A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$, as shown in figure. The radius of curvature of the left and right spherical surface are equal.  $f$ is the focal length of the lens when  $n_1=n_2=n$. The focal length is $f+\Delta f$  when $n_1=n$ and $n_2=n+\mathrm{\Delta }n$ . Assuming $\Delta n<<\left(n-1\right)$ and $1<n<2$ , the correct statement(s) is/are,

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

In 2nd case,

$\frac{1}{f+df}=\frac{n-1}{R}+\frac{\left(n+dn\right)-1}{R}$

 $\Rightarrow \frac{1}{f+df}=\frac{1}{f}+\frac{dn}{R}$     
Clearly, if  $dn<0,$ R.H.S value is less than$\frac{1}{f}$    

 $\therefore df>0$i.e (1)  is correct

Option(2)
$f=\frac{R}{2\left(1.5-1\right)}$

$\Rightarrow R=20cm$    

Now, $\frac{1}{f+df}=\frac{1}{f}+\frac{dn}{R}$
$\Rightarrow \frac{1}{20+df}=\frac{1}{20}+\frac{dn}{20}$

$\Rightarrow \frac{1}{20\left(1+\frac{df}{20}\right)}=\frac{1}{20}(1+dn)$
$\Rightarrow 1-\frac{df}{20}=1+dn$

$\Rightarrow df=-20dn$    
$\therefore \left|df\right|=0.02cm$  (2) is correct
Option(3): If convex surfaces are replaced with concave surfaces, sign of R will change
$\frac{1}{f\left(1+\frac{df}{f}\right)}=\frac{1}{f}+\frac{dn}{R}=\frac{1}{f}+\frac{dn}{2f\left(n-1\right)}$

$Since, f=\frac{R}{2\left(n-1\right)}$

$\Rightarrow \frac{1}{R}=\frac{1}{2f\left(n-1\right)}$   
$\therefore \frac{1-\frac{df}{f}}{f}=\frac{1}{f}\left[1+\frac{dn}{2n\left(1-\frac{1}{n}\right)}\right]$

$\Rightarrow -\frac{df}{f}=\left(\frac{dn}{n}\right)\frac{1}{2\left(1-\frac{1}{n}\right)}$     

As this is independent of R, (3) is correct. 

Option (4): $\left|\frac{df}{f}\right|=\left|\frac{dn}{n}\right|\frac{1}{2\left(1-\frac{1}{n}\right)}$ 

For the range 1 < n < 2,  $2\left(1-\frac{1}{n}\right)$ is always less than 1 i.e, $\left|\frac{df}{f}\right|>\left|\frac{dn}{n}\right|$  (4) is wrong