and principal axis coincides with the $\text{x-axis}$. An object$\left(S\right)$ and a plane mirror are arranged as shown in the figure. Find value of $d$ $\left(\text{in m}\right)$ so that $\text{y-coordinate}$ of the image (after refraction and reflection) is $0.3\text{m}$.
(Assume that $\tan \theta =0.3)$

• A. $4\text{m}$
• B. $3\text{m}$
• C. $5\text{m}$
• D. $6\text{m}$

Answer 1

The correct option is C $5\text{m}$

For the lens,
$u=-2\text{m}$, $f=+1.5\text{m}$

Applying the lens formula,

$\frac{1}{v}-\frac{1}{(-2)}=\frac{1}{(+1.5)}$

$\Rightarrow v=6\text{m}$

Now, since object $S$ is above the principal axis of lens, we can treat it as an extended object with height $h_o=0.1\text{cm}$

Then, magnification,

$m=\frac{v}{u}=\frac{h_i}{h_o}$

$\Rightarrow m=\frac{6}{-2}=\frac{h_i}{h_o}$

$\Rightarrow h_i=-3\times 0.1$

$\therefore h_i=-0.3\text{m}$ (below principal axis)

Therefore, $\text{y-coordinate}$ of image formed by lens is $-0.3\text{m}$.


This image $\left(I_1\right)$ will act as the object for the plane mirror. The final image formed by the mirror is shown in the above figure.

From geometry in $\Delta PM{I}_1$,

$\tan \theta =\frac{M{I}_1}{PM}$

$\Rightarrow 0.3=\frac{0.3}{PM}$

$\Rightarrow PM=1\text{m}$

$\Rightarrow d=6-PM$

$\therefore d=6-1=5\text{m}$

Hence, correct answer is option (c).