A thin convex lens is made of two materials with refractive indices...

Given that the lens is made of two plano-convex lenses. Let it be ${L}_{1} or {L}_{2}$ .

Let focal length of ${L}_{1} = {f}_{1},$
An focal length of ${L}_{2} = {f}_{2}$
$∴$ Using lens maker's formula,
$\dfrac{1}{{f}_{1}} = \left({n}_{1} - 1\right)\left(\dfrac{1}{R} - \dfrac{1}{\infty}\right),$ and $\dfrac{1}{{f}_{2}} = \left({n}_{2} - 1\right)\left(\dfrac{1}{\infty} - \dfrac{1}{\left(- R\right)}\right)$
$⇒ \dfrac{1}{{f}_{1}} = \dfrac{{n}_{1} - 1}{R}$
$⇒ \dfrac{1}{{f}_{2}} = \dfrac{{n}_{2} - 1}{R}$
Now, for the first situation, given that,
${n}_{1} = {n}_{2} = n$ and focal length of combination is $‘f’$
$⇒ \dfrac{1}{f} = \dfrac{1}{{f}_{1}} + \dfrac{1}{{f}_{2}} = \dfrac{2\left(n - 1\right)}{R}$ ....(1)
In the second situation, ${n}_{1} = n, {n}_{2} = n + \Deltan$ and focal length= $f + \Deltaf$
$⇒ \dfrac{1}{f + \Deltaf} = \dfrac{n - 1}{R} + \dfrac{n + \Del\tan^{-1}}{R}$
$⇒ \dfrac{1}{f + \Deltaf} = \dfrac{2n + \Deltan - 2}{R}$ .....(2)
Nos dividing equation (1)/(2) we get
$\dfrac{f + \Deltaf}{f} = \dfrac{2\left(n - 1\right)}{2n + \Deltan - 2} ⇒ 1 + \dfrac{\Deltaf}{f} = \dfrac{\left(2n - 2 + \Deltan\right) - \Deltan}{2n + \Deltan - 2}$
$⇒ 1 + \dfrac{\Deltaf}{f} = 1 - \dfrac{\Deltan}{2n + \Deltan - 2} ⇒ \dfrac{\Deltaf}{f} = \dfrac{- \Deltan}{2n + \Deltan - 2}$
Now, as $′\Delta{n}^{′}$ is very small $⇒ \dfrac{\Deltaf}{f} = \dfrac{- \Deltan}{2\left(n - 1\right)} \ldots . . \left(3\right)$
Now, option $( A )$
$n = 1.5, \Deltan = {10}^{- 3}and f = 20cm$
$⇒ | \Deltaf | = \dfrac{f ⋅ ( \Deltan )}{2 ( n - 1 )} = \dfrac{20 \times {10}^{- 3}}{2 \times ( 1.5 - 1 )} = 0.02cm ⇒ ( A )$ is correct.
Option $( B )$
The relation between $\dfrac{\Deltaf}{f}$ or $\dfrac{\Deltan}{n}$ is dependent of $‘R’$
$⇒$ Remain unchanged for convex or concave surfaces
$⇒ ( B )$ is correct.
Option \left(C\right)$<br >As$n < 2 ⇒ 2\left(n - 1\right) < n$<br >$⇒ \dfrac{\Deltaf}{f} = \dfrac{1}{2}\left|\dfrac{\Deltan}{n - 1}\right| > \dfrac{\Deltan}{n}$<br >$⇒ \dfrac{\Deltaf}{f} > \dfrac{\Deltan}{n} ⇒ ( C )$is incorrect.<br >Option \left(D\right)$
If $\dfrac{\Deltan}{n} < 0,$ then $\dfrac{\Deltaf}{f} > 0,$ using equation $( 3 )$
$⇒ ( D )$ is correct.