Question

In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m. the minimum distance between the slits S1 and S2 is ______ mm.

Answer 1

Correct Answer: 0.2

In a double slit experiment, the condition for a dark fringe can be stated as:
d sin θ = (m + 0.5)λ.
Given that the dark fringe is observed at point P, we can approximate sin θ ≈ θ ≈ x/D, where x is the distance of the fringe from the center, and D is the distance to the screen. 

Using the given data:
Wavelength λ = 400 nm = 400 × 10^-9 m,
Distance D = 0.2 m.

The first dark fringe corresponds to m = 0, so:
d(0) + 0.5(400 × 10^-9) = d.

Since x is approximately D:
d = λ/2.

Calculating:
d = (400 × 10^-9)/2 = 200 × 10^-9 m = 0.2 mm.

The minimum distance between the slits is 0.2 mm, which falls within the range [0.2, 0.2] mm.

Answer 2

Step1. Path Difference Condition for Minima: - For a dark fringe (minima) at P, the path difference should be:

\[ 2\sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2} \]

- Rearrange:

\[ \sqrt{D^2 + d^2} - D = \frac{\lambda}{4} \]

- Therefore:

\[ \sqrt{D^2 + d^2} = D + \frac{\lambda}{4} \]

Step 2. Square Both Sides:

\[ D^2 + d^2 = D^2 + D\lambda + \frac{\lambda^2}{16} \]

- Simplify to solve for d:

\[ d^2 = D\lambda + \frac{\lambda^2}{16} \]

Step 3. Substitute Given Values: - \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\), \(D = 0.2 \, \text{m}\)

\[ d^2 = 0.2 \times 400 \times 10^{-9} + \frac{(400 \times 10^{-9})^2}{16} \]

- Calculate:

\[ d^2 \approx 4 \times 10^{-8} \, \text{m}^2 \]

\[ d \approx 2 \times 10^{-4} \, \text{m} = 0.20 \, \text{mm} \]

So, the correct answer is: \(d = 0.20 \, \text{mm}\)