Problem 60 - Master Resource Book in JEE Main Physics

Question

In Young's double slit experiment, 12 fringes are obtained in a certain segment of the screen when light of wavelength \(600 \mathrm{~nm}\), is used. If the wavelength of light is changed to \(400 \mathrm{~nm}\), number of fringes observed in the same segment of the screen is given by (a) 12 (b) 18 (c) 24 (d) 30

Step-by-Step Solution

Answer

The number of fringes is 18 (option b).

Step 1: Understand the Concept

In Young's double-slit experiment, the number of fringes on the screen depends on the wavelength of light. The formula for the number of fringes is derived from the path difference condition for constructive interference, which leads to the appearance of bright fringes.

Step 2: Set the Initial Condition

With a wavelength \( \lambda_1 = 600 \, \mathrm{nm} \), the experimental setup yields 12 fringes. This implies that for this wavelength, the path difference condition allows for 12 constructive interferences within the considered segment of the screen.

Step 3: Calculate Ratio of Wavelengths

Find the ratio of the two wavelengths. Here, \[ \text{Ratio} = \frac{\lambda_2}{\lambda_1} = \frac{400 \, \mathrm{nm}}{600 \, \mathrm{nm}} = \frac{2}{3} \]

Step 4: Determine the New Number of Fringes

The number of fringes is inversely proportional to the wavelength of light. Therefore, if the wavelength is reduced, the number of fringes increases proportionally. Given that the number of fringes for \( \lambda_1 \) is 12, calculate the new number of fringes: \[ \text{New Number of Fringes} = \frac{\lambda_1}{\lambda_2} \times \text{Old Number of Fringes} = \frac{600}{400} \times 12 = \frac{3}{2} \times 12 = 18\]

Step 5: Choose the Correct Answer

The calculated new number of fringes is 18, which corresponds to option (b). Therefore, the new number of fringes observed in the same segment when the wavelength is changed to \(400 \, \mathrm{nm}\) is 18.

Key Concepts

Interference Fringes

Young's double-slit experiment is a fascinating demonstration of the wave nature of light, giving rise to what we call interference fringes.
When light passes through two closely spaced slits, it splits into two sets of coherent waves. These waves overlap and interfere with each other on the screen behind the slits.
The points on the screen where the waves meet constructively (i.e., their crests and troughs overlap) appear as bright bands or fringes, known as interference fringes.

• These fringes result from constructive interference and are spaced at regular intervals.
• The pattern consists of alternating bright and dark bands, with the bright bands being the constructive interference points.

The number of these bright fringes observed along a segment of the screen depends on several factors, including the wavelength of the light used.

Wavelength Dependence

The wavelength of light directly impacts the appearance and number of interference fringes in Young's double-slit experiment.
This is because the wavelength determines how far apart the interference fringes are on the screen.
Changing the wavelength alters the distance between fringes and thus the number of fringes within a given segment.

• For longer wavelengths, the fringes spread out more, leading to fewer bands in a specific section.
• Conversely, shorter wavelengths result in more closely packed fringes, increasing the number of visible bands in the same area.

Essentially, there's an inverse relationship: as the wavelength decreases, the number of fringes increases and vice versa.

Constructive Interference

Constructive interference occurs when waves overlap in such a way that their amplitudes add up to produce brighter light, manifesting as the bright fringes in the interference pattern.
In Young's experiment, the path length difference between the two light waves reaching a point on the screen must be an integral multiple of the wavelength for constructive interference to occur.

• This results in the distinct bright bands that we're trying to understand and quantify.
• The condition is expressed mathematically as: \[ ext{Path difference} = m \cdot \lambda \] where \( m \) is an integer and \( \lambda \) is the wavelength.

When the path lengths differ by this prescribed amount, the waves reinforce each other, leading to the brightest possible points on the screen.

Path Difference

The concept of path difference is crucial to understanding why and where constructive interference leads to bright fringes in Young's double-slit experiment.
Path difference refers to the difference in distance traveled by two waves from the slits to a point on the screen. For the waves to interfere constructively, this path difference must align perfectly with the conditions of constructive interference.

• If the path difference is zero or an integer multiple of the wavelength, the crests and troughs of the waves align precisely, resulting in bright fringes.
• Conversely, when the path difference is an odd multiple of half-wavelengths, destructive interference occurs, causing dark fringes.

Understanding and calculating path difference is therefore key to predicting where the bright and dark stripes will occur on the interference pattern.