A lens having focal length `f` and aperture of diameter `d` forms an image of intensity `I`. Aperture of diameter `d//2` in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

Options

A. f and `I/4`
B. `(3f)/(4)` and ` I/2`
C. f and `(3I)/(4)`
D. `(f)/(2)` and ` I/2`

Explanation

To solve the problem, we need to analyze the effect of covering the central region of a lens with a black paper, which reduces the aperture size. We will determine how this affects the focal length of the lens and the intensity of the image formed.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- We have a lens with a focal length \( f \) and an aperture of diameter \( d \).
- The lens forms an image with intensity \( I \).

2. **Effect of Reducing the Aperture:**
- When the central region of the lens with diameter \( d/2 \) is covered, the effective aperture diameter is reduced to \( d/2 \).
- The radius of the aperture also changes; it becomes \( r = \frac{d}{4} \) (since radius is half of the diameter).

3. **Calculating the Area of the Aperture:**
- The area of the original aperture (diameter \( d \)) is given by:
\[ A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
- The area of the new aperture (diameter \( d/2 \)) is:
\[ A' = \pi \left(\frac{d/2}{2}\right)^2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16} \]

4. **Comparing Areas:**
- The new area \( A' \) is one-fourth of the original area \( A \):
\[ A' = \frac{A}{4} \]

5. **Effect on Intensity:**
- Intensity \( I \) is proportional to the area of the aperture. Therefore, if the area is reduced to one-fourth, the intensity will also reduce to one-fourth:
\[ I' = \frac{I}{4} \]

6. **Final Intensity Calculation:**
- The new intensity \( I' \) is:
\[ I' = \frac{I}{4} \]

7. **Effect on Focal Length:**
- The focal length \( f \) of the lens does not depend on the size of the aperture. It is determined by the lens material and its curvature.
- Therefore, the focal length remains unchanged:
\[ f' = f \]

### Conclusion:
- The new focal length of the lens remains \( f \).
- The new intensity of the image becomes \( \frac{I}{4} \).

### Final Answer:
- Focal length: \( f \)
- Intensity: \( \frac{I}{4} \)