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.

Answer 2

The problem asks for the percentage change in the focal length of an equiconvex lens when it is immersed in water.

Given:

• Refractive index of the glass lens, \( n_g = \frac{3}{2} \)
• Initial focal length in air is \( f_{air} \). (Refractive index of air, \( n_a \approx 1 \))
• The lens is immersed in water. Refractive index of water, \( n_w = \frac{4}{3} \).
• Let the new focal length in water be \( f_{water} \).

We use the Lens Maker's formula: \[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( f \) is the focal length, \( n_{lens} \) is the refractive index of the lens material, \( n_{medium} \) is the refractive index of the surrounding medium, and \( R_1, R_2 \) are the radii of curvature of the lens surfaces. For an equiconvex lens, \( R_1 = R \) and \( R_2 = -R \).

Case 1: Lens in air Here, \( n_{lens} = n_g = \frac{3}{2} \) and \( n_{medium} = n_a = 1 \). \[ \frac{1}{f_{air}} = \left( \frac{n_g}{n_a} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{air}} = \left( \frac{3/2}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{air}} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{air}} = \left( \frac{1}{2} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \quad (1) \]

Case 2: Lens in water Here, \( n_{lens} = n_g = \frac{3}{2} \) and \( n_{medium} = n_w = \frac{4}{3} \). \[ \frac{1}{f_{water}} = \left( \frac{n_g}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{water}} = \left( \frac{3/2}{4/3} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{water}} = \left( \frac{3}{2} \times \frac{3}{4} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{water}} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ \frac{1}{f_{water}} = \left( \frac{1}{8} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \quad (2) \]

Now, divide equation (1) by equation (2): \[ \frac{1/f_{air}}{1/f_{water}} = \frac{(1/2) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{(1/8) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \] \[ \frac{f_{water}}{f_{air}} = \frac{1/2}{1/8} = \frac{1}{2} \times 8 = 4 \] \[ f_{water} = 4 f_{air} \]

The focal length in water is 4 times the focal length in air.

Percentage change in focal length: \[ \text{Percentage Change} = \frac{\text{Final Focal Length} - \text{Initial Focal Length}}{\text{Initial Focal Length}} \times 100\% \] \[ \text{Percentage Change} = \frac{f_{water} - f_{air}}{f_{air}} \times 100\% \] \[ \text{Percentage Change} = \frac{4 f_{air} - f_{air}}{f_{air}} \times 100\% \] \[ \text{Percentage Change} = \frac{3 f_{air}}{f_{air}} \times 100\% \] \[ \text{Percentage Change} = 3 \times 100\% = 300\% \]

Since \( f_{water} > f_{air} \), the focal length has increased. The percentage change is a 300% increase.

Final Answer: The final answer is (C): \(300\% \ increase\)