If a equiconvex lens of focal length f is cut into two halves by a plane perpendicular to the principal axis, then

Options

A. The focal length of each half becomes` f/2`
B. The focal length of each half becomes 2f
C. The focal length of each half remains f
D. The focal length of each half becomes f4

Answer

To solve the problem of determining the focal length of each half of an equiconvex lens when it is cut into two halves by a plane perpendicular to the principal axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Lens Properties**:
- An equiconvex lens is a lens with two outwardly curved surfaces and equal radii of curvature. The focal length (f) of a complete equiconvex lens can be determined using the lens maker's formula.

2. **Lens Maker's Formula**:
- The focal length (f) of a lens is given by the formula:
\[ \frac{1}{f} = \left( n - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For an equiconvex lens, \( R_1 = R \) (positive) and \( R_2 = -R \) (negative), leading to:
\[ \frac{1}{f} = \left( n - 1 \right) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(n - 1)}{R} \]

3. **Cutting the Lens**:
- When the lens is cut into two halves, each half will still have the same radius of curvature (R) but will now behave as a separate lens.

4. **Focal Length of Each Half**:
- Each half of the lens can be considered as a plano-convex lens, where one side is flat (the cut side) and the other side remains curved.
- The focal length (f') of a plano-convex lens can be calculated using:
\[ \frac{1}{f'} = (n - 1) \left( \frac{1}{R} \right) \]
- Substituting the values, we find:
\[ \frac{1}{f'} = (n - 1) \left( \frac{1}{R} \right) \]
- Since the original focal length of the complete lens is \( f \), and we know that \( f = \frac{R}{2(n - 1)} \), we can derive that:
\[ f' = 2f \]

5. **Conclusion**:
- Therefore, the focal length of each half of the equiconvex lens after being cut is \( 2f \).

### Final Answer:
The focal length of each half becomes \( 2f \).

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