Problem 34 - Master Resource Book in JEE Main Physics

Question

In an experiment, the two slits are \(0.5 \mathrm{~mm}\) apart and the fringes are observed to \(100 \mathrm{~cm}\) from the plane of the slits. The distance of the 11 th bright fringe from the 1st bright fringe is \(9.72 \mathrm{~mm}\). The wavelength is (a) \(4.86 \times 10^{-5} \mathrm{~cm}\) (b) \(5.72 \times 10^{-4} \mathrm{~cm}\) (c) \(5.87 \times 10^{-4} \mathrm{~cm}\) (d) \(3.25 \times 10^{-4} \mathrm{~cm}\)

Step-by-Step Solution

Answer

The wavelength is \(4.86 \times 10^{-5} \text{ cm}\), which corresponds to option (a).

Step 1: Understanding the Problem

We are dealing with a double-slit experiment. We know the distance between the slits, the distance to the screen, and the separation between certain bright fringes. We need to determine the wavelength of the light used.

Step 2: Determine Fringe Separation Formula

In a double-slit experiment, the position of the bright fringes is given by the formula: \( y = \frac{m \cdot \lambda \cdot L}{d} \), where \(y\) is the fringe separation for the \(m\)th bright fringe, \(\lambda\) is the wavelength, \(L\) is the distance to the screen, and \(d\) is the slit separation. Here, \(m\) is the order of the fringe.

Step 3: Calculate Fringe Separation

The fringe separation \( \Delta y \) between the 11th bright fringe and the 1st bright fringe is given as \(9.72 \text{ mm}\). This corresponds to 10 fringe widths, so \( \Delta y = 10 \cdot y_1 \). Therefore, \( y_1 = \frac{9.72 \text{ mm}}{10} = 0.972 \text{ mm} \).

Step 4: Apply the Formula to Find Wavelength

Substitute the known values into the fringe separation formula to solve for \(\lambda\): \( y = \frac{\lambda \cdot L}{d} \) implies \( \lambda = \frac{y \cdot d}{L} \). Substituting \( y = 0.972 \times 10^{-3} \text{ m} \), \( d = 0.5 \times 10^{-3} \text{ m} \), and \( L = 1 \text{ m} \), we calculate: \( \lambda = \frac{0.972 \times 10^{-3} \cdot 0.5 \times 10^{-3}}{1} \).

Step 5: Calculate and Convert Wavelength

Calculate the wavelength: \( \lambda = 0.486 \times 10^{-6} \text{ m} \), which converted to cm is \( \lambda = 4.86 \times 10^{-5} \text{ cm} \).

Step 6: Identify the Correct Option

The calculated wavelength \( 4.86 \times 10^{-5} \text{ cm} \) corresponds to option (a).

Key Concepts

Interference Pattern

In the double-slit experiment, the interference pattern arises because light waves move through each slit and spread out, then overlap on a screen. This overlapping creates a pattern of bright and dark bands called fringes. When waves overlap constructively (their crests align), they form bright fringes, while destructive overlap (crests align with troughs) leads to dark fringes.
This phenomenon occurs because of the wave nature of light. The pattern is a direct demonstration of light behaving like a wave, with bright fringes representing areas where the waves are 'in phase'. The classic look of the interference pattern provides direct evidence for wave interference.
Understanding this pattern is key, as it forms the basis for conclusions about the light's properties like wavelength. The formula used to describe the positioning of these fringes is instrumental in story-telling the physics behind the scene, which is further elaborated when calculating fringe separation and light's wavelength.

Fringe Separation

Fringe separation is critical when determining properties like wavelength in a double-slit experiment. It is the distance between consecutive bright (or dark) fringes on the interference pattern seen on the screen. Calculating this separation involves understanding fringe positions using the formula:

• \( y = \frac{m \cdot \lambda \cdot L}{d} \)

where:

• \( y \) is the distance from the central maximum to the \( m^{th} \) fringe
• \( m \) is the fringe order
• \( \lambda \) is the wavelength of light
• \( L \) is the distance from the slits to the screen
• \( d \) is the separation between the slits

When the distance between multiple fringes is given, such as from the 1st to the 11th bright fringe, it reflects cumulative fringe widths. For example, in a given problem, 10 fringe widths separate 11 bright or dark fringes. Finding one fringe width then becomes straightforward through division. This step is necessary to prepare for any wavelength calculation.

Wavelength Calculation

The final step in analyzing the double-slit experiment involves calculating the light's wavelength, one of the most intriguing aspects of the experiment. Using the measure of fringe separation derived earlier and substituting it back into the formula for fringe positions, we can solve for the wavelength \( \lambda \).
Given:

• Fringe width \( y_1 = \frac{9.72 \text{ mm}}{10} = 0.972 \text{ mm} \)
• Slit separation \( d = 0.5 \text{ mm} \)
• Screen distance \( L = 100 \text{ cm} \)

Substitute these values into the equation:

• \( \lambda = \frac{y \cdot d}{L} \)

After conversion to compatible units (meters), the calculated wavelength \( \lambda \) comes out as \( 4.86 \times 10^{-5} \text{ cm} \). This result shows how precise measurements lead to discovering fundamental properties like wavelength, emphasizing the importance of data collection and calculation accuracy in physics experiments. It also reinforces understanding of the wave-particle duality of light.