Question

When a convex lens is immersed in two different liquids of refractive indices 1.25 and 1.5, the ratio of the focal lengths of the lens is 5:16. The refractive index of the material of the lens is:

• A. 1.55
• B. 1.5
• C. 1.65
• D. 1.6

Hint:

Use the relation between the focal lengths in different media to determine the refractive index of the lens material when immersed in different liquids.

Answer 1

The Correct Option is C

Determining the Refractive Index of the Lens 

Given:

• Refractive index of liquid 1, $n_1 = 1.25$
• Refractive index of liquid 2, $n_2 = 1.5$
• Ratio of focal lengths, $\frac{f_1}{f_2} = \frac{5}{16}$

Step 1: Lens Maker's Formula

The lens maker's formula for a lens immersed in a medium is:

$\frac{1}{f} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

Step 2: Focal Lengths in Different Media

For liquid 1:

$\frac{1}{f_1} = \left( \frac{n_{\text{lens}}}{1.25} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

For liquid 2:

$\frac{1}{f_2} = \left( \frac{n_{\text{lens}}}{1.5} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

Step 3: Ratio of Focal Lengths

Given $\frac{f_1}{f_2} = \frac{5}{16}$, we have:

$\frac{f_2}{f_1} = \frac{16}{5}$

Using the lens maker's formula:

$\frac{f_2}{f_1} = \frac{\left( \frac{n_{\text{lens}}}{1.25} - 1 \right)}{\left( \frac{n_{\text{lens}}}{1.5} - 1 \right)} = \frac{16}{5}$

Step 4: Solving for $n_{\text{lens}}$

Let $n = n_{\text{lens}}$. Then:

$\frac{\frac{n}{1.25} - 1}{\frac{n}{1.5} - 1} = \frac{16}{5}$

Simplify the equation:

$\frac{\frac{4n}{5} - 1}{\frac{2n}{3} - 1} = \frac{16}{5}$

Cross-multiply:

$5 \left( \frac{4n}{5} - 1 \right) = 16 \left( \frac{2n}{3} - 1 \right)$

Simplify:

$4n - 5 = \frac{32n}{3} - 16$

Multiply through by 3 to eliminate the fraction:

$12n - 15 = 32n - 48$

Rearrange:

$-20n = -33$

Solve for $n$:

$n = \frac{33}{20} = 1.65$

Final Answer:

$\boldsymbol{1.65}$ (Option 3)