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 l/4`
B. `(3f)/4and (l)/(2)`
C. `fand (3l)/4`
D. `f/2andl/2`

Answer

To solve the problem step by step, we need to analyze the effects of covering part of the lens on both the focal length and the intensity of the image formed by the lens.

### Step 1: Understanding the problem
We have a lens with a focal length \( f \) and an aperture of diameter \( d \). The lens forms an image with intensity \( I \). When we cover the central region of the lens with a diameter of \( d/2 \) using black paper, we need to determine how this affects the focal length and the intensity of the image.

### Step 2: Effect on Focal Length
The focal length of a lens is determined by its curvature and the refractive index of the material. Covering part of the lens does not change its curvature or the material, hence the focal length remains unchanged. Therefore, the focal length after covering the lens is still:

\[ f' = f \]

### Step 3: Effect on Intensity
The intensity of light passing through a lens is proportional to the area of the aperture that is not covered. The intensity \( I \) is related to the amplitude \( A \) of the light wave, and the relationship can be expressed as:

\[ I \propto A^2 \]

The area of the full aperture is given by:

\[ A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]

When we cover the central region of the lens with a diameter of \( d/2 \), the area of the covered region is:

\[ A_{covered} = \pi \left(\frac{d/2}{2}\right)^2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16} \]

The area of the remaining uncovered region is:

\[ A_2 = A_1 - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{4\pi d^2}{16} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16} \]

### Step 4: Calculating New Intensity
Using the relationship between intensity and area, we can express the new intensity \( I_2 \) as:

\[ \frac{I_2}{I} = \frac{A_2}{A_1} \]

Substituting the areas we calculated:

\[ \frac{I_2}{I} = \frac{\frac{3\pi d^2}{16}}{\frac{\pi d^2}{4}} = \frac{3}{4} \]

Thus, the new intensity \( I_2 \) is:

\[ I_2 = \frac{3}{4} I \]

### Final Answer
Therefore, the focal length of the lens remains unchanged, and the new intensity of the image is \( \frac{3}{4} I \).

- **Focal Length:** \( f' = f \)
- **Intensity:** \( I_2 = \frac{3}{4} I \)