Question

An object $AB$ of height $5\text{mm}$ is kept to the left and on the axis of a converging lens of focal length $10\text{cm}$ as shown in the figure. A plane mirror $M$ is placed inclined at ${45}^{\circ }$ to the lens axis at a distance of $10\text{cm}$ to the right of the lens.
$\left(a\right)$ Find the size of the image of object $AB$.
$\left(b\right)$ Find the coordinates of the image of points $A$ & $B$ formed by this combination (take the centre of the lens as the origin and the lens axis as the positive x-axis).
Consider the image after one refraction from lens and then one reflection from mirror only.

• A. $\left(a\right)10\text{mm}$
  $\left(b\right)(+10\text{cm},-20\text{cm}),(+11\text{cm},-20\text{cm})$
• B. $\left(a\right)10\text{mm}$
  $\left(b\right)(-10\text{cm},-20\text{cm}),(-11\text{cm},-20\text{cm})$
• C. $\left(a\right)20\text{mm}$
  $\left(b\right)(-10\text{cm},-10\text{cm}),(-11\text{cm},-10\text{cm})$
• D. $\left(a\right)10\text{mm}$
  $\left(b\right)(+10\text{cm},+10\text{cm}),(+11\text{cm},+10\text{cm})$

Answer 1

The correct option is A $\left(a\right)10\text{mm}$

$\left(b\right)(+10\text{cm},-20\text{cm}),(+11\text{cm},-20\text{cm})$

Given :- $u=-5\text{cm}$
$f=10\text{cm}$
Applying lens formula :-
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$
$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}=\frac{-1}{10}$
$v=-10\text{cm}$
Now, magnification :-
$m=\frac{v}{u}=\frac{-10}{-5}=2$
Also, $m=2=\frac{h_i}{h_0}=\frac{h_i}{5\text{mm}}$
$\Rightarrow h_i=10\text{mm}$

This image will act as an object for the mirror. Plane mirror creates image of the object of same size as that of the object.

Therefore, the height of the final image will be $=10\text{mm}=1\text{cm}$

Now, diagram for the image formation due to the mirror will be :-Here $A^{\prime }{B}^{\prime }$ is image created by the lens $A^{\prime \prime }{B}^{\prime \prime }$ is the image created by the mirror.

From the figure it is clear that coordinates of points $A^{\prime \prime },B^{\prime \prime }$ will be :-
$A^{\prime \prime }=(+10\text{cm},-20\text{cm})$
$B^{\prime \prime }=(+11\text{cm},-20\text{cm})$