Problem 45 - Master Resource Book in JEE Main Physics

Question

In a double slit interference experiment, the distance between the slits is \(0.05 \mathrm{~cm}\) and screen is \(2 \mathrm{~m}\) away from the slits. The wavelength of light is \(6000 \AA\). The distance between the fringes is (a) \(0.24 \mathrm{~cm}\) (b) \(0.12 \mathrm{~cm}\) (c) \(1.24 \mathrm{~cm}\) (d) \(2.28 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The distance between the fringes is (a) 0.24 cm.

Step 1: Understand the Formula

In a double slit experiment, the distance between the fringes, also known as fringe spacing or fringe width, is given by the formula: \( \beta = \frac{\lambda L}{d} \), where \( \lambda \) is the wavelength of light, \( L \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

Step 2: Convert Units

The wavelength of light is given as \(6000 \ \text{Å} = 6000 \times 10^{-10} \ \text{m} = 6 \times 10^{-7} \ \text{m} \). The distance between the slits \( d = 0.05 \ \text{cm} = 0.0005 \ \text{m} \). The distance to the screen \( L = 2 \ \text{m} \).

Step 3: Substitute Values into Formula

Substitute the converted values into the formula: \[ \beta = \frac{6 \times 10^{-7} \times 2}{0.0005} \].

Step 4: Calculate Fringe Spacing

Calculate the value: \( \beta = \frac{12 \times 10^{-7}}{5 \times 10^{-4}} = 2.4 \times 10^{-3} \ \text{m} \). Convert this value from meters to centimeters: \( 2.4 \times 10^{-3} \times 100 = 0.24 \ \text{cm} \).

Step 5: Choose the Correct Answer

Compare the calculated fringe spacing of \( 0.24 \ \text{cm} \) to the given options and select the correct one. Option (a) \( 0.24 \ \text{cm} \) is the right answer.

Key Concepts

Interference Pattern

The double slit experiment is a classic demonstration of light's wave properties, particularly through what we call an **Interference Pattern**. This occurs when waves overlap and combine, creating regions of constructive and destructive interference. Imagine you have two sets of water waves that meet; where the peaks of one wave align with the peaks of the other, you have constructive interference, resulting in higher amplitude waves. Conversely, when the peak of one wave aligns with the trough of another, you get destructive interference, effectively cancelling each other out.

In a double slit experiment with light, you'll see a pattern of bright and dark fringes on a screen. This results from the different paths light waves take from the two slits to a particular point on the screen, causing them to interfere. Bright spots, or fringes, appear where the light waves are constructively interfering, and dark spots occur where they destructively interfere.

This pattern not only provides insight into the wave nature of light but also allows us to measure and calculate physical quantities like wavelengths and fringe spacing.

Fringe Spacing

The distance between consecutive bright fringes or dark fringes in an interference pattern is known as **Fringe Spacing**. In the context of the double slit experiment, this measurement is crucial as it helps reveal the properties of the light being used. The formula for fringe spacing is given by: \[ \beta = \frac{\lambda L}{d} \] where:

• \( \beta \) is the fringe spacing.
• \( \lambda \) is the wavelength of the light.
• \( L \) is the distance from the slits to the screen.
• \( d \) is the distance between the slits.

The formula illustrates how fringe spacing increases with the wavelength and the screen distance but decreases with greater slit separation. In our example, given the light wavelength of 6000 Angstroms, screen distance of 2 meters, and a slit separation of 0.05 cm, calculating the fringe spacing provides crucial insight into the interaction of light waves.

Wavelength Calculation

When engaged in experiments such as the double slit, knowing or deducing the **Wavelength** of light is essential. The wavelength (\( \lambda \)) of a wave is the distance over which the wave's shape repeats, playing a critical role in determining wave properties and interference patterns.

In our sample problem, the wavelength was given in Angstroms—a common unit for light wavelength. First, it's necessary to convert this into meters for compatibility with the formula for fringe spacing. We do this as follows: \[ 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} \]This conversion is crucial as using compatible units ensures accuracy in calculations.

Accurate wavelength measurements allow the calculation of fringe spacing and contrast, aiding in both classroom experiments and professional scientific research. Understanding these conversions and calculations provides a solid foundation in wave optics and the principles at play in even the simplest light experiments.