Question

A biconvex lens has a radius of curvature of magnitude $20cm$. Which of the following options describes best the image formed of an object of height $2cm$ placed $30cm$ from the lens? $\left( {\mu = 1.5} \right)$
(A) Virtual, upright , height $= 1cm$
(B) Virtual, upright , height $= 0.5cm$
(C) Real, inverted, height $= 4cm$
(D) Real, inverted, height $= 1cm$

Answer 1

Hint
We need to first calculate the focal length of the biconvex lens using the radius of curvature and the given value of $\mu$. Next we find the distance at which the image is formed, using which we can detect if the image is real or virtual. We find the magnification next which will give us whether the image is upright or inverted. And from the magnification, we can find the height.
In this solution, we will be using the following formula,
$\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
where $f$ is the focal length, $\mu$ is the refractive index and ${R_1}$ and ${R_2}$ are the radius of curvatures of the 2 surfaces of the biconvex lens.
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ where $v$ and $u$ are the image and object distances from the lens respectively.
and the magnification $m = \dfrac{f}{{f + u}} = \dfrac{{{h_i}}}{{{h_o}}}$
where ${h_i}$ is the height of the image and ${h_o}$ is the height of the object.

Complete step by step answer
In the given problem, we are given the radius of curvature of the 2 surfaces of the biconvex lens as , ${R_1} = 20cm$ and ${R_2} = - 20cm$. The refractive index of the glass is given by, $\mu = 1.5$. Now the formula for the focal length of the biconvex lens is given by,
$\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
So substituting the values we get,
$\Rightarrow \dfrac{1}{f} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{20}} - \left( { - \dfrac{1}{{20}}} \right)} \right]$
Hence we get on calculating and taking the LCM as 20,
$\Rightarrow \dfrac{1}{f} = 0.5 \times \left[ {\dfrac{{1 + 1}}{{20}}} \right]$
To find the focal length we invert the fraction, and hence get
$\Rightarrow f = \dfrac{{20}}{{2 \times 0.5}}$
On calculating this we get,
$\Rightarrow f = 20cm$
Now in the question we are given the object distance as $u = - 30cm$. So using the object distance and the focal length we can find the image distance from the formula,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Substituting we get,
$\Rightarrow \dfrac{1}{{20}} = \dfrac{1}{v} - \left( { - \dfrac{1}{{30}}} \right)$
Hence on taking $\dfrac{1}{v}$ to one side,
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} - \dfrac{1}{{30}}$
The LCM will be 60. So we have
$\Rightarrow \dfrac{1}{v} = \dfrac{{3 - 2}}{{60}}$
We reciprocate to get $v$ so, $v = 60cm$. The positive value of the image distance shows that the image is real.
Now magnification, $m = \dfrac{f}{{f + u}}$. So substituting the values,
$\Rightarrow m = \dfrac{{20}}{{20 + \left( { - 30} \right)}}$
On calculating we get $m = - \dfrac{{20}}{{10}} = - 2$. The negative value of magnification shows that the image is inverted. Again the magnification is equal to, $m = \dfrac{{{h_i}}}{{{h_o}}}$ where we are given ${h_o} = 2cm$. So we get
$\Rightarrow {h_i} = m{h_o}$
Substituting we get,
$\Rightarrow {h_i} = - 2 \times 2 = - 4cm$
So the image height is $4cm$
Therefore, the image will be real, inverted, height $= 4cm$
So the correct option is (C).

Note
Bi convex lenses are simple lenses which have 2 spherical convex surfaces that are joined together. The radius of curvature of both the surfaces is generally the same. They have a wide range of applications like focusing laser beams, quality imaging and many other kinds of optical instruments.