Question

A point object $\text{O}$ is placed on the principal axis of a convex lens of focal length $10\text{cm}$ at $12\text{cm}$ from the lens. When object is displaced $1\text{mm}$ along the principal axis away from lens, magnitude of displacement of image is $x_1$ .When the lens is displaced by $1\text{mm}$ perpendicular to the principal axis, displacement of image is $x_2$ in magnitude. Find the value of $\left|\frac{x_1}{x_2}\right|$ upto 2 decimal places..

Answer 1

Given, object distance,$u=-12\text{cm}$
focal length, $f=10\text{cm}$
Let the image distance is $v$.

From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

we have,
$\frac{1}{v}=\frac{1}{10}+\frac{1}{-12}$

$\Rightarrow \frac{1}{v}=\frac{6-5}{60}$

$\Rightarrow \frac{1}{v}=\frac{1}{60}$

$\therefore v=60\text{cm}$

Now, let the image distance is $v_1$ when object is displaced $1\text{mm}$ along principal axis.
The new object distance will be,
$u_1=12+0.1=12.1\text{cm}$
Applying lens formula we get,

$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f}$

Substituting the known values from above,

$\frac{1}{v_1}-\frac{1}{-12.1}=\frac{1}{10}$

$\Rightarrow \frac{1}{v_1}=\frac{1}{10}-\frac{1}{12.1}$

$\Rightarrow \frac{1}{v_1}=\frac{2.1}{121}$

$\therefore v_1=57.62\text{cm}$

Thus image is displaced by ,$x_1=60-57.62=2.38\text{cm}$

Now, when the lens is displaced by $1\text{mm}$ perpendicular to the principal axis, the displacement of the image is $x_2$.

We know that magnification, $M$
$M=\frac{h_i}{h_o}=\frac{v}{u}.........1$

Where, $h_i$=image size
$h_o=1\text{mm}$=object size


Substituting the known values in equation $\left(1\right)$, we have

$\frac{h_i}{1\text{mm}}=\frac{60}{-12}$

$\therefore h_i=-5\text{mm}$

When the the lens is displaced by $1\text{mm}$, image is displaced by, $x_2=5+1=6\text{mm}$.

Now, $\frac{x_1}{x_2}=\frac{2.38}{0.6}$

$\therefore \frac{x_1}{x_2}=3.97$