Question

In a double slit experiment, the two slits are 1.5 mm apart. The slits are illuminated by a mixture of lights of wavelengths of 600 nm and 400 nm and the interference pattern is observed on a screen 1.5 m away from the slits. Find the minimum distance of the point from the central maximum at which bright fringes of the interference patterns of the two wavelengths coincide.

Hint:

Bright fringes coincide when \(n_1\lambda_1 = n_2\lambda_2\); take \(n_1 = 2\), \(n_2 = 3\), then \(y = n_1\lambda_1 D/d = 1.2\) mm.

Answer 1

Step 1: Note the data:
Slit separation \(d = 1.5\ \text{mm} = 1.5\times10^{-3}\ \text{m}\), screen distance \(D = 1.5\ \text{m}\), \(\lambda_1 = 600\) nm and \(\lambda_2 = 400\) nm.

Step 2: Write the condition for coinciding bright fringes:
The \(n\)-th bright fringe of wavelength \(\lambda\) is at \(y = \dfrac{n\lambda D}{d}\).
The fringes coincide when \(n_1\lambda_1 = n_2\lambda_2\), that is \(\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{400}{600} = \dfrac{2}{3}\).

Step 3: Take the smallest whole numbers:
The smallest whole numbers in the ratio \(2:3\) are \(n_1 = 2\) and \(n_2 = 3\). This gives the nearest point to the centre where both bright fringes meet.

Step 4: Find the distance:
\[ y = \frac{n_1\lambda_1 D}{d} = \frac{2\times 600\times10^{-9}\times 1.5}{1.5\times10^{-3}} = 1.2\times10^{-3}\ \text{m} \]
Check with the other wavelength: \(\dfrac{3\times 400\times10^{-9}\times 1.5}{1.5\times10^{-3}} = 1.2\times10^{-3}\) m, the same value.

Final Answer:
The bright fringes of the two wavelengths first coincide at \[ \boxed{y = 1.2\ \text{mm}} \]