Question

Interference fringes produced by a double slit arrangement using a monochromatic light of wave length 5,890Å have an angular fringe width \( 0.28^\circ \). If the entire arrangement is immersed in water, the new angular fringe width will be (\( ^a\mu_w = 4/3 \)):

• A. \( 0.24^\circ \)
• B. \( 0.21^\circ \)
• C. \( 0.18^\circ \)
• D. \( 0.36^\circ \)

Hint:

Immersing a YDSE setup in a medium increases the effective refractive index, causing the fringe width to decrease by a factor of \( \mu \).

Answer 1

The Correct Option is B

Step 1: Formula for angular fringe width.
In a double-slit setup the angular fringe width is $\beta_\theta = \dfrac{\lambda}{d}$, where $\lambda$ is the wavelength of light and $d$ the slit separation.
Step 2: What immersion changes.
Dipping the whole setup in water of index $\mu$ leaves $d$ unchanged but shortens the wavelength to $\lambda' = \dfrac{\lambda}{\mu}$.
Step 3: New fringe width.
So the new angular width is \[ \beta_\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\beta_\theta}{\mu}. \]
Step 4: The fringe width simply scales down by $\mu$.
$\beta_\theta' = \dfrac{\beta_\theta}{\mu}$. The wavelength used to state the original width does not need to be plugged in.
Step 5: Put in the numbers.
With $\beta_\theta = 0.28^\circ$ and $\mu = \dfrac{4}{3}$, \[ \beta_\theta' = \frac{0.28}{4/3} = 0.28 \times \frac{3}{4}. \]
Step 6: Evaluate.
$0.28 \times 0.75 = 0.21^\circ$.
\[ \boxed{0.21^\circ} \]