In young's double slit experiment, distance between the slit `S_1 and S_2` is d and the distance between slit and screen is D . Then longest wavelength that will be missing on the screen in front of `S_1` is

Options

A. `d^2/D`
B. `d^2/(2D)`
C. `D/d^2`
D. None of these

Answer

To solve the problem regarding the longest wavelength that will be missing on the screen in front of \( S_1 \) in Young's double slit experiment, we can follow these steps:

### Step 1: Understanding the Setup
In the Young's double slit experiment, we have two slits \( S_1 \) and \( S_2 \) separated by a distance \( d \). The distance from the slits to the screen is \( D \). We are interested in finding the longest wavelength that will be missing at a point directly in front of \( S_1 \).

### Step 2: Condition for Destructive Interference
For destructive interference to occur at point \( P \) (which is directly in front of \( S_1 \)), the path difference between the light coming from \( S_1 \) and \( S_2 \) must be equal to an odd multiple of half wavelengths. The condition for destructive interference is given by:

\[ \Delta x = \left(2n + 1\right) \frac{\lambda}{2} \]

where \( n = 0, 1, 2, \ldots \)

### Step 3: Calculate the Path Difference
The path difference \( \Delta x \) can be approximated for small angles as follows:

\[ \Delta x = S_1P - S_2P \]

Using geometry, we can express \( S_1P \) and \( S_2P \):

- \( S_1P \) is approximately \( D \).
- \( S_2P \) can be approximated using the Pythagorean theorem:

\[ S_2P = \sqrt{D^2 + d^2} \approx D + \frac{d^2}{2D} \quad \text{(for small } d \text{ compared to } D\text{)} \]

Thus, the path difference becomes:

\[ \Delta x = D - \left(D + \frac{d^2}{2D}\right) = -\frac{d^2}{2D} \]

### Step 4: Set Up the Equation for Destructive Interference
For the first minimum (\( n = 0 \)):

\[ \frac{d^2}{2D} = \frac{\lambda}{2} \]

### Step 5: Solve for the Wavelength
Rearranging the above equation gives:

\[ \lambda = \frac{d^2}{D} \]

### Conclusion
Thus, the longest wavelength that will be missing on the screen in front of \( S_1 \) is:

\[ \lambda = \frac{d^2}{D} \]