Problem 79 - Master Resource Book in JEE Main Physics

Question

Plane microwaves are incident on a long slit having a width of \(5 \mathrm{~cm}\). The wavelength of the microwaves if the first minimum is formed at \(30^{\circ}\) is (a) \(2.5 \mathrm{~cm}\) (b) \(2 \mathrm{~cm}\) (c) \(25 \mathrm{~cm}\) (d) \(2 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The wavelength of the microwaves is \(2.5 \mathrm{~cm}\), which is option (a).

Step 1: Understanding the Condition for Minimum Diffraction

For a single slit diffraction, the first minimum occurs when \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the first minimum, \( m \) is the order of the minimum, and \( \lambda \) is the wavelength. For the first minimum, \( m = 1 \).

Step 2: Substitute Known Values

We know \( \theta = 30^{\circ} \), and \( a = 5 \mathrm{~cm} \). Substitute these values into the formula: \( 5 \sin(30^{\circ}) = 1 \times \lambda \).

Step 3: Calculate \( \sin(30^{\circ}) \)

\( \sin(30^{\circ}) = 0.5 \). Substitute it back into the equation: \( 5 \times 0.5 = \lambda \).

Step 4: Solve for Wavelength \( \lambda \)

Calculate \( \lambda \) : \( 5 \times 0.5 = \lambda \) gives \( \lambda = 2.5 \mathrm{~cm} \).

Step 5: Select the Correct Option

The calculated wavelength \( \lambda = 2.5 \mathrm{~cm} \) matches option (a).

Key Concepts

Single Slit Diffraction

In the fascinating world of wave physics, diffraction refers to the bending of waves around obstacles or through small openings. When it comes to single slit diffraction, the wave, such as light or microwaves, encounters a narrow slit, resulting in a pattern of alternating dark and bright spots on a screen. This phenomenon is due to the interference of waves going through different parts of the slit.
For a single slit, the condition for observing the first minimum (or dark spot) in a diffraction pattern is expressed as \( a \sin \theta = m \lambda \), where:

• \( a \) is the slit width.
• \( \theta \) is the angle of the first diffraction minimum.
• \( m \) is the order of the minimum, with \( m = 1 \) for the first minimum.
• \( \lambda \) is the wavelength of the wave, like light or microwaves.

The reasoning is that portions of the wave that pass through the slit interfere destructively at this angle, canceling each other out and creating a point of low intensity or darkness.

Wavelength Calculation

Wavelength calculation is a crucial aspect of understanding wave phenomena, especially in diffraction. In the context of this problem, calculating the wavelength of microwaves involves using the formula derived from the condition for minimum diffraction. We have:
\[ \lambda = \frac{a \sin \theta}{m} \]
For our exercise, we set \( a = 5 \mathrm{~cm} \) and \( \theta = 30^{\circ} \) with \( m = 1 \). Therefore, the equation becomes:
\[ 5 \times \sin(30^{\circ}) = \lambda \]
Since \( \sin(30^{\circ}) = 0.5 \), our calculation simplifies to:
\[ \lambda = 5 \times 0.5 = 2.5 \mathrm{~cm} \]
Through this straightforward approach, we find that the wavelength is \( 2.5 \mathrm{~cm} \), which is one of the potential answers in the given options.

Microwave Physics

Microwave physics delves into the fascinating features of microwaves, a type of electromagnetic wave primarily used for communication, radar, and cooking!
Microwaves have wavelengths in the range of 1 mm to 1 meter, corresponding to frequencies from about 300 MHz (1 meter) to 300 GHz (1 mm). This makes them quite suitable for various practical applications. However, in the context of physics and experiments, they help us understand wave behavior, such as diffraction through slits.
The concepts of diffraction and wavelength, discussed in our exercise, are central to microwave applications. Diffraction tells us how microwaves can be manipulated, for instance, how they can bend around obstructions. Understanding the wavelength allows us to design systems that efficiently transmit or receive these waves.
Looking beyond everyday applications, microwave physics extends into areas like studying the cosmic microwave background, offering insights into the early universe.