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 determine the minimum number of 480 nm bright fringes that align with a bright fringe of 600 nm light in a Young's double-slit experiment, we use the condition for bright fringes: \(d \sin \theta = n \lambda\). Here, \(d\) is the slit separation, \(\theta\) is the diffraction angle, \(n\) is the fringe order, and \(\lambda\) is the wavelength.

For fringe coincidence, the path difference must be equal for both wavelengths:

\(n_1 \lambda_1 = n_2 \lambda_2\)

Given wavelengths:

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

We seek the smallest \(n_1\) for fringe coincidence:

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

Rearranging the equation gives the ratio of fringe orders:

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

This implies \(n_1 = 5k\) and \(n_2 = 4k\), where \(k\) is an integer. The smallest positive integer value for \(k\) is 1.

Substituting \(k = 1\) yields \(n_1 = 5 \times 1 = 5\).

Therefore, the least number of 480 nm bright fringes that coincide with a 600 nm bright fringe is 5.

The correct answer is: 5