The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

Options

A. three
B. five
C. infinite
D. zero

Explanation

To determine the maximum number of possible interference maxima in Young's double-slit experiment when the slit separation \( d \) is equal to twice the wavelength \( \lambda \), we can follow these steps:

### Step 1: Understand 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 \( d \) is the slit separation, \( \theta \) is the angle of the maxima, \( n \) is an integer (0, ±1, ±2, ...), and \( \lambda \) is the wavelength.

### Step 2: Substitute the Given Slit Separation
Given that the slit separation \( d \) is equal to \( 2\lambda \):
\[ 2\lambda \sin \theta = n \lambda \]

### Step 3: Simplify the Equation
We can simplify the equation by dividing both sides by \( \lambda \) (assuming \( \lambda \neq 0 \)):
\[ 2 \sin \theta = n \]

### Step 4: Determine the Range of \( \sin \theta \)
Since \( \sin \theta \) can only take values between -1 and 1, we can set up the inequality:
\[ -1 \leq \sin \theta \leq 1 \]
Substituting this into our equation gives:
\[ -1 \leq \frac{n}{2} \leq 1 \]

### Step 5: Solve for \( n \)
Multiplying the entire inequality by 2:
\[ -2 \leq n \leq 2 \]
This means \( n \) can take the integer values: -2, -1, 0, 1, and 2.

### Step 6: Count the Possible Values of \( n \)
The possible integer values of \( n \) are:
- -2
- -1
- 0
- 1
- 2

This gives us a total of 5 possible values for \( n \).

### Conclusion
Thus, the maximum number of possible interference maxima for a slit separation equal to twice the wavelength in Young's double-slit experiment is **5**.

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