In Young's double slit experiment, the intensity of central maximum is `I`. What will be the intensity at the same place if one slit is closed ?

Options

A. same as
B. twice
C. 1/4 times
D. half times

Answer

To solve the problem, we need to determine the intensity at the central maximum when one of the slits in Young's double slit experiment is closed.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- In Young's double slit experiment, when both slits are open, the intensity at the central maximum is given as \( I \).
- The intensity \( I \) at the central maximum is related to the amplitudes of the waves coming from the two slits.

2. **Amplitude Calculation**:
- Let the amplitude of the wave from each slit be \( A \).
- When both slits are open, the total amplitude at the central maximum (where the waves from both slits constructively interfere) is \( A_1 = A + A = 2A \).

3. **Intensity Relation**:
- The intensity \( I_1 \) when both slits are open is given by:
\[ I_1 \propto (A_1)^2 = (2A)^2 = 4A^2 \]
- Thus, we can express the intensity as:
\[ I_1 = k(2A)^2 = 4kA^2 \]
- Where \( k \) is a proportionality constant.

4. **Closing One Slit**:
- Now, if we close one slit (let's say slit S2), the amplitude from the remaining open slit (S1) will be \( A_2 = A \).
- The intensity \( I_2 \) when one slit is closed is given by:
\[ I_2 \propto (A_2)^2 = A^2 \]
- Therefore, we can express this intensity as:
\[ I_2 = kA^2 \]

5. **Finding the Relationship Between Intensities**:
- We already have \( I_1 = 4kA^2 \) and \( I_2 = kA^2 \).
- To find the relationship between \( I_1 \) and \( I_2 \):
\[ \frac{I_1}{I_2} = \frac{4kA^2}{kA^2} = 4 \]
- This implies:
\[ I_2 = \frac{I_1}{4} \]

6. **Substituting the Value of \( I_1 \)**:
- Since \( I_1 = I \) (the intensity at the central maximum when both slits are open), we can substitute:
\[ I_2 = \frac{I}{4} \]

### Final Answer:
The intensity at the same place when one slit is closed will be \( \frac{I}{4} \).