Consider two palne convex lanse of same radius of curvature and refrective index `n_(1)` and `n_(2)` respectively. Now consider two cases :

Case - I : When `n_(1) = n_(2)=n`, then equivalent focal length of length is `f_(0)`
Case - II : When `n_(1) = n, n_(2) = n+ Delta n`, then equilivant focal length of lens is `f = f_(0) + Delta f_(0)`
Then correct options are :

Options

A. If `Delta n//n gt 0`, then `Delta f_(0)//f_(0) lt 0`
B. `|Delta f_(0)//f_(0)| lt |Delta n// n|`
C. If `n = 1.5, Delta n = 10^(-3)` and `f_(0) = 20` cm then `|Delta f_(0)|=0.02` cm
D. If `Delta n//n lt 0`, then `Delta f_(0)//f_(0) gt 0`

Answer

A, C

Explanation

`(1)/(f_(1))=(n-1)((1)/(R )-(1)/(oo)) rArr (1)/(f_(0)) = (2(n-1))/(R )` …..(1)
`(1)/(f_(2)) = (n+Delta n-1)((1)/(R )-(1)/(oo))`
`(1)/(f_(0)+Delta f_(0)) = ((n-1))/(R )+(n+Delta n-1)((1)/(R ))`
`(1)/(f_(0)+Delta f_(0)) = (2n+Delta n-2)/(R )` .....(2)
`(1)//(2) rArr (f_(0)+Delta f_(0))/(f_0)=((2(n-1))/(R ))/((2n+Delta n-2)/(R ))`
`1 + (Delta f_(0))/(f_(0)) = (2(n-1))/(2n+Delta n - 2)`
`(Delta f_(0))/(f_(0)) = (-Delta n)/((2n+Delta n-2))`
`(Delta f_(0))/(20)= -(10^(-3))/(3+10^(-3)-2)`
`rArr Delta f_(0) = - 2xx10^(-2)`
`|Delta f_(0)|=0.02 cm`