The maximum intensity of fringes in Young's experiment is I. If one of the slit is closed, then the intensity at that place becomes `I_o`. Which of the following relation is true?

Answer

To solve the problem, we need to analyze the situation in Young's double-slit experiment where we have two slits and the intensity of light at a point on the screen.

### Step-by-Step Solution:

1. **Understanding Maximum Intensity (I)**:
- In Young's experiment, when both slits are open, the maximum intensity (I) on the screen is given by the formula:
\[ I = (A_1 + A_2)^2 \]
- Here, \( A_1 \) and \( A_2 \) are the amplitudes of light waves coming from slit 1 and slit 2, respectively.

2. **Expressing Maximum Intensity**:
- The maximum intensity can also be expressed in terms of the amplitudes:
\[ I = 4A_1^2 \]
- This is derived from the fact that if both slits are identical, \( A_1 = A_2 \), then:
\[ I = (A_1 + A_1)^2 = (2A_1)^2 = 4A_1^2 \]

3. **When One Slit is Closed**:
- If one of the slits is closed (let's say slit 2 is closed), the intensity at that point becomes \( I_0 \).
- In this case, the intensity is given by:
\[ I_0 = A_1^2 \]

4. **Relating Intensities**:
- From the previous steps, we have:
\[ A_1^2 = I_0 \]
- We also know from the maximum intensity condition that:
\[ A_1^2 = \frac{I}{4} \]
- Setting these two equations equal gives:
\[ I_0 = \frac{I}{4} \]

5. **Final Relation**:
- Rearranging the equation provides the final relation:
\[ I = 4I_0 \]

### Conclusion:
The correct relation between the maximum intensity \( I \) when both slits are open and the intensity \( I_0 \) when one slit is closed is:
\[ I = 4I_0 \]