In Young's double-slit experiment, the separation between the slits is d, distance between the slit and screen is `D (D gt gt d)`, In the interference pattern, there is a maxima exactly in front of each slit. Then the possilbe wavelength(s) used in the experiment are

Options

A. `d^(2) // D, d^(2) // 2D, d^(2) // 3D`
B. `d^(2) // D, d^(2) // 3d, d^(2) // 5D`
C. `d^(2) // 2D, d^(2) // 4D, d^(2) // 6D`
D. none of these

Explanation

To solve the problem, we need to analyze the conditions for maxima in Young's double-slit experiment. Here’s a step-by-step solution:

### Step 1: Understand the setup
In Young's double-slit experiment, we have two slits separated by a distance \( d \), and the screen is at a distance \( D \) from the slits. The condition for constructive interference (maxima) at a point on the screen is given by the equation:
\[ y = \frac{n \lambda D}{d} \]
where \( n \) is the order of the maxima, \( \lambda \) is the wavelength of the light used, and \( y \) is the distance from the central maximum to the \( n^{th} \) maximum.

### Step 2: Condition for maxima in front of each slit
For there to be a maxima exactly in front of each slit, the positions of the maxima must coincide with the positions of the slits. This means that at \( y = 0 \) (the central maximum), and at \( y = \frac{d}{2} \) (the position of the first maximum in front of the first slit) and \( y = -\frac{d}{2} \) (the position of the first maximum in front of the second slit), we need to find the corresponding wavelengths.

### Step 3: Fringe width
The fringe width \( \beta \) is defined as:
\[ \beta = \frac{\lambda D}{d} \]
According to the problem, the fringe width is given as \( \frac{d}{2} \). Therefore, we can set up the equation:
\[ \frac{d}{2} = \frac{\lambda D}{d} \]

### Step 4: Solve for \( \lambda \)
Rearranging the equation gives us:
\[ \lambda = \frac{d^2}{2D} \]

### Step 5: Determine possible wavelengths
Now we can find the possible wavelengths by substituting different integer values for \( n \) (the order of the maxima):
- For \( n = 1 \):
\[ \lambda_1 = \frac{d^2}{2D} \]
- For \( n = 2 \):
\[ \lambda_2 = \frac{d^2}{4D} \]
- For \( n = 3 \):
\[ \lambda_3 = \frac{d^2}{6D} \]

### Step 6: Check options
Now we can check the options given in the question to see which wavelengths match the derived expressions. The correct option will be the one that corresponds to one of the wavelengths calculated.

### Conclusion
After checking the options, we find that option C corresponds to one of the calculated wavelengths, confirming it as the correct answer.

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