Consider the situation shown in figure. The two slite `S_1 and S_2`
placed symmetrically around the centre line are illuminated by a monochromatic
light of wavelength lambda. The separation between the slits is d. The light transmitted
by the slits falls on a screen `M_1` placed at a distance D from the slits. The slit `S_3` is
at the centre line and the slit `S_4` is at a distance y form `S_3`. Another screen `M_2` is
placed at a further distance D away from `M_1`. Find the ration of the maximum to
minimum intensity observed on `M_2` if y is equal to `(dltltD)`.
`

Explanation

To solve the problem, we need to find the ratio of the maximum to minimum intensity observed on screen M2, given the conditions of the slits and the distances involved. Let's break down the solution step by step:

### Step 1: Understanding the Setup
We have two slits S1 and S2 that are illuminated by monochromatic light of wavelength λ. The distance between the slits is d, and the distance from the slits to screen M1 is D. There are two additional slits S3 and S4, with S3 at the center line and S4 at a distance y from S3. Another screen M2 is placed at a further distance D from M1.

**Hint:** Visualize the arrangement of the slits and screens to understand the path differences that will affect the interference pattern.

### Step 2: Path Difference at S4
The path difference between the light coming from S1 and S2 to point S4 on screen M2 can be calculated. Since S4 is at a distance y from S3, the path difference (Δ) can be approximated as:
\[ \Delta = \frac{d \cdot y}{D} \]
This is valid under the condition that \( y \) is much smaller than \( D \) (y << D).

**Hint:** Remember that the path difference is crucial for determining the interference pattern.

### Step 3: Conditions for Maximum and Minimum Intensity
1. **Maximum Intensity (I_max)** occurs when the path difference is an integer multiple of the wavelength:
\[ \Delta = n\lambda \quad (n = 0, 1, 2, \ldots) \]
2. **Minimum Intensity (I_min)** occurs when the path difference is an odd multiple of half the wavelength:
\[ \Delta = (n + \frac{1}{2})\lambda \quad (n = 0, 1, 2, \ldots) \]

**Hint:** Use the conditions for constructive and destructive interference to find the respective intensities.

### Step 4: Calculating Intensities
The intensity at any point on the screen can be expressed in terms of the amplitudes of the waves coming from the slits. If we denote the amplitude from each slit as A, the resultant amplitude at S4 can be calculated for both maximum and minimum conditions.

1. For maximum intensity:
\[ I_{max} = (A_1 + A_2)^2 = (2A)^2 = 4A^2 \]
2. For minimum intensity:
\[ I_{min} = (A_1 - A_2)^2 = (0)^2 = 0 \quad \text{(when they cancel each other)} \]

**Hint:** Remember that the intensity is proportional to the square of the amplitude.

### Step 5: Finding the Ratio
The ratio of maximum to minimum intensity can be calculated as:
\[ \frac{I_{max}}{I_{min}} = \frac{4A^2}{0} \quad \text{(undefined, since minimum intensity cannot be zero)} \]
However, if we consider the case where the amplitudes do not cancel completely, we can express the ratio in terms of the amplitudes:
\[ \frac{I_{max}}{I_{min}} = \frac{(2A)^2}{(A - A)^2} \]
This leads to a more general case where we can find a finite value based on the specific amplitudes involved.

**Hint:** Be cautious with the minimum intensity; it should not be zero in practical scenarios.

### Final Answer
The ratio of the maximum to minimum intensity observed on M2 can be expressed as:
\[ \frac{I_{max}}{I_{min}} = \text{finite value based on specific amplitudes} \]