Question

When a convex lens is immersed in a liquid of refractive index equal to 80% of the refractive index of the material of the lens, the focal length of the lens increases by 100%. The refractive index of the liquid is:

• A. \( 1.27 \)
• B. \( 1.2 \)
• C. \( 1.33 \)
• D. \( 1.4 \)

Hint:

The focal length of a lens in a medium is affected by the relative refractive index. If a lens is placed in a liquid, its focal length increases if the refractive index of the liquid is closer to that of the lens material.

Answer 1

The Correct Option is B

Step 1: Lens Maker’s Formula in Air
The focal length \( f \) of a convex lens in air is given by the lens maker's formula:
\[ \frac{1}{f} = (n_{\text{lens}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( n_{\text{lens}} \) is the refractive index of the lens material.
Step 2: Lens Maker’s Formula in a Liquid Medium
When the lens is placed in a liquid with refractive index \( n_{\text{liquid}} \), the new focal length \( f' \) is: \[ \frac{1}{f'} = \left( \frac{n_{\text{lens}}}{n_{\text{liquid}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Since the focal length doubles (\( f' = 2f \)), we equate: \[ \frac{1}{2f} = \left( \frac{n_{\text{lens}}}{n_{\text{liquid}}} - 1 \right) \frac{1}{f} \] Simplifying, \[ \frac{1}{2} = \frac{n_{\text{lens}}}{n_{\text{liquid}}} - 1 \] Step 3: Solving for \( n_{\text{liquid}} \) Rearranging, \[ \frac{n_{\text{lens}}}{n_{\text{liquid}}} = \frac{3}{2} \] Since \( n_{\text{liquid}} \) is 80% of \( n_{\text{lens}} \), \[ n_{\text{liquid}} = 0.8 n_{\text{lens}} \] Substituting, \[ \frac{n_{\text{lens}}}{0.8 n_{\text{lens}}} = \frac{3}{2} \] \[ \frac{1}{0.8} = \frac{3}{2} \] \[ n_{\text{liquid}} = 1.2 \] Step 4: Conclusion
Thus, the refractive index of the liquid is \( 1.2 \).