Question

An equiconvex lens made of glass of refractive index \(\frac{3}{2}\) has focal length f in air. It is completely immersed in water of refractive index \(\frac{4}{3}\). The percentage change in the focal length is

• A. 300% decrease
• B. 400% decrease
• C. 300% increase
• D. 400% increase

Answer 1

The Correct Option is C

Given Information: 
Refractive index of glass, \( \mu_g = \frac{3}{2} \)
Refractive index of water, \( \mu_w = \frac{4}{3} \)

Step-by-Step Explanation:

Step 1: Focal length of the lens in air (\( f_{air} \))

For an equiconvex lens (radius of curvature \( R \) same for both sides), the focal length in air is given by the lens maker's formula:

\[ \frac{1}{f_{air}} = (\mu_g - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = 2\frac{(\mu_g - 1)}{R} \]

Substitute the value of \( \mu_g = \frac{3}{2} \):

\[ \frac{1}{f_{air}} = 2\frac{\left(\frac{3}{2} - 1\right)}{R} = 2 \times \frac{1}{2R} = \frac{1}{R} \]

Thus,

\[ f_{air} = R \]

Step 2: Focal length of lens in water (\( f_{water} \)):

When immersed in water, the lens formula becomes:

\[ \frac{1}{f_{water}} = (\frac{\mu_g}{\mu_w} - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = 2\frac{\left(\frac{\mu_g}{\mu_w} - 1\right)}{R} \]

Substitute the given values:

\[ \frac{\mu_g}{\mu_w} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2}\times\frac{3}{4} = \frac{9}{8} \]

Thus,

\[ \frac{1}{f_{water}} = 2\left(\frac{9}{8}-1\right)\frac{1}{R} = 2\left(\frac{1}{8}\right)\frac{1}{R} = \frac{1}{4R} \]

Hence, the new focal length in water is:

\[ f_{water} = 4R \]

Step 3: Calculate the percentage change in focal length clearly:

The initial focal length was \( f_{air} = R \), and it becomes \( f_{water} = 4R \).

Change in focal length:

\[ \Delta f = f_{water} - f_{air} = 4R - R = 3R \]

Percentage increase in focal length:

\[ \% \text{change} = \frac{\Delta f}{f_{air}} \times 100 = \frac{3R}{R}\times 100 = 300\% \]

Thus, the focal length increases by 300% when immersed in water.