Question

The width of a fringe is $0.5 \text{ mm}$ in Young’s double slit experiment for a light of wavelength $500\text{ nm}$. If the wave length of light alone is changed to $600 \text{ nm}$, the width of the fringe becomes}

• A. $0.4 \text{ mm}$
• B. $0.3 \text{ mm}$
• C. $0.2 \text{ mm}$
• D. $0.6 \text{ mm}$
• E. $0.55 \text{ mm}$

Hint:

Increasing the wavelength increases the fringe width. A $20\%$ increase in wavelength ($500$ to $600$) leads to a $20\%$ increase in fringe width ($0.5$ to $0.6$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), the fringe width is directly proportional to the wavelength of the light used, assuming the experimental setup (screen distance and slit separation) remains unchanged.
Step 2: Key Formula or Approach:
Fringe width equation: \(\beta = \frac{\lambda D}{d}\).
Since \(D\) and \(d\) are constant, \(\beta \propto \lambda\).
Therefore, \(\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}\).
Step 3: Detailed Explanation:
Given values:
Initial wavelength \(\lambda_1 = 500 \text{ nm}\)
Initial fringe width \(\beta_1 = 0.5 \text{ mm}\)
Final wavelength \(\lambda_2 = 600 \text{ nm}\)
Substitute into the proportional relationship:
\[ \frac{\beta_2}{0.5} = \frac{600}{500} \] \[ \frac{\beta_2}{0.5} = \frac{6}{5} = 1.2 \] Multiply by 0.5:
\[ \beta_2 = 1.2 \times 0.5 = 0.6 \text{ mm} \] Step 4: Final Answer:
The width of the fringe becomes 0.6 mm.