A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4D, the power of a cut lens will be

Options

A. A. 2D
B. B. 3D
C. C. 4D
D. D. 5D

Answer

To solve the problem, we need to determine the power of a cut lens when a symmetric double convex lens is cut into two equal parts. The power of the original lens is given as 4D (diopters).

### Step-by-Step Solution:

1. **Understand the Power of a Lens**:
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{F} \]
where \( F \) is the focal length of the lens.

2. **Power of the Original Lens**:
The power of the original double convex lens is given as \( P = 4D \).

3. **Cutting the Lens**:
When the lens is cut into two equal parts along a plane perpendicular to the principal axis, each half will behave like a new lens.

4. **Analyzing the Cut Lens**:
Each half of the lens will have a different effective focal length. The original lens has two curved surfaces, and when cut, one side will remain curved while the other side will become flat (plane surface).

5. **Determine the New Radius of Curvature**:
For the cut lens:
- The radius of curvature \( R_1 \) of the curved surface remains \( R \).
- The radius of curvature \( R_2 \) of the flat surface is considered to be infinite (\( R_2 = \infty \)).

6. **Using the Lens Maker's Formula**:
The power of the new lens (cut lens) can be calculated using the lens maker's formula:
\[ P' = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting \( R_1 = R \) and \( R_2 = \infty \):
\[ P' = \mu - 1 \left( \frac{1}{R} - 0 \right) = \frac{\mu - 1}{R} \]

7. **Relating to the Original Power**:
From the original lens, we know:
\[ P = \mu - 1 \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) = \frac{2(\mu - 1)}{R} \]
Since \( P = 4D \):
\[ 4 = \frac{2(\mu - 1)}{R} \]
Rearranging gives:
\[ \mu - 1 = \frac{4R}{2} = 2R \]

8. **Substituting Back to Find New Power**:
Now substituting \( \mu - 1 = 2R \) into the equation for \( P' \):
\[ P' = \frac{2R}{R} = 2D \]

9. **Conclusion**:
The power of each cut lens is \( 2D \).

### Final Answer:
The power of the cut lens will be \( 2D \).