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 finding the focal length of each half of an equiconvex lens after it has been cut into two halves by a plane perpendicular to the principal axis, we can follow these steps:

### Step 1: Understand the Lens Properties
An equiconvex lens has equal radii of curvature on both sides. Let the focal length of the original lens be \( f \). The lens maker's formula is given by:

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

where:
- \( \mu \) is the refractive index of the lens material,
- \( r_1 \) and \( r_2 \) are the radii of curvature of the two surfaces of the lens.

For an equiconvex lens, we have \( r_1 = r \) and \( r_2 = -r \) (the negative sign indicates that the second surface is convex towards the incoming light).

### Step 2: Apply the Lens Maker's Formula
Substituting the values into the lens maker's formula, we get:

\[ \frac{1}{f} = \mu \left( \frac{1}{r} - \left(-\frac{1}{r}\right) \right) = \mu \left( \frac{1}{r} + \frac{1}{r} \right) = \mu \left( \frac{2}{r} \right) \]

Thus, we can express the focal length as:

\[ \frac{1}{f} = \frac{2\mu}{r} \]

### Step 3: Cut the Lens in Half
When the lens is cut into two halves by a plane perpendicular to the principal axis, we need to analyze one of the halves. For one half of the lens:
- One surface remains with radius of curvature \( r \).
- The other surface becomes flat (infinite radius of curvature), hence \( r_2 = \infty \).

### Step 4: Apply the Lens Maker's Formula Again
Now, applying the lens maker's formula for the half lens, we have:

\[ \frac{1}{f'} = \mu \left( \frac{1}{r} - \frac{1}{\infty} \right) = \mu \left( \frac{1}{r} - 0 \right) = \frac{\mu}{r} \]

### Step 5: Relate the Focal Lengths
From the previous steps, we have:

\[ \frac{1}{f'} = \frac{\mu}{r} \]

We know from the earlier step that \( \frac{1}{f} = \frac{2\mu}{r} \). Therefore, we can relate \( f' \) to \( f \):

\[ \frac{1}{f'} = \frac{1}{2} \cdot \frac{1}{f} \]

This implies:

\[ f' = 2f \]

### Conclusion
The focal length of each half of the lens after being cut is \( 2f \).