Question

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4 \mu \text{m}$. If the experiment is carried out in another medium having refractive index 1.2, the fringe width will be _______ $\mu \text{m}$.

• A. 1.2
• B. 2
• C. 2.4
• D. 2.88

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The fringe width in YDSE depends on the wavelength of light. When the medium changes, the speed of light and therefore its wavelength changes, while frequency remains constant.
Step 2: Key Formula or Approach:
1. Fringe width $\beta = \frac{\lambda D}{d}$.
2. Wavelength in medium $\lambda_{med} = \frac{\lambda_{air}}{\mu}$.
Step 3: Detailed Explanation:
Given $\beta_{air} = 2.4 \mu \text{m}$ and $\mu = 1.2$.
The new fringe width $\beta_{med}$ is:
\[ \beta_{med} = \frac{\lambda_{med} D}{d} = \frac{(\lambda_{air}/\mu) D}{d} = \frac{1}{\mu} \left( \frac{\lambda_{air} D}{d} \right) \]
\[ \beta_{med} = \frac{\beta_{air}}{\mu} \]
\[ \beta_{med} = \frac{2.4}{1.2} = 2 \mu \text{m} \]
Step 4: Final Answer:
The fringe width in the medium is 2 $\mu$m.