Question

The angular width of interference fringes in Young's double-slit experiment depends on:

• A. distance between the slits and the screen only
• B. wavelength of light used only
• C. both wavelength of light used and the slits separation
• D. slits separation only

Hint:

Angular fringe width is the fringe width beta = lambda D / d divided by D, so the screen distance cancels.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Young's double-slit experiment, bright and dark fringes form on a screen.
The fringe width \(\beta\) is the distance between two neighbouring bright fringes. The angular fringe width \(\theta\) is the angle this width makes at the slits.

Step 2: Key Formula or Approach:
\[ \beta = \frac{\lambda D}{d} \]
Here \(\lambda\) is the wavelength, \(D\) is the distance from the slits to the screen and \(d\) is the slit separation.
Angular width is the fringe width divided by the screen distance: \(\theta = \dfrac{\beta}{D}\).

Step 3: Detailed Explanation:
Divide the fringe width by \(D\).
\[ \theta = \frac{\beta}{D} = \frac{\lambda}{d} \]
The screen distance \(D\) cancels out. So the angular width depends only on \(\lambda\) and \(d\).

Step 4: Check the options:
Option (A) says the slit-to-screen distance only. But \(D\) cancels out, so this is wrong.
Option (B) says wavelength only. The slit separation \(d\) also appears in the formula, so this is incomplete.
Option (D) says slit separation only. The wavelength is also needed, so this is incomplete.
Option (C) says both wavelength and slit separation, and this matches \(\theta = \lambda/d\).

Final Answer:
The angular width of the fringes depends on both the wavelength and the slit separation, which is option (C). \[ \boxed{\theta = \frac{\lambda}{d}} \]