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 ($\beta$) is directly proportional to the wavelength ($\lambda$) of the light used, assuming the distance between slits ($d$) and the distance to the screen ($D$) remain constant.
Key Formula or Approach:
Fringe width formula: \( \beta = \frac{\lambda D}{d} \).
Since $D$ and $d$ are constant, \( \frac{\beta_1}{\lambda_1} = \frac{\beta_2}{\lambda_2} \).

Step 2: Detailed Explanation:
Given:
Initial fringe width $\beta_1 = 0.5 \text{ mm}$.
Initial wavelength $\lambda_1 = 500 \text{ nm}$.
Final wavelength $\lambda_2 = 600 \text{ nm}$.
Using the proportionality:
\[ \beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} \]
\[ \beta_2 = 0.5 \times \frac{600}{500} \]
\[ \beta_2 = 0.5 \times 1.2 \]
\[ \beta_2 = 0.6 \text{ mm} \]

Step 3: Final Answer:
The new fringe width is $0.6 \text{ mm}$.