Question

In Young's double slit experiment, with a source of light having wavelength $6300\ \text{\AA}$, the first maxima will occur when the

• A. path difference is $9200\ \text{\AA}$
• B. phase difference is $n$ radian
• C. phase difference is $\frac{\pi}{2}$ radian
• D. path difference is $6300\ \text{\AA}$

Hint:

Always associate "maxima" with full wavelength steps ($1\lambda, 2\lambda, 3\lambda$) and "minima" with half wavelength steps ($0.5\lambda, 1.5\lambda, 2.5\lambda$). The first maximum is exactly $1\lambda$ away from the center.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to identify the condition for the first maximum (bright fringe) in a Young's double slit experiment (YDSE) for a given wavelength.

Step 2: Key Formula or Approach:
In YDSE, the condition for constructive interference (maxima) is that the path difference ($\Delta x$) must be an integer multiple of the wavelength ($\lambda$):
$$\Delta x = n\lambda \quad \text{where } n = 0, 1, 2, \ldots$$
The condition for phase difference ($\Delta \phi$) for maxima is an even multiple of $\pi$:
$$\Delta \phi = 2n\pi \quad \text{where } n = 0, 1, 2, \ldots$$

Step 3: Detailed Explanation:
The central maximum corresponds to $n = 0$ (path difference = 0).
The "first maxima" refers to the first-order bright fringe, which corresponds to $n = 1$.
Using the path difference formula for $n = 1$:
$$\Delta x = 1 \cdot \lambda = \lambda$$
We are given the wavelength $\lambda = 6300\ \text{\AA}$.
Therefore, the path difference for the first maximum is exactly $6300\ \text{\AA}$.
Looking at the phase difference options: the phase difference for the first maximum would be $2\pi$ radians, which is not listed correctly in any of the choices.

Step 4: Final Answer:
The first maxima occurs when the path difference is $6300\ \text{\AA}$, matching option (D).