A thin lens made of glass of refractive index `mu = 1.5` has a focal length equal to 12 cm in air. It is now immersed in water (`mu=4/3`). Its new focal length is

Options

A. 48 cm
B. 36 cm
C. 24 cm
D. 12 cm

Explanation

To find the new focal length of a thin lens when it is immersed in water, we can use the lensmaker's formula. Let's go through the solution step by step.

### Step 1: Understand the Lensmaker's Formula
The lensmaker's formula relates the focal length of a lens to the refractive indices of the lens material and the surrounding medium, as well as the radii of curvature of the lens surfaces. The formula is given by:

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

Where:
- \( f \) is the focal length of the lens.
- \( \mu \) is the refractive index of the lens material.
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Apply the Formula for Air
In air, the refractive index of the lens (\( \mu_g \)) is 1.5, and the focal length (\( f_a \)) is given as 12 cm. We can write the equation for air as:

\[ \frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the known values:

\[ \frac{1}{12} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{12} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Rearranging the Equation
From the above equation, we can express \( \frac{1}{R_1} - \frac{1}{R_2} \):

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{12 \times 0.5} = \frac{1}{6} \]

### Step 4: Apply the Formula for Water
Now, when the lens is immersed in water (\( \mu_w = \frac{4}{3} \)), we can write the lensmaker's formula for water as:

\[ \frac{1}{f_w} = \left( \mu_g - \mu_w \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the values:

\[ \frac{1}{f_w} = \left( 1.5 - \frac{4}{3} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Calculating \( \mu_g - \mu_w \):

\[ 1.5 - \frac{4}{3} = 1.5 - 1.3333 = 0.1667 \text{ (or } \frac{1}{6} \text{)} \]

### Step 5: Substitute the Value of \( \frac{1}{R_1} - \frac{1}{R_2} \)
Now substitute \( \frac{1}{R_1} - \frac{1}{R_2} \) from Step 3:

\[ \frac{1}{f_w} = 0.1667 \times \frac{1}{6} \]

### Step 6: Calculate the New Focal Length
Now we can find \( f_w \):

\[ \frac{1}{f_w} = \frac{0.1667}{6} = \frac{1}{36} \]

Thus, the new focal length \( f_w \) is:

\[ f_w = 36 \text{ cm} \]

### Final Answer
The new focal length of the lens when immersed in water is **36 cm**.

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