1. Lens maker formula for the original lens

The lens is made of two halves, each contributing half the optical power because the interface between the two materials is plane and hence has zero power.

If both parts have refractive index $n$, and the two outer spherical surfaces have equal radius $R$, then the total power is

$P=\frac{n-1}{R}+\frac{n-1}{R}=\frac{2(n-1)}{R}$

So,

$\frac{1}{f}=\frac{2(n-1)}{R}$

or

$f=\frac{R}{2(n-1)}$

2. Lens when $n_1=n$ and $n_2=n+\Delta n$

Now the left half contributes power

$P_1=\frac{n-1}{R}$

and the right half contributes

$P_2=\frac{(n+\Delta n)-1}{R}=\frac{n-1+\Delta n}{R}$

Hence total power is

$P'=\frac{n-1}{R}+\frac{n-1+\Delta n}{R}=\frac{2(n-1)+\Delta n}{R}$

Thus,

$\frac{1}{f+\Delta f}=\frac{2(n-1)+\Delta n}{R}$

Using

$\frac{1}{f}=\frac{2(n-1)}{R}$

we get

$\frac{1}{f+\Delta f}=\frac{1}{f}\left(1+\frac{\Delta n}{2(n-1)}\right)$

For small $\Delta n$,

$\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$

This is the required relation.

3. Check option A

If

$\frac{\Delta n}{n}<0$

then $\Delta n<0$ since $n>0$. From

$\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$

and $2(n-1)>0$, we get

$\frac{\Delta f}{f}>0$

So A is correct.

4. Check option B

Given:

$n=1.5,\quad \Delta n=10^{-3},\quad f=20\text{ cm}$

Using

$\left|\frac{\Delta f}{f}\right|=\frac{|\Delta n|}{2(n-1)}$

we get

$\left|\frac{\Delta f}{f}\right|=\frac{10^{-3}}{2(0.5)}=10^{-3}$

Hence,

$|\Delta f|=f\times 10^{-3}=20\times 10^{-3}=0.02\text{ cm}$

So B is correct.

5. Check option C

We have

$\left|\frac{\Delta f}{f}\right|=\frac{|\Delta n|}{2(n-1)}$

Compare with

$\left|\frac{\Delta n}{n}\right|$

Option C claims

$\left|\frac{\Delta f}{f}\right|<\left|\frac{\Delta n}{n}\right|$

That would require

$\frac{1}{2(n-1)}<\frac{1}{n}$

which gives

$n<2n-2\quad \Rightarrow\quad n>2$

This is not always true. For example, for $n=1.5$,

$\left|\frac{\Delta f}{f}\right|=|\Delta n|,$ $\left|\frac{\Delta n}{n}\right|=\frac{|\Delta n|}{1.5}$

so actually

$\left|\frac{\Delta f}{f}\right|>\left|\frac{\Delta n}{n}\right|$

Hence C is false.

6. Check option D

If both convex surfaces are replaced by concave surfaces of same radius magnitude, then each surface power changes sign, so total power becomes negative:

$\frac{1}{f}= -\frac{2(n-1)}{R}$

and after change in refractive index,

$\frac{1}{f+\Delta f}= -\frac{2(n-1)+\Delta n}{R}$

The same fractional relation follows on linearization:

$\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$

So the relation between $\dfrac{\Delta f}{f}$ and $\dfrac{\Delta n}{n}$ remains unchanged.

Hence D is correct.

7. Final answer

The correct options are:

$\boxed{A,\ B,\ D}$

This matches the stored correct answer.