A symmetric double convex lens is cut in two equal parts by a plane perpendiculr to the pricipal axis. If the power of the original lens was 4D, the power of a cut lens will be
a.2D b.3D c.4D d.5D

Options

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

Explanation

To solve the problem step by step, we will analyze the situation of cutting a symmetric double convex lens and how it affects the power of the lens.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- The original lens is a symmetric double convex lens with a power of \( P = 4D \).
- The lens is cut into two equal parts by a plane perpendicular to the principal axis.

2. **Recall the Formula for Power:**
- The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length of the lens.

3. **Determine the Focal Length of the Original Lens:**
- From the power of the original lens:
\[ 4 = \frac{1}{f} \implies f = \frac{1}{4} \text{ meters} = 0.25 \text{ meters} \]

4. **Understand the Effect of Cutting the Lens:**
- When the lens is cut in half, the focal length of each half changes. The new focal length \( f' \) of each half can be derived from the original focal length:
- The focal length of the cut lens is given by:
\[ \frac{1}{f'} = \mu - 1 \left( \frac{1}{R} - \frac{1}{\infty} \right) \]
- Since \( \frac{1}{\infty} = 0 \), we have:
\[ \frac{1}{f'} = \mu - 1 \cdot \frac{1}{R} \]

5. **Relate the Focal Lengths:**
- For the original lens, we had:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) = (\mu - 1) \cdot \frac{2}{R} \]
- Therefore, when the lens is cut, the new focal length \( f' \) becomes:
\[ f' = 2f \]

6. **Calculate the Power of the Cut Lens:**
- The power of the cut lens \( P' \) can now be calculated as:
\[ P' = \frac{1}{f'} = \frac{1}{2f} \]
- Since we know \( f = 0.25 \) meters, we have:
\[ P' = \frac{1}{2 \times 0.25} = \frac{1}{0.5} = 2D \]

7. **Final Answer:**
- The power of the cut lens is \( 2D \).

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