A convex lens of focal length `12 cm` is made of glass of `mu=(3)/(2)`. What will be its focal length when immersed in liquid of `mu=(5)/(4)` ?

Options

A. 6cm
B. 12cm
C. 24cm
D. 30cm

Answer

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

\[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

Where:
- \( f \) is the focal length of the lens.
- \( \mu_2 \) is the refractive index of the lens material.
- \( \mu_1 \) is the refractive index of the surrounding medium.
- \( r_1 \) and \( r_2 \) are the radii of curvature of the lens surfaces.

### Step 1: Determine the initial conditions
In the first case, the lens is in air:
- Focal length \( f = 12 \, \text{cm} \)
- Refractive index of the lens \( \mu_2 = \frac{3}{2} \)
- Refractive index of air \( \mu_1 = 1 \)

### Step 2: Apply the Lensmaker's formula for the first case
Using the Lensmaker's formula:

\[ \frac{1}{12} = \left(\frac{3/2}{1} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

Calculating \( \frac{3/2}{1} - 1 \):

\[ \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]

Now substituting this back into the equation:

\[ \frac{1}{12} = \frac{1}{2} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

### Step 3: Solve for \( \frac{1}{r_1} - \frac{1}{r_2} \)
Multiplying both sides by 2:

\[ \frac{1}{6} = \frac{1}{r_1} - \frac{1}{r_2} \]

This gives us our first equation:

\[ \frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{6} \quad \text{(Equation 1)} \]

### Step 4: Determine the conditions when the lens is immersed in liquid
Now, we consider the second case where the lens is immersed in a liquid with:

- Refractive index of the liquid \( \mu_1 = \frac{5}{4} \)

### Step 5: Apply the Lensmaker's formula for the second case
Using the Lensmaker's formula again:

\[ \frac{1}{f'} = \left(\frac{3/2}{5/4} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

Calculating \( \frac{3/2}{5/4} \):

\[ \frac{3}{2} \cdot \frac{4}{5} = \frac{12}{10} = \frac{6}{5} \]

Now substituting this into the equation:

\[ \frac{1}{f'} = \left(\frac{6}{5} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

Calculating \( \frac{6}{5} - 1 \):

\[ \frac{6}{5} - \frac{5}{5} = \frac{1}{5} \]

Now substituting \( \frac{1}{r_1} - \frac{1}{r_2} \) from Equation 1:

\[ \frac{1}{f'} = \frac{1}{5} \cdot \frac{1}{6} \]

### Step 6: Solve for \( f' \)
This simplifies to:

\[ \frac{1}{f'} = \frac{1}{30} \]

Thus, the focal length \( f' \) is:

\[ f' = 30 \, \text{cm} \]

### Final Answer
The focal length of the lens when immersed in the liquid is \( 30 \, \text{cm} \).

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