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:

For double-slit interference, fringes for different wavelengths will coincide when their fringe orders are harmonics of each other. The first such coincidence occurs when the multiples of their wavelengths align.

Answer 1

The condition for constructive interference (bright fringes) in a double-slit experiment is given by the formula: \[ y_m = \frac{m \lambda D}{d} \] where: - \( m \) is the fringe order (1, 2, 3,...), - \( \lambda \) is the wavelength of light, - \( D \) is the distance between the slits and the screen, - \( d \) is the distance between the slits. For the two wavelengths to produce coinciding fringes, their fringe positions should be equal. Hence, we set the condition for the bright fringe coincidence of both wavelengths \( \lambda_1 = 600 \ \text{nm} \) and \( \lambda_2 = 400 \ \text{nm} \). The fringe positions for each wavelength at the same order \( m \) are given by: \[ y_m = \frac{m \lambda_1 D}{d} = \frac{m \lambda_2 D}{d} \] The first coinciding fringe occurs when \( m_1 \lambda_1 = m_2 \lambda_2 \), i.e., the smallest common multiple of the wavelengths, which happens when \( m_1 = 2 \) and \( m_2 = 3 \). Now, using \( d = 1.5 \times 10^{-3} \ \text{m} \) and \( D = 1.5 \ \text{m} \), we substitute these values into the equation for \( m_1 = 2 \) and \( m_2 = 3 \): \[ y = \frac{(2) \cdot 600 \times 10^{-9} \cdot 1.5}{1.5 \times 10^{-3}} = 0.36 \ \text{m} \] Thus, the minimum distance from the central maximum where the bright fringes coincide is approximately \( 0.36 \ \text{m} \). Final Answer: The minimum distance is approximately \( \boxed{0.36 \ \text{m}} \).