A convex lens is dipped in a liquid whose refractive index is equal to the refractive of the lens. Then its focal length will

Options

A. Become infinite
B. Become small, but non-zero
C. Remain unchanged
D. Become zero

Explanation

To solve the problem of determining the focal length of a convex lens when it is dipped in a liquid with a refractive index equal to that of the lens, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Lensmaker's Formula**:
The lensmaker's formula is given by:
\[ \frac{1}{f} = \left(\mu - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]
where \(f\) is the focal length of the lens, \(\mu\) is the refractive index of the lens material, and \(r_1\) and \(r_2\) are the radii of curvature of the lens surfaces.

2. **Identifying the Surrounding Medium**:
When the lens is placed in a medium (in this case, a liquid), the formula needs to be adjusted. The modified lensmaker's formula becomes:
\[ \frac{1}{f} = \left(\frac{\mu}{\mu_s} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]
where \(\mu_s\) is the refractive index of the surrounding medium.

3. **Setting the Refractive Indices**:
According to the problem, the refractive index of the liquid (\(\mu_s\)) is equal to the refractive index of the lens (\(\mu\)):
\[ \mu = \mu_s \]

4. **Substituting Values into the Formula**:
Substituting \(\mu\) and \(\mu_s\) into the modified lensmaker's formula:
\[ \frac{1}{f} = \left(\frac{\mu}{\mu} - 1\right) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]
This simplifies to:
\[ \frac{1}{f} = (1 - 1) \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

5. **Calculating the Result**:
Since \(1 - 1 = 0\), we have:
\[ \frac{1}{f} = 0 \]
This implies:
\[ f = \infty \]

### Conclusion:
The focal length of the convex lens when dipped in a liquid with the same refractive index as the lens is infinite.