White light is used to illuminate the two slits in a Young's
double slit experiment. The separation between the slits is b and the screen is at
a distance `d(gtgtb)` from the slits. At a point on the screen directly in front of
one of the slits, certain wavelengths are missing. Some of these missing
wavelength are

Options

A. `lambda = b^2//d`
B. `lambda = 2b^2//d`
C. ` lambda = b^2//3d`
D. `lambda = 2b^2//3d`

Explanation

To solve the problem regarding the missing wavelengths in the Young's double slit experiment with white light, we will follow these steps:

### Step 1: Understand the Setup
We have two slits separated by a distance \( b \) and a screen located at a distance \( d \) from the slits. The point of interest on the screen is directly in front of one of the slits.

### Step 2: Identify the Condition for Missing Wavelengths
At the point directly in front of one of the slits, certain wavelengths are missing due to destructive interference. For destructive interference to occur, the path difference between the light waves from the two slits must be an odd multiple of half the wavelength.

### Step 3: Calculate the Path Difference
The path difference \( \Delta \) between the two waves arriving at the point on the screen can be calculated using the formula:
\[ \Delta = \frac{b \cdot x}{d} \]
where \( x \) is the horizontal distance from the midpoint between the slits to the point on the screen. Since we are considering the point directly in front of one slit, \( x = 0 \), and thus the path difference simplifies to:
\[ \Delta = \frac{b \cdot 0}{d} = 0 \]
However, we need to consider the scenario where we look at points slightly off-center, leading to a path difference of:
\[ \Delta = \frac{b}{d} \cdot \left(\frac{b}{2}\right) = \frac{b^2}{2d} \]
This represents the path difference when considering the symmetry of the setup.

### Step 4: Set Up the Condition for Destructive Interference
The condition for destructive interference is given by:
\[ \Delta = \left(n - \frac{1}{2}\right) \lambda \]
where \( n \) is an integer (1, 2, 3, ...).

### Step 5: Equate the Path Difference to the Condition for Destructive Interference
Setting the two expressions for path difference equal gives:
\[ \frac{b^2}{2d} = \left(n - \frac{1}{2}\right) \lambda \]
Rearranging this gives:
\[ \lambda = \frac{b^2}{d} \cdot \frac{1}{2n - 1} \]

### Step 6: Identify the Missing Wavelengths
The wavelengths that are missing correspond to different values of \( n \):
- For \( n = 1 \):
\[ \lambda_1 = \frac{b^2}{d} \]
- For \( n = 2 \):
\[ \lambda_2 = \frac{b^2}{3d} \]
- For \( n = 3 \):
\[ \lambda_3 = \frac{b^2}{5d} \]
- And so on...

### Step 7: Conclude the Missing Wavelengths
The missing wavelengths at the point directly in front of one of the slits are:
\[ \lambda = \frac{b^2}{d}, \frac{b^2}{3d}, \frac{b^2}{5d}, \ldots \]