A lens is made of flint glass (refractive index `=1.5`). When the lens is immersed in a liquid of refractive index `1.25` , the focal length:

Options

A. increases by a factor of `1.25`
B. increases by a factor of `2.5`
C. increase by a factor of `1.2`
D. decrease by a factor of `1.2`

Explanation

To solve the problem of finding the new focal length of a lens made of flint glass when it is immersed in a liquid, we can follow these steps:

### Step 1: Understand the formula for focal length
The formula for the focal length \( f \) of a lens in air is given by:
\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
where \( \mu \) is the refractive index of the lens material, and \( r_1 \) and \( r_2 \) are the radii of curvature of the lens.

### Step 2: Write the focal length in air
For the lens made of flint glass with a refractive index \( \mu = 1.5 \):
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
This simplifies to:
\[ \frac{1}{f} = 0.5 \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Step 3: Write the focal length in the liquid
When the lens is immersed in a liquid with a refractive index of \( \mu_{\text{liquid}} = 1.25 \), the new focal length \( f_1 \) can be expressed as:
\[ \frac{1}{f_1} = \left( \frac{\mu}{\mu_{\text{liquid}}} - 1 \right) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
Substituting \( \mu = 1.5 \) and \( \mu_{\text{liquid}} = 1.25 \):
\[ \frac{1}{f_1} = \left( \frac{1.5}{1.25} - 1 \right) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Step 4: Simplify the expression for \( f_1 \)
Calculating \( \frac{1.5}{1.25} \):
\[ \frac{1.5}{1.25} = 1.2 \]
Thus,
\[ \frac{1}{f_1} = (1.2 - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = 0.2 \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Step 5: Relate \( f_1 \) to \( f \)
Now we can relate \( f_1 \) and \( f \):
\[ \frac{1}{f_1} = \frac{0.2}{0.5} \cdot \frac{1}{f} \]
This simplifies to:
\[ \frac{1}{f_1} = \frac{2}{5} \cdot \frac{1}{f} \]
Taking the reciprocal gives:
\[ f_1 = \frac{5}{2} f \]

### Step 6: Conclusion
The new focal length \( f_1 \) is \( 2.5 \) times the original focal length \( f \).

### Final Answer
The focal length when the lens is immersed in the liquid is \( 2.5f \).

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