In a Young's double slit experiment minimum intensity is found to be non-zero. If one of the slits is covered by a transparent film which absorbs 10% of light energy passing through it, then

Options

A. Intensity at maxima must decrease
B. Intensity at maxima may decrease
C. Intensity at maxima may increase
D. Intensity at maxima may decrease

Answer

To solve the problem, we will analyze the situation in a Young's double slit experiment where one of the slits is covered by a transparent film that absorbs 10% of the light energy passing through it.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
In a Young's double slit experiment, the intensity at a point on the screen due to two coherent sources (slits) is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \(I_1\) and \(I_2\) are the intensities from the two slits, and \(\phi\) is the phase difference.

2. **Minimum Intensity is Non-Zero**:
The problem states that the minimum intensity is non-zero, which implies that \(I_1\) and \(I_2\) are not equal. Thus, we can denote:
\[ I_1 \neq I_2 \]

3. **Effect of the Transparent Film**:
When one of the slits (say slit 1) is covered with a transparent film that absorbs 10% of the light energy, the intensity from that slit becomes:
\[ I_1' = 0.9 I_1 \]
The intensity from the other slit remains unchanged:
\[ I_2' = I_2 \]

4. **New Maximum Intensity Calculation**:
The new maximum intensity when both slits are open again (with the film on slit 1) can be calculated as follows:
\[ I'_{\text{max}} = I_1' + I_2 + 2\sqrt{I_1' I_2} \]
Substituting \(I_1' = 0.9 I_1\):
\[ I'_{\text{max}} = 0.9 I_1 + I_2 + 2\sqrt{0.9 I_1 I_2} \]

5. **Comparison with Previous Maximum Intensity**:
The previous maximum intensity (without the film) was:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
We need to compare \(I'_{\text{max}}\) with \(I_{\text{max}}\).

6. **Analyzing the Change in Intensity**:
Since \(I'_{\text{max}} = 0.9 I_1 + I_2 + 2\sqrt{0.9 I_1 I_2}\), we can see that:
- The term \(0.9 I_1\) is less than \(I_1\).
- The term \(2\sqrt{0.9 I_1 I_2}\) is less than \(2\sqrt{I_1 I_2}\) because \(\sqrt{0.9} < 1\).

Therefore, it can be concluded that:
\[ I'_{\text{max}} < I_{\text{max}} \]

7. **Conclusion**:
The intensity at the maximum point after covering one slit with the film will decrease.

### Final Answer:
The intensity at the maximum point must decrease.