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) \(6 \pi\).

Key concepts

Constructive Interference

Constructive interference is the phenomenon where two or more waves superimpose to produce a wave of greater amplitude. In Young's double slit experiment, this occurs when the path difference between the two waves from the slits is a whole number multiple of the wavelength.

This means the waves meet in phase and their crests and troughs align perfectly. As a result, the light intensity at this point is maximized, creating what's known as a bright fringe on the interference pattern. The condition for constructive interference can be expressed mathematically as:

• \[ n\lambda = d\sin\theta \]

Where \(n\) is the order of the fringe, \(\lambda\) is the wavelength of the light, and \(\theta\) is the angle of the fringe from the central maximum. This is why understanding constructive interference is crucial in comprehending the formation of bright fringes on the screen.

Path Difference

The path difference in the context of Young's double slit experiment is the difference in the distances two light waves travel from the two slits to a point on the screen. This difference directly influences whether we see constructive or destructive interference.

For constructive interference and a bright fringe to appear, the path difference must be an integer multiple of the wavelength, as shown in the equation:

• \[ n\lambda = d\sin\theta \]

This equation tells us that if the path difference is equal to \(n\lambda\), waves will arrive in phase and constructive interference will occur, leading to a bright spot on the screen. For the third bright fringe in the given problem, this path difference becomes \(3\lambda\), ensuring that the light waves from the slits combined perfectly at that point.

Phase Difference

Phase difference refers to the shift between the peaks and troughs of two waves arriving at a point. In Young's double slit experiment, it is essential to understand because it determines the type of interference observed.

The phase difference \(\delta\) is calculated using the formula:

• \[ \delta = \frac{2\pi}{\lambda} \cdot \text{path difference} \]

This relationship outlines that the phase difference is proportional to the path difference. For a bright fringe, the phase difference must be an even multiple of \(\pi\) (like \(2\pi, 4\pi, 6\pi\), etc.).

In our problem, the phase difference for the third bright fringe results in \(6\pi\), confirming that waves reach in constructive interference, forming a bright fringe.

Bright Fringe

A bright fringe is an area on the screen in Young's double slit experiment where constructive interference occurs, resulting in increased light intensity. Every bright fringe corresponds to a point where the path difference between waves from the slits is an integer multiple of the wavelength.

This constructive interference results in the light waves aligning perfectly, enhancing brightness. It's why understanding bright fringes helps in analyzing the interference pattern of light.

• For example:
  • The central bright fringe is the result of zero path difference.
  • Subsequent bright fringes occur at path differences of \(\lambda, 2\lambda, 3\lambda\), and so on.

The appearance of these bright fringes can be predicted using the formula \[n\lambda = d\sin\theta\], putting into practical use the mathematical relationship between path difference, wavelength, and angle.