Question

In Young's double slit experiment, how many maxima can be seen on a screen (including central maxima) if d = \(\frac{5\lambda}{2}\) (where \(\lambda\) is wavelength of light and d is distance between the two slits).

• A. 5
• B. 4
• C. 7
• D. 1

Hint:

To find the total number of maxima, use the formula \(2 \times \lfloor d/\lambda \rfloor + 1\). If \(d/\lambda\) is exactly an integer, the maxima at \(\pm 90^{\circ}\) are usually not "seen" on a flat screen at a finite distance, but here the value is 2.5, so 5 is the clear result.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the total number of interference maxima possible in a YDSE setup given the slit separation in terms of the wavelength.
Step 2: Key Formula or Approach:
The condition for a maximum in YDSE is:
\[ d \sin \theta = n \lambda \]
where \(n\) is the order of the maximum and \(\theta\) is the angular position.
Step 3: Detailed Explanation:
The maximum possible value for \(\sin \theta\) is 1 (corresponding to \(\theta = 90^{\circ}\) at infinity on the screen).
So, the maximum order \(n\) is constrained by:
\[ |n| \le \frac{d}{\lambda} \]
Given \(d = \frac{5\lambda}{2} = 2.5 \lambda\):
\[ |n| \le \frac{2.5 \lambda}{\lambda} \Rightarrow |n| \le 2.5 \]
Since \(n\) must be an integer, the possible values for \(n\) are:
\[ n = 0, \pm 1, \pm 2 \]
- \(n = 0\) is the central maximum.
- \(n = 1, 2\) are maxima on one side of the center.
- \(n = -1, -2\) are maxima on the other side.
Total number of maxima = \(2 \times (\text{maximum integer value of } n) + 1 = (2 \times 2) + 1 = 5\).
Step 4: Final Answer:
A total of 5 maxima can be seen on the screen.