Problem 81 - Master Resource Book in JEE Main Physics

Question

A plane wave of wavelength \(6250 \mathrm{~A}\) is incident normally on a slit of width \(2 \times 10^{-2} \mathrm{~cm}\). The width of the principal maximum on a screen distant \(50 \mathrm{~cm}\) will be (a) \(312.5 \times 10^{-3} \mathrm{~cm}\) (b) \(312.5 \times 10^{-4} \mathrm{~cm}\) (c) \(312 \mathrm{~cm}\) (d) \(312.5 \times 10^{-5} \mathrm{~cm}\)

Step-by-Step Solution

Answer

(a) 312.5 \times 10^{-3} \mathrm{~cm}

Step 1: Convert Units

First, we need to ensure all measurements are in consistent units. The wavelength of the plane wave is given as \(6250 \mathrm{~A}\), which can be converted to centimeters: \(1 \mathrm{~A}=10^{-8} \mathrm{~cm}\). So, \(6250 \mathrm{~A}=6250 \times 10^{-8} \mathrm{~cm}=6.25 \times 10^{-5} \mathrm{~cm}\).

Step 2: Calculate Angular Width of Principal Maximum

The angular width of the principal maximum in a diffraction pattern is given by \( \theta = \frac{\lambda}{a} \) for single-slit diffraction, where \( \lambda \) is the wavelength and \( a \) is the slit width. Substitute \( \lambda = 6.25 \times 10^{-5} \mathrm{~cm}\) and \( a = 2 \times 10^{-2} \mathrm{~cm}\) into the formula, \( \theta = \frac{6.25 \times 10^{-5}}{2 \times 10^{-2}} = 3.125 \times 10^{-3} \mathrm{~radians}\).

Step 3: Find Width of the Principal Maximum

The width \( W \) of the principal maximum on the screen can be found using the formula \( W = 2 L \theta \), where \( L = 50 \mathrm{~cm} \) is the distance from the slit to the screen. Substitute \( L = 50 \mathrm{~cm} \) and \( \theta = 3.125 \times 10^{-3} \mathrm{~radians} \) into the formula, \( W = 2 \cdot 50 \cdot 3.125 \times 10^{-3} = 0.3125 \mathrm{~cm} \). Convert this to \(312.5 \times 10^{-3} \mathrm{~cm}\) for direct comparison with the given options.

Key Concepts

Wavelength Conversion

Wavelength conversion is an essential step when solving problems in physics, especially when dealing with different unit systems. In this specific exercise, the wavelength of a plane wave is provided in angstroms (Å). It is crucial to convert this measurement into a consistent unit before proceeding to calculations. Here, 1 angstrom (\(\mathrm{~A}\)) is equal to \(10^{-8} \mathrm{~cm}\). Thus, converting \(6250\mathrm{~A}\) becomes a simple multiplication:

• Multiply \(6250 \mathrm{~A}\) by \(10^{-8} \mathrm{~cm/A}\)
• Result: \(6250 \times 10^{-8} = 6.25 \times 10^{-5} \mathrm{~cm}\)

This conversion is crucial for ensuring the units align with other given dimensions in centimeters and simplifies further calculations involving wavelength in the diffraction formulas.

Angular Width

The angular width of the principal maximum is a fundamental concept in single-slit diffraction. It refers to the angle in which the maximum intensity occurs on a screen due to diffraction through the slit. Calculating this requires understanding the relationship between the wavelength (\(\lambda\)) and the slit width (\(a\)). The formula is:

• Angular width: \(\theta = \frac{\lambda}{a}\)

With \(\lambda = 6.25 \times 10^{-5} \mathrm{~cm}\) and \(a = 2 \times 10^{-2} \mathrm{~cm}\), you substitute these values into the formula:

• \(\theta = \frac{6.25 \times 10^{-5}}{2 \times 10^{-2}} = 3.125 \times 10^{-3} \mathrm{~radians}\)

The angular width indicates the spread of the light as it passes through the slit, affecting how clearly the principal maximum appears on the screen.

Principal Maximum

The principal maximum in a diffraction pattern is the brightest fringe, where constructive interference is at its peak. In single-slit diffraction, understanding how to find the width of this maximum on a screen is vital for interpreting diffraction patterns. The width (\(W\)) of the principal maximum can be calculated using the formula:

• Width: \(W = 2L\theta\)

Where \(L\) is the distance from the slit to the screen, and \(\theta\) is the angular width calculated earlier. In this exercise, \(L\) is \(50 \mathrm{~cm}\):

• Substitute: \(W = 2 \times 50 \mathrm{~cm} \times 3.125 \times 10^{-3}\)
• Result: \(W = 0.3125 \mathrm{~cm}\), or equivalently \(312.5 \times 10^{-3} \mathrm{~cm}\)

This calculation shows the width of the principal maximum on the screen, helping us understand the effect of diffraction through the slit size and distance from the screen on the visibility of this maximum.