Question

In young's double slit experiment, phase difference between light waves reaching 3rd bright fringe from central fringe with, is \((\lambda=5000 \AA)\) (A) zero (B) \(2 \pi\) (C) \(4 \pi\) (D) \(6 \pi\)

Short answer

The phase difference between the light waves reaching the 3rd bright fringe from the central fringe in Young's double slit experiment with a given wavelength of \(\lambda = 5000 \AA\) can be calculated using the relationship \(\delta = \frac{2\pi}{\lambda} \cdot \text{path difference}\). For the 3rd bright fringe, the path difference is \(15 \times 10^{-7}\) m. Substituting the given values, we get the phase difference as \(\delta = 6 \pi\). Therefore, the correct answer is (D) \(6 \pi\).

Step-by-step solution

Recall the condition for a bright fringe

In Young's double slit experiment, a bright fringe is formed when the constructive interference occurs due to equal path difference, which is equal to an integral multiple of the wavelength: \[n\lambda = d\sin\theta\] Where \(n\) is the order of the bright fringe, \(\lambda\) is the wavelength, \(d\) is the distance between slits and \(\theta\) is the angle between the central fringe and the observed fringe.

Calculate the path difference

The path difference for the 3rd bright fringe can be calculated using the formula: \[n\lambda = d\sin\theta\] Here, we are given the wavelength \(\lambda = 5000 \AA = 5 \times 10^{-7} m\), and \(n = 3\) (since it's the 3rd bright fringe). The path difference is thus: \[n\lambda = 3(5 \times 10^{-7}) = 15 \times 10^{-7} m\]

Calculate the phase difference

To calculate the phase difference \(\delta\), recall the relationship between path difference and phase difference: \[\delta = \frac{2\pi}{\lambda} \cdot \text{path difference}\] Here, the path difference is \(15 \times 10^{-7}\) m and \(\lambda = 5\times 10^{-7}\) m. Substituting the values, we get: \[\delta = \frac{2\pi}{5 \times 10^{-7}} \cdot (15 \times 10^{-7})\]

Simplify the expression

Now, simplify the expression for the phase difference: \[\delta = 2\pi \cdot \frac{15 \times 10^{-7}}{5 \times 10^{-7}}\] Cancel the common factors: \[\delta = 2\pi \cdot 3\] Hence, the phase difference is: \[\delta = 6 \pi\] The correct answer is (D) \(6 \pi\).