A thin lens focal length `f` and its aperture has diameter `d`. It forms an image of intensity `I`. Now the central part of the aperture up to diameter`(d)/(2)` is blocked by an opaque paper. The focal length and image intensity will change to

Options

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

Answer

To solve the problem step by step, we will analyze the effects of blocking the central part of the lens aperture on both the focal length and the intensity of the image formed by the lens.

### Step 1: Understand the Initial Conditions
We start with a thin lens that has:
- Focal length \( f \)
- Aperture diameter \( d \)
- Image intensity \( I \)

### Step 2: Determine the Effect of Blocking the Aperture
When the central part of the aperture (up to diameter \( \frac{d}{2} \)) is blocked, we need to find out how this affects the focal length and intensity.

### Step 3: Analyze the Focal Length
The focal length of a thin lens is determined by its curvature and the refractive index of the material. Since the curvature and the refractive index remain unchanged when only a part of the aperture is blocked, the focal length remains the same:
- New focal length \( f' = f \)

### Step 4: Calculate the Area of the Aperture
The area of the original aperture is given by:
\[ A_{\text{original}} = \frac{\pi d^2}{4} \]

### Step 5: Calculate the Area of the Blocked Section
The area of the blocked section (the central part with diameter \( \frac{d}{2} \)) is:
\[ A_{\text{blocked}} = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi d^2}{16} \]

### Step 6: Calculate the Remaining Area of the Aperture
The remaining area after blocking the central part is:
\[ A_{\text{remaining}} = A_{\text{original}} - A_{\text{blocked}} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} \]
To simplify this:
\[ A_{\text{remaining}} = \frac{4\pi d^2}{16} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16} \]

### Step 7: Relate the New Area to the Original Area
The new area can be expressed in terms of the original area:
\[ A_{\text{remaining}} = \frac{3}{4} \cdot A_{\text{original}} \]

### Step 8: Determine the Effect on Intensity
Intensity \( I \) is directly proportional to the area of the aperture. Thus, the new intensity \( I' \) can be calculated as:
\[ I' = \frac{3}{4} I \]

### Conclusion
- The new focal length remains the same: \( f' = f \)
- The new intensity becomes \( I' = \frac{3}{4} I \)

### Final Answer
- Focal length: \( f' = f \)
- Image intensity: \( I' = \frac{3}{4} I \)