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 topic of this question is Wave Optics, specifically interference in Young's Double Slit Experiment (YDSE).
We are requested to calculate how the fringe width changes when the system is moved into a medium with a higher refractive index.
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:
We know that the fringe width is directly proportional to the wavelength of light.
When light moves into a medium with refractive index $\mu$, its wavelength shrinks by a factor of $\mu$.
Therefore, the fringe width in the medium is:
\[ \beta_{med} = \frac{\beta_{air}}{\mu} \]
Substituting the given values:
\[ \beta_{med} = \frac{2.4 \ \mu\text{m}}{1.2} = 2 \ \mu\text{m} \]
Step 4: Final Answer:
The fringe width in the medium is 2 $\mu$m.