Question

Consider two plane convex lens of same radius of curvature and refractive index $n_1$ and $n_2$ respectively. Now consider two cases:

Case – $I$: When $n_1=n_2=n$, then equivalent focal length of lens is $f_0$.

Case – $II$: When $n_1=n$, $n_2=n+∆n$, then equivalent focal length of lens is $f=f_0+∆f_0$ . The correct options are:

• A. If $\frac{∆n}{n}>0$, then $\frac{∆f_0}{f_0}<0$
• B. $\left|\frac{∆f_0}{f_0}\right|<\left|\frac{∆n}{n}\right|$
• C. If $n=1.5,∆n={10}^{-3}$ and $f_0=20cm$ then $\left|∆f_0\right|=0.02cm$
• D. If $\frac{∆n}{n}<0$, then $\frac{∆f_0}{f_0}>0$

Answer 1

The correct option is C

If $n=1.5,∆n={10}^{-3}$ and $f_0=20cm$ then $\left|∆f_0\right|=0.02cm$

Step 1. Given Data:

Refractive index of two plane convex lens of same radius of curvature are $n_1$ and $n_2$, respectively

Step 2. The focal length of lens:

For a lens of refractive index, $n$ and radius of curvature, $R_{1}and{R}_2$, then the focal length, $f$

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Step 3. Solve for Case 1:

Now considering the first case, $n_1=n_2=n$and focal length, $f_0$

$\frac{1}{f_0}=\left(n-1\right)\left(\frac{1}{R}-\left(\frac{1}{R}\right)\right)$

$\frac{1}{f_0}=\frac{2\left(n-1\right)}{R}$ [$eq\left(1\right)$]

Step 4. Solve for Case 2:

Now, considering the second case, $n_1=n,n_2=n+∆n,f=f_0+∆f_0$

$\frac{1}{f_0+∆f_0}=\frac{\left(n-1\right)}{R}+\frac{n+∆n-1}{R}$

$\frac{1}{f_0+∆f_0}=\frac{2n+∆n-2}{R}$ $eq\left(2\right)$

Step 5. Find the focal length:

Dividing both the equations, we get

$\frac{{\displaystyle \frac{1}{f_0}}}{{\displaystyle \frac{1}{f_0+∆f_0}}}=\frac{{\displaystyle \frac{2\left(n-1\right)}{R}}}{{\displaystyle \frac{2n+∆n-2}{R}}}$

$1+\frac{∆f_0}{f_0}=\frac{2n-2}{2n+∆n-2}$

$\frac{∆f_0}{f_0}=-\frac{∆n}{2n+∆n-2}$ [$eq\left(3\right)$]

$\frac{∆f_0}{20}=-\frac{{10}^{-3}}{2\times 1.5+{10}^{-3}-2}$ [As given in option $n=1.5,∆n={10}^{-3},f_0=20cm$]

$∆f_0=-0.02cm$

$\left|f_0\right|=0.02cm$

From, $eq\left(3\right)$

If $\frac{∆n}{n}>0$, then $\frac{∆f_0}{f_0}<0$

Hence, the correct option is $\left(A\right)\&\left(C\right)$.