Question

In Young's double slit experiment, the fringe width is \(2.4\) mm with light of wavelength \(9000\) A.U. If the light of wavelength \(5400\) A.U. is used, with no other change in the experimental set up. The change in fringe width is

• A. \(0.86\) mm
• B. \(0.96\) mm
• C. \(0.86\) cm
• D. \(0.96\) cm

Hint:

Fringe width is directly proportional to wavelength when the setup is unchanged, so scale the width by the ratio of wavelengths.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In Young's experiment the fringe width is \(\beta=\dfrac{\lambda D}{d}\).
With the setup unchanged, \(\beta\propto\lambda\).

Step 2: New Fringe Width
\(\dfrac{\beta_2}{\beta_1}=\dfrac{\lambda_2}{\lambda_1}=\dfrac{5400}{9000}=0.6\)
\(\beta_2=0.6\times2.4=1.44\) mm

Step 3: Calculation
Change in fringe width \(=\beta_1-\beta_2=2.4-1.44=0.96\) mm.

Step 4: Compare the options
The value \(0.96\) mm is option (B). The other mm option is 0.86 mm, and the cm options are ten times larger.

Final Answer:
The fringe width decreases by 0.96 mm. \[ \boxed{\text{(B) }\text{0.96 mm}} \]