Problem 65 - Master Resource Book in JEE Main Physics

Question

In Young's double slit experiment, distance between source is \(1 \mathrm{~mm}\) and distance between the screen and source is \(1 \mathrm{~m}\). If the fringe width on the screen is \(0.06 \mathrm{~cm}\), then \(\lambda\) is (a) \(6000 \dot{A}\) (b) \(4000 \hat{A}\) (c) \(1200 \mathrm{~A}\) (d) \(2400 \mathrm{~A}\)

Step-by-Step Solution

Answer

The wavelength \(\lambda\) is \(6000 \dot{A}\) (option a).

Step 1: Identify Given Values

First, identify the given values in the problem. We know: \(d = 1 \text{ mm} = 0.1 \text{ cm}\), \(D = 1 \text{ m} = 100 \text{ cm}\), and the fringe width \(\beta = 0.06 \text{ cm}\).

Step 2: Understand the Fringe Width Formula

In Young's double slit experiment, the fringe width \( \beta \) is given by the formula \( \beta = \frac{\lambda D}{d} \). Here, \(\lambda\) is the wavelength, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits.

Step 3: Rearrange the Formula for Wavelength

Rearrange the fringe width formula to solve for \(\lambda\): \( \lambda = \frac{\beta d}{D} \).

Step 4: Substitute Values into Formula

Substitute the known values into the formula: \( \lambda = \frac{0.06 \times 0.1}{100} \).

Step 5: Calculate the Wavelength

Perform the calculation: \( \lambda = \frac{0.006}{100} = 6 \times 10^{-5} \text{ cm} \). Convert to angstroms: \(\lambda = 6000 \dot{A}\).

Step 6: Select the Correct Option

The calculated wavelength is \(6000 \dot{A}\), corresponding to option (a).

Key Concepts

Fringe Width Calculation

Young's double slit experiment is a fundamental method in optics for observing wave properties of light. One of the key calculations involved is determining the fringe width, which refers to the spacing between adjacent bright or dark fringes on the screen. In this experiment, the fringe width, denoted by \( \beta \), can be calculated using the formula:

• \( \beta = \frac{\lambda D}{d} \)

where \( \lambda \) is the wavelength of the light used, \( D \) is the distance from the slits to the screen, and \( d \) is the separation between the slits.
The fringe width is crucial because it represents how much the light spreads as it passes through the slits and creates an interference pattern on the screen.
Understanding the relationship between these parameters helps to predict and measure the behavior of waves under different conditions. In the exercise, we are given the fringe width as 0.06 cm, the slit separation \( d = 0.1 \text{ cm} \), and the distance to the screen \( D = 100 \text{ cm} \). These help calculate the wavelength, which demonstrates how light interacts in this classic physics setup.

Wavelength Determination

Determining the wavelength of light using Young’s double slit experiment is a key outcome of the procedure. Once the fringe width \( \beta \) is known, the wavelength \( \lambda \) can be calculated from rearranging the formula for fringe width:

• \( \lambda = \frac{\beta d}{D} \)

This equation shows that the wavelength is directly proportional to the fringe width and the separation between the slits, while being inversely proportional to the distance from the slits to the screen.
In the provided exercise, substituting the known values \( \beta = 0.06 \text{ cm} \), \( d = 0.1 \text{ cm} \), and \( D = 100 \text{ cm} \) into the formula gives:

• \( \lambda = \frac{0.06 \times 0.1}{100} = 6 \times 10^{-5} \text{ cm} \)

Converting this result to Angstroms (\( 1 \text{ cm} = 10^8 \text{ Angstroms} \)) yields \( \lambda = 6000 \dot{A} \), confirming option (a) as the correct answer.
This method highlights how controlled experimental setups allow physicists to measure fundamental properties like wavelength, which are linked to the color and energy of the light.

Optics in Physics

Optics is a significant area of study in physics, focused on the behavior and properties of light. Young’s double slit experiment is one of the cornerstone experiments in optics because it visually demonstrates the wave nature of light.
The experiment involves light passing through two narrow slits, causing interference that results in a pattern of bright and dark bands on a screen. Each bright fringe represents areas of constructive interference where waves add together, while dark fringes correspond to destructive interference where waves cancel out.

• Constructive interference: Waves in phase increase intensity.
• Destructive interference: Waves out of phase decrease intensity.

The setup is not only used to measure wavelengths but also to explore other wave phenomena such as the effects of changing slit widths, distances, and light sources.
In a broader context, these principles are applied in technology and various scientific fields, from designing lenses and microscopes to understanding complex wave interactions in quantum mechanics. Hence, the study of optics through such experiments lays critical foundational knowledge for both technological innovations and theoretical advancements.