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

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: $d = 0.20 \, \text{mm}$