Problem 82 - Master Resource Book in JEE Main Physics

Question

Light of wavelength \(6000 \AA\) is incident on a single slit. The first minimum of the diffraction pattern is obtained at \(4 \mathrm{~mm}\) from the centre. The screen is at a distance of \(2 \mathrm{~m}\) from the slit. The slit width will be (a) \(0.3 \mathrm{~mm}\) (b) \(0.2 \mathrm{~mm}\) (c) \(0.15 \mathrm{~mm}\) (d) \(0.1 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The slit width is \( 0.3 \text{ mm} \) (option a).

Step 1: Understand the Problem

We have light with a wavelength of \( \lambda = 6000 \text{ Å} \) (where 1 Å = \( 10^{-10} \text{ m} \)) incident on a single slit. The diffraction pattern's first minimum (dark spot) occurs at a position 4 mm from the center on a screen 2 m away. We need to determine the slit width, \( a \).

Step 2: Apply the Diffraction Formula

For a single slit, the position of the first minimum is given by the formula \( a \sin \theta = m \lambda \), where \( m = \pm 1 \) for the first minimum. Here, \( \theta \) is the angle between the central maximum and the first minimum.

Step 3: Use Small Angle Approximation

Since the angle \( \theta \) is small, we can use the small angle approximation \( \sin \theta \approx \tan \theta = \frac{y}{L} \), where \( y = 4 \text{ mm} = 4 \times 10^{-3} \text{ m} \) is the distance from the center to the first minimum, and \( L = 2 \text{ m} \) is the distance to the screen.

Step 4: Express the Slit Width

Substitute the small angle approximation into the diffraction formula: \( a \times \frac{y}{L} = \lambda \). Rearrange for the slit width: \( a = \frac{\lambda L}{y} \).

Step 5: Calculate the Slit Width

Convert the wavelength to meters: \( \lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} \). Now substitute the values into the expression: \( a = \frac{(6 \times 10^{-7}) \times 2}{4 \times 10^{-3}} \).

Step 6: Simplification and Conclusion

Calculate \( a = \frac{12 \times 10^{-7}}{4 \times 10^{-3}} = 3 \times 10^{-4} \text{ m} \), which is equivalent to \( 0.3 \text{ mm} \). Thus, the slit width is \( 0.3 \text{ mm} \).

Key Concepts

Diffraction Pattern

When light passes through a single slit, it bends around the edges, creating a diffraction pattern. This pattern consists of alternating bands of light and dark. The bright central band is the most intense part of the pattern, known as the central maximum. On either side of this band, the intensity decreases, creating a series of dark and light areas.
These alternating bands result from the wave nature of light. As light waves pass through the slit, they interfere with each other. When waves meet in-phase, they produce bright bands (constructive interference). When they meet out-of-phase, they cancel each other out, creating dark bands (destructive interference).
In the context of the problem, the first dark band is referred to as the first minimum. This is the point where waves interfere completely destructively, and it occurs at a specific distance from the central maximum.

Wavelength

Wavelength, denoted as \( \lambda \), is a fundamental concept in understanding diffraction patterns. It represents the distance between two consecutive points that are in phase, such as crests or troughs, in a wave. In simple terms, it's the length of one complete wave cycle.
In the exercise, the wavelength is given as 6000 Å, which needs conversion for calculation purposes since Ångströms are not standard in most physics equations. We convert 6000 Å to meters as \( 6000 \times 10^{-10} \text{ m} \) or \( 6 \times 10^{-7} \text{ m} \).
This wavelength helps determine how the light diffracts through the slit. Wavelength and slit size together dictate the nature of the resulting diffraction pattern. Smaller wavelengths tend to produce narrower diffraction patterns, while larger wavelengths create more spread-out patterns.

Small Angle Approximation

The small angle approximation is a crucial tool in simplifying calculations involving diffraction. When the angle \( \theta \) involved in viewing or analyzing a diffraction pattern is small, which is typical in these cases, we can approximate \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \).
Here, \( y \) is the distance from the central maximum to the first minimum on the screen, and \( L \) is the distance from the slit to the screen. By using this approximation, we can sidestep the need for precise trigonometric calculations, making the math much simpler.
In the problem, \( y = 4 \times 10^{-3} \text{ m} \) and \( L = 2 \text{ m} \). This allows us to express the sine of the angle \( \theta \) as \( \sin \theta = \frac{y}{L} \), and use it in the diffraction formula \( a \sin \theta = \lambda \) to solve for the slit width \( a \). This makes it easier to find the slit width without needing advanced mathematical methods.