Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be

Options

A. `10 cm`
B. `2.5 cm`
C. `5 cm`
D. `7.5 cm`

Answer

To solve the problem, we will use the lens maker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces. The formula is given by:

\[ \frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( n_1 \) is the refractive index of the medium in which the lens is placed,
- \( n_2 \) is the refractive index of the lens material,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 1: Find the radii of curvature of the lens in air.

Given:
- Focal length of the lens in air, \( f = 2 \) cm
- Refractive index of the lens, \( n_2 = 1.5 \)
- Refractive index of air, \( n_1 = 1 \)

Using the lens maker's formula:

\[ \frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the known values:

\[ \frac{1}{2} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Calculating the left side:

\[ \frac{1}{2} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

\[ \frac{1}{2} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Dividing both sides by 0.5:

\[ 1 = \frac{1}{R_1} - \frac{1}{R_2} \quad \text{(Equation 1)} \]

### Step 2: Find the focal length of the lens when immersed in a liquid.

Now, when the lens is immersed in a liquid with a refractive index \( n_1 = 1.25 \):

Using the same lens maker's formula:

\[ \frac{1}{f_{\text{water}}} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the new values:

\[ \frac{1}{f_{\text{water}}} = \left( \frac{1.5}{1.25} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Calculating the left side:

\[ \frac{1}{f_{\text{water}}} = \left( 1.2 - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

\[ \frac{1}{f_{\text{water}}} = 0.2 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \text{(Equation 2)} \]

### Step 3: Relate the two equations.

Now, we have two equations:

1. \( 1 = \frac{1}{R_1} - \frac{1}{R_2} \) (from air)
2. \( \frac{1}{f_{\text{water}}} = 0.2 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

From Equation 1, we can substitute \( \frac{1}{R_1} - \frac{1}{R_2} = 1 \) into Equation 2:

\[ \frac{1}{f_{\text{water}}} = 0.2 \times 1 \]

\[ \frac{1}{f_{\text{water}}} = 0.2 \]

### Step 4: Calculate the focal length in water.

Taking the reciprocal to find \( f_{\text{water}} \):

\[ f_{\text{water}} = \frac{1}{0.2} = 5 \text{ cm} \]

### Final Answer:
The focal length of the lens when immersed in a liquid of refractive index 1.25 is **5 cm**.

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