If one of the two slits of Young's double-slit experiment is painted so that it transmits half the light intensity as the second slit, then

Options

A. finge pattern disappears
B. bright fringes become brighten and dark ones become darker
C. dark and bright fringes get fainter
D. dark fringes get brighter and bright fringes get darker

Explanation

To solve the problem, we will analyze the situation in Young's double-slit experiment where one of the slits transmits half the light intensity of the other.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- Let the intensity of light from the first slit be \( I_0 \).
- Since the second slit transmits half the intensity, its intensity will be \( \frac{I_0}{2} \).

2. **Calculating Maximum Intensity**:
- The maximum intensity \( I_{\text{max}} \) in Young's double-slit experiment is given by the formula:
\[ I_{\text{max}} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 \]
- Here, \( I_1 = I_0 \) and \( I_2 = \frac{I_0}{2} \).
- Thus, we have:
\[ I_{\text{max}} = \left( \sqrt{I_0} + \sqrt{\frac{I_0}{2}} \right)^2 \]
- Simplifying this:
\[ I_{\text{max}} = \left( \sqrt{I_0} + \frac{\sqrt{I_0}}{\sqrt{2}} \right)^2 = \left( \sqrt{I_0} \left( 1 + \frac{1}{\sqrt{2}} \right) \right)^2 \]
- This can be further simplified:
\[ I_{\text{max}} = I_0 \left( 1 + \frac{1}{\sqrt{2}} \right)^2 = I_0 \left( 1 + \frac{2}{\sqrt{2}} + \frac{1}{2} \right) = I_0 \left( \frac{3}{2} + \sqrt{2} \right) \]
- This value will be less than \( 4I_0 \) (the maximum when both slits have equal intensity).

3. **Calculating Minimum Intensity**:
- The minimum intensity \( I_{\text{min}} \) is given by:
\[ I_{\text{min}} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2 \]
- Substituting the values:
\[ I_{\text{min}} = \left( \sqrt{I_0} - \sqrt{\frac{I_0}{2}} \right)^2 = \left( \sqrt{I_0} - \frac{\sqrt{I_0}}{\sqrt{2}} \right)^2 \]
- Simplifying this:
\[ I_{\text{min}} = \left( \sqrt{I_0} \left( 1 - \frac{1}{\sqrt{2}} \right) \right)^2 = I_0 \left( 1 - \frac{1}{\sqrt{2}} \right)^2 \]
- This value will be greater than 0 since \( 1 - \frac{1}{\sqrt{2}} \) is positive.

4. **Conclusion**:
- Initially, with equal intensities, the maximum intensity was \( 4I_0 \) and the minimum was \( 0 \).
- After painting one slit, the maximum intensity is less than \( 4I_0 \) and the minimum intensity is greater than \( 0 \).
- Thus, the bright fringes become darker and the dark fringes become brighter.

### Final Answer:
The dark fringes get brighter and the bright fringes get darker.

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