Question

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position and nature of the image formed.

• A. Position of image = 60 cm, image is virtual and erect
• B. Position of image = 60 cm, image is real and inverted
• C. Position of image = 12 cm, image is real and inverted
• D. Position of image = - 60 cm, image is virtual and erect

Answer 1

The correct option is B Position of image = 60 cm, image is real and inverted

Given object size $h$ = 5 cm
Object distance from lens $u$ = − 30 cm
Focal length $f$ = 20 cm (as per sign conventions)
We have to find image distance $v$ and the nature of the image formed.
Let $h_i$ be the height of the image.
Using the lens formula
$\frac{1}{f}$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{1}{v}$ - <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{1}{u}$
Putting the values we have,
$\frac{1}{20}$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{1}{v}$ - <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{1}{-30}$
$\frac{1}{20}$ - $\frac{1}{30}$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{1}{v}$
$\frac{1}{v}$ = $\frac{3-2}{60}$
or $v=60cm$.
We know, magnification, $M=\frac{h_i}{h}=\frac{v}{u}$ $=\frac{h_i}{5}=\frac{60}{-30}\Rightarrow h_i=-10cm$
As per sign conventions, since the image is formed towards right side of lens, and height of the image is $-10cm$, the image is real and inverted.