In Young's double slit experiment, how many maxima can be obtained on a screen (including central maxima). If `d=(5lambda)/2` (where `lambda` is the wavelength of light)?

Options

A. 5
B. 4
C. 7
D. 1

Answer

To determine the total number of maxima (bright fringes) that can be obtained on a screen in Young's double slit experiment, we start with the given information:

1. **Given**: The distance between the slits \( d = \frac{5\lambda}{2} \), where \( \lambda \) is the wavelength of light.

2. **Understanding the condition for maxima**: In Young's double slit experiment, the condition for constructive interference (maxima) is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is an integer (0, ±1, ±2, ...), and \( \theta \) is the angle at which the maxima occur.

3. **Finding the maximum value of \( n \)**: The maximum value of \( \sin \theta \) is 1. Therefore, we can set up the equation:
\[ d = n \lambda \]
Substituting the value of \( d \):
\[ \frac{5\lambda}{2} = n \lambda \]
Dividing both sides by \( \lambda \) (assuming \( \lambda \neq 0 \)):
\[ \frac{5}{2} = n \]
This implies that the maximum integer value of \( n \) is 2.

4. **Counting the maxima**: The values of \( n \) can be:
- Central maximum: \( n = 0 \)
- First maxima on either side: \( n = ±1 \)
- Second maxima on either side: \( n = ±2 \)

Thus, the possible values of \( n \) are \( -2, -1, 0, +1, +2 \).

5. **Total maxima**: Counting these values gives us:
- \( n = -2 \) (1st order maximum on one side)
- \( n = -1 \) (2nd order maximum on one side)
- \( n = 0 \) (central maximum)
- \( n = +1 \) (1st order maximum on the other side)
- \( n = +2 \) (2nd order maximum on the other side)

Therefore, the total number of maxima is:
\[ 5 \text{ (including the central maximum)} \]

**Final Answer**: The total number of maxima on the screen is 5.

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