Question

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 surfaces are equal. / 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 $n_2=n+\Delta n$. Assuming $\Delta n<<(n—1)$ and $1<n<2$, the correct statement(s) is/are,

• A. $\left|\frac{\Delta f}{f}\right|<\left|\frac{\Delta n}{n}\right|$
• B. For $n=1.5,\Delta n=10^{-3}$ and $f=20$ cm, the value of $\left|\Delta f\right|$ will be 0.02 cm (round off to $2^{nd}$ decimal place)
• C. If $\frac{\Delta n}{n}<0$ then $\frac{\Delta f}{f}>0$
• D. The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature

Answer 1

Answer: (B), (C), (D)

$\frac{1}{f}=\left(n_1-1\right)\left(\frac{1}{R}-\frac{1}{\infty }\right)+\left(n_2-1\right)\left(\frac{1}{\infty }-\frac{1}{-R}\right)$
$\frac{1}{f}=\frac{\left(n_1-1\right)}{R}+\frac{n_2-1}{R}=\frac{\left(n_1+n_2-2\right)}{R}$
Now $\frac{\Delta f}{f^2}=\frac{\Delta n}{R}$
$\frac{\Delta f}{f}=\frac{\Delta n}{\left(n_1+n_2-2\right)}=\frac{\Delta n}{\left[2n+\Delta n-2\right]}$
For $n_1=n_2=1.5\Delta n=10^{-3},f=20cmthenR=20cm$
and $\Delta f=\frac{10^{-3}\times 20}{\left(2\times 1.5-2+10^{-3}\right)}=0.02cm.$
If $\frac{\Delta n}{n}<0\left(Diversingnatureincreases\right)\therefore \frac{\Delta f}{f}>0$
If the surfaces are replaced by concave surfaces of same radius, focal length changes the sign withsame magnitude.
$\therefore \frac{\Delta f}{f}=\frac{\Delta n}{\left(2n+\Delta n-2\right)}$ (remain unchanged).