Question

In Young's double slit experiment, the band width of the fringes observed is \( \beta \), when light of wavelength \( \lambda \) is used. With same experimental set up, to double the band width of the fringes, the wavelength of light required is

• A. \( \lambda \)
• B. \( \frac{\lambda}{2} \)
• C. \( 2\lambda \)
• D. \( \frac{\lambda}{4} \)
• E. \( \frac{\lambda}{8} \)

Hint:

Fringe width directly proportional to wavelength.

Answer 1

The Correct Option is C

Step 1: Understanding Fringe Width in YDSE:
In a Young's Double Slit Experiment (YDSE), the fringe width or band width (\(\beta\)) is the distance between the centers of two consecutive bright fringes or two consecutive dark fringes. It is a measure of how spread out the interference pattern is.
Step 2: Key Formula or Approach:
The formula for the fringe width (\(\beta\)) is given by:
\[ \beta = \frac{\lambda D}{d} \] where:
\(\lambda\) = wavelength of the light used
\(D\) = distance from the slits to the screen
\(d\) = separation between the two slits
Step 3: Detailed Explanation:
The problem states that the "same experimental set up" is used, which means the distances \(D\) and \(d\) are kept constant.
From the formula, we can see that the fringe width \(\beta\) is directly proportional to the wavelength \(\lambda\).
\[ \beta \propto \lambda \] Let the initial condition be \(\beta_1 = \beta\) and \(\lambda_1 = \lambda\).
The desired final condition is a doubled fringe width, \(\beta_2 = 2\beta\). We need to find the new wavelength, \(\lambda_2\).
Using the proportionality:
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \] Substituting the values:
\[ \frac{2\beta}{\beta} = \frac{\lambda_2}{\lambda} \] \[ 2 = \frac{\lambda_2}{\lambda} \] \[ \lambda_2 = 2\lambda \] Step 4: Final Answer:
To double the band width, the wavelength of the light must also be doubled. The required wavelength is \(2\lambda\).