If a convex lens is immersed in water, then its focal length

Options

A. increases
B. decreases
C. remains unchanged
D. may increase or decrease depending upon the level of water

Answer

To determine how the focal length of a convex lens is affected when it is immersed in water, we can use the lens maker's formula:

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

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: Identify the refractive indices
In air, the refractive index \( n_1 \) is approximately 1.0, and for glass (the lens material), \( n_2 \) is approximately 1.5. When the lens is immersed in water, the refractive index \( n_1 \) becomes approximately 1.33.

### Step 2: Calculate the relative refractive index
When the lens is in water, we need to calculate the relative refractive index \( n_{21} \):

\[ n_{21} = \frac{n_2}{n_1} = \frac{1.5}{1.33} \approx 1.127 \]

### Step 3: Analyze the effect on the focal length
From the lens maker's formula, we see that the focal length \( f \) is inversely proportional to \( (n_2 - n_1) \):

\[ \frac{1}{f} \propto (n_2 - n_1) \]

In air, \( n_{21} \) was 1.5, but now in water, it has decreased to approximately 1.127. Since \( n_{21} \) is decreasing, the term \( (n_2 - n_1) \) is also decreasing.

### Step 4: Conclusion on focal length
Since \( (n_2 - n_1) \) is decreasing, it implies that the value of \( \frac{1}{f} \) is decreasing. Therefore, the focal length \( f \) must increase.

### Final Answer
Thus, when a convex lens is immersed in water, its focal length increases.

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