Question

A double slit experiment is immersed in water of refractive index 1.33. The slit separation is 1 mm and the distance between slit and screen is 1.33 m. The slits are illuminated by a light of wavelength \(6300 \, \text{Å}\). The fringe width is

• A. \(4.9 \times 10^{-4} \, \text{m}\)
• B. \(6.3 \times 10^{-4} \, \text{m}\)
• C. \(8.6 \times 10^{-4} \, \text{m}\)
• D. \(5.8 \times 10^{-4} \, \text{m}\)

Hint:

When the entire setup is immersed in a medium, the wavelength decreases by factor μ, but the geometry (d, D) remains the same. The fringe width β ∝ λ, so β becomes β₀/μ. Here β₀ = λ₀ D / d = 6.3e-7 * 1.33 / 1e-3 = 8.379e-4 m, divided by 1.33 gives 6.3e-4 m.

Answer 1

The Correct Option is B

Step 1: Note the data.
A Young's double slit setup is placed under water of refractive index $\mu = 1.33$. Slit separation $d = 1\ \text{mm} = 10^{-3}\ \text{m}$, screen distance $D = 1.33\ \text{m}$, and the air wavelength is $\lambda_0 = 6300\ \text{\AA} = 6.3 \times 10^{-7}\ \text{m}$. We want the fringe width in water.

Step 2: Adjust the wavelength for water.
Light slows in water and its wavelength shrinks to $\lambda = \frac{\lambda_0}{\mu}$.

Step 3: Write the fringe width.
Fringe width is $\beta = \frac{\lambda D}{d} = \frac{\lambda_0 D}{\mu d}$.

Step 4: Put in the numbers.
$\beta = \frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}$.

Step 5: Cancel the matching factors.
The $1.33$ on the top and bottom cancel neatly, leaving $\beta = \frac{6.3 \times 10^{-7}}{10^{-3}}$.

Step 6: Finish the division.
$\beta = 6.3 \times 10^{-4}\ \text{m}$.
\[ \boxed{\beta = 6.3 \times 10^{-4}\ \text{m}} \]