In Young's double slit experiment, the two slits acts as coherent sources of equal amplitude A and wavelength `lambda`. In another experiment with the same set up the two slits are of equal amplitude A and wavelength `lambda` but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is

Options

A. `4:1`
B. `1:1`
C. `2:1`
D. `1:4`

Explanation

To solve the problem, we will analyze the two cases of Young's double slit experiment: one with coherent sources and the other with incoherent sources. We will then calculate the intensity of light at the midpoint of the screen for both cases and find the ratio.

### Step-by-Step Solution:

1. **Understanding Coherent Sources**:
- In the first case, the two slits act as coherent sources of equal amplitude \( A \) and wavelength \( \lambda \). Coherent sources have a constant phase difference, which is zero in this case.
- The wave equations for the two sources can be represented as:
\[ X_1 = A \sin(\omega t) \quad \text{and} \quad X_2 = A \sin(\omega t) \]

2. **Resultant Amplitude for Coherent Sources**:
- The resultant amplitude \( X_R1 \) at the midpoint of the screen is the sum of the two amplitudes:
\[ X_{R1} = X_1 + X_2 = A \sin(\omega t) + A \sin(\omega t) = 2A \sin(\omega t) \]
- The intensity \( I_{R1} \) is proportional to the square of the amplitude:
\[ I_{R1} \propto (X_{R1})^2 = (2A)^2 = 4A^2 \]

3. **Understanding Incoherent Sources**:
- In the second case, the two slits are incoherent but still have equal amplitude \( A \) and wavelength \( \lambda \). Incoherent sources have a random phase difference, which we can assume to be \( 90^\circ \) (or \( \frac{\pi}{2} \)).
- The wave equations for the two sources can be represented as:
\[ X_1 = A \sin(\omega t) \quad \text{and} \quad X_2 = A \cos(\omega t) \]

4. **Resultant Amplitude for Incoherent Sources**:
- The resultant amplitude \( X_{R2} \) is given by the vector sum of the two amplitudes:
\[ X_{R2} = \sqrt{(A \sin(\omega t))^2 + (A \cos(\omega t))^2} = \sqrt{A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t)} = \sqrt{A^2} = A \]
- The intensity \( I_{R2} \) is:
\[ I_{R2} \propto (X_{R2})^2 = A^2 \]

5. **Finding the Ratio of Intensities**:
- Now, we can find the ratio of the intensities at the midpoint of the screen for the two cases:
\[ \text{Ratio} = \frac{I_{R1}}{I_{R2}} = \frac{4A^2}{A^2} = 4 \]

### Final Answer:
The ratio of the intensity of light at the mid-point of the screen in the first case (coherent sources) to that in the second case (incoherent sources) is **4:1**.

---