Question

Calculate the image distance for an object of height $12mm$ at a distance of $0.20m$ from a concave lens of focal length $0.30m$, and state the nature and size of the image.

Answer 1

Hint: A concave lens is a diverging lens, which means that it spreads out light rays that get refracted through it. The image distance of an object is the distance between the object and its image. To solve the given problem, write down the provided physical quantities and then apply the lens formula to find the required correct answer.

Formula used:
${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v}}$

Complete step-by-step answer:
From the question we get to know that $u=-0.20m$ and $f=0.30m$; where $u$ is object distance and $f$ is focal length.
Using lens formula, we can get the image distance $v$,
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}+{\displaystyle \frac{1}{u}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{(0.30)}}-{\displaystyle \frac{1}{(-0.20)}}$ $=10\times ({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2}})={\displaystyle \frac{-1}{0.60}}$
We get, $v=-0.60m$
Now to find the height of the image first we need to find the magnification.
Magnification, $m={\displaystyle \frac{v}{u}}={\displaystyle \frac{-0.60}{-0.20}}=3$
So, height of the image, $h_i=3\times h_0=36mm$
Since, an image is formed on the same side of the concave lens. Thus, the image formed will be a virtual image and the magnification is positive so it is erect.
Hence, the image distance for the image is, $v=-0.60m$, height of the image is $h_i=36mm$ and the nature of the image is virtual, erect and magnified.

Additional Information:
According to Cartesian sign convention:
1) All distances are measured from the optical centre of the lens.
2) Distances measured in the direction opposite to the direction of incident light are considered negative.
3) Height measured upwards and perpendicular to the principal axis are considered positive.
4) Height measured downwards and perpendicular to the principal axis are considered negative.

Note: According to Cartesian sign convention, we took the object distance $(u)$ negative because the object is placed to the left of the lens. This sign convention is very crucial to be followed for solving every optics numerical, because otherwise it would result in wrong answers if not applied correctly.