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

Options

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

Explanation

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

### Step 1: Understand the condition for interference maxima
The condition for interference maxima in Young's double-slit experiment is given by the equation:
\[ d \sin \theta = n \lambda \]
where:
- \( d \) = slit separation
- \( \theta \) = angle of the maxima
- \( n \) = order of the maxima (an integer)

### Step 2: Substitute the given value of slit separation
According to the problem, the slit separation \( d \) is equal to \( 2\lambda \). Therefore, we can substitute this into the equation:
\[ 2\lambda \sin \theta = n \lambda \]

### Step 3: Simplify the equation
Dividing both sides of the equation by \( \lambda \) (assuming \( \lambda \neq 0 \)):
\[ 2 \sin \theta = n \]
This can be rearranged to find \( \sin \theta \):
\[ \sin \theta = \frac{n}{2} \]

### Step 4: Determine the range of \( \sin \theta \)
The sine function has a range of values between -1 and 1. Therefore, we have:
\[ -1 \leq \sin \theta \leq 1 \]
Substituting our expression for \( \sin \theta \):
\[ -1 \leq \frac{n}{2} \leq 1 \]

### Step 5: Solve for \( n \)
Multiplying the entire inequality by 2 gives:
\[ -2 \leq n \leq 2 \]
This indicates that \( n \) can take integer values from -2 to 2.

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

Thus, there are a total of 5 possible values for \( n \).

### Conclusion
The maximum number of possible interference maxima is **5**.

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