Question

The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is:

• A. 4
• B. 8
• C. 6
• D. 5

Hint:

In Young's double-slit experiment, the condition for the first coincidence of fringes of two different wavelengths can be determined by solving the equation \( m_1 \lambda_1 = m_2 \lambda_2 \), where \( m_1 \) and \( m_2 \) are the number of bright fringes.

Answer 1

The Correct Option is D

To solve this problem, we need to determine the least number of bright fringes of 480 nm light that will coincide with a bright fringe of 600 nm light in a Young's double slit interference pattern.

The condition for bright fringes in a double slit experiment is given by the formula:

\(d \sin \theta = n \lambda\) 

where:

• \(d\) is the distance between the slits,
• \(\theta\) is the angle of diffraction,
• \(n\) is the order of the fringe,
• \(\lambda\) is the wavelength of the light.

To find the coincidence of bright fringes, the path difference for both wavelengths should be the same. That is:

\(n_1 \lambda_1 = n_2 \lambda_2\)

Given:

• \(\lambda_1 = 480 \text{ nm}\)
• \(\lambda_2 = 600 \text{ nm}\)

Let's find the smallest \(n_1\) for which a coincidence occurs:

\(n_1 \cdot 480 = n_2 \cdot 600\)

Rearranging gives:

\(\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}\)

This implies \(n_1 = 5k\) and \(n_2 = 4k\) where \(k\) is an integer. The smallest integer \(k\) that satisfies this is \(k = 1\).

Substituting back gives \(n_1 = 5 \times 1 = 5\).

Hence, the least number of bright fringes of 480 nm light required to coincide with a bright fringe of 600 nm light is 5.

Therefore, the correct answer is: 5

Answer 2

Given:

• The wavelengths of the light are 480 nm and 600 nm.
• We are asked to find the least number of bright fringes of 480 nm light required for the first coincidence with the bright fringes formed by 600 nm light.

Step 1: Understanding the interference pattern

In Young's double-slit experiment, the position of the bright fringes is given by the formula: \[ y_m = \frac{m \lambda D}{d}, \] where: - \( y_m \) is the position of the \( m \)-th bright fringe, - \( m \) is the fringe number (an integer), - \( \lambda \) is the wavelength of light, - \( D \) is the distance from the slits to the screen, - \( d \) is the distance between the slits. For the first coincidence of bright fringes formed by two different wavelengths (\( \lambda_1 = 480 \, \text{nm} \) and \( \lambda_2 = 600 \, \text{nm} \)), the condition is that the fringe positions for both wavelengths must coincide. This means the positions of the bright fringes formed by both wavelengths should be equal. \[ y_m(\lambda_1) = y_n(\lambda_2), \] where \( m \) and \( n \) are the fringe numbers for the two wavelengths.

Step 2: Setting up the condition for first coincidence

For the first coincidence, the least number of bright fringes of 480 nm light required to coincide with the bright fringes of 600 nm light occurs when the following condition holds: \[ m \lambda_1 = n \lambda_2. \] Substituting the values for \( \lambda_1 = 480 \, \text{nm} \) and \( \lambda_2 = 600 \, \text{nm} \): \[ m \times 480 = n \times 600. \] Simplifying: \[ \frac{m}{n} = \frac{600}{480} = \frac{5}{4}. \] Therefore, \( m = 5 \) and \( n = 4 \) is the first integer solution where the fringes coincide.

Final Answer:

The least number of bright fringes of 480 nm light required for the first coincidence with the bright fringes formed by 600 nm light is \( \boxed{5} \).