Question

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?
A. Virtual, upright, height = 1 cm
B. Virtual, upright, height = 0.5 cm
C. Real, inverted, height = 4 cm
D. Real, inverted, height = 1 cm

Answer 1

For an object placed in front of a convex lens beyond the radius of curvature, image formed is real and inverted. For an inverted image, linear magnification is negative. By sign convention, distances above the principal axis are taken as positive and below as negative.

Formula used:
From the Lens Maker’s formula, the focal length of a lens is given by
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Where, f denotes the focal length of the lens, $\mu$ is the refractive index of the lens material with respect to the incident medium, and ${{R}_{1}}$, ${{R}_{2}}$ are the radii of curvature of the spherical surfaces of the lens.

Formula used:
Thin lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Where, f denotes the focal length of the lens, u is the object distance from the lens and v is the image distance.
Magnification produced by the lens is given by:
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}=\dfrac{v}{u}$
Where ${{h}_{o}}$ and ${{h}_{i}}$ denotes the object height and image height respectively.

Complete step by step answer:
Given:
The radii of curvature, ${{R}_{1}}={{R}_{2}}=20\text{ cm}$
Object distance, $u=30\text{ cm}$(the object is placed on the left side, so u is negative)
Height of the object, ${{h}_{o}}=2\text{ cm}$
Let us consider the light incident from air on a glass lens, so $\mu$=$\dfrac{3}{2}$ . Also, for a biconvex lens, ${{R}_{1}}$ is positive and ${{R}_{2}}$is negative.
Therefore, in case of biconvex lens,

$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow \dfrac{1}{f}=\left( \dfrac{3}{2}-1 \right)\left[ \dfrac{1}{20}-\left( -\dfrac{1}{20} \right) \right]$
$\Rightarrow \dfrac{1}{f}=\left( \dfrac{3-2}{2} \right)\left( \dfrac{1}{20}+\dfrac{1}{20} \right)$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{2}\times \left( \dfrac{2}{20} \right)$
$\therefore \dfrac{1}{f}=\dfrac{1}{20}$
So,$f=20$cm.
Now, calculate image distance by substituting the values of focal length and object distance in thin lens formula:
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
$\Rightarrow \dfrac{1}{20}=\dfrac{1}{v}-\dfrac{1}{-30}$
$\Rightarrow \dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{30}$
$\Rightarrow \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{30}$
$\Rightarrow \dfrac{1}{v}=\dfrac{3-2}{60}$
$\therefore\dfrac{1}{v}=\dfrac{1}{60}$
So, $v=60$cm.
So, the image distance is 60 cm from the lens and the image is formed on its right side. Hence, the image is real in nature.
Now magnification of the object by the lens,
$m=\dfrac{-{{h}_{i}}}{{{h}_{o}}}=\dfrac{-v}{u}$
${{h}_{i}}=\dfrac{v}{u}\times {{h}_{o}}$
$\Rightarrow {{h}_{i}}=\dfrac{60}{-30}\times 2$
$\therefore {{h}_{i}}=-4\text{ cm}$
The negative sign of the image height indicates that the image is inverted.

Therefore, option C is the correct answer.

Note: Refractive index value is not given, hence we assume that the light travels from air to glass.
‘Coordinate geometry sign convention’ is used to solve the problem. By this convention,
1. Light rays fall on the lens surface from the left side
2. All distances measured in the direction of incident light are taken with positive sign and those measured in the direction opposite to the direction of incident light are taken with negative sign.
3. The lengths of the object and the image measured upwards perpendicular to the principal axis are taken as positive, while those measured downwards are taken as negative.