Problem 50 - Master Resource Book in JEE Main Physics

Question

Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance of \(0.14 \mathrm{~m}\). The frequency of transmitter is (a) \(1.5 \times 10^{10} \mathrm{~Hz}\) (b) \(10^{10} \mathrm{~Hz}\) (c) \(3 \times 10^{10} \mathrm{hz}\) (d) \(6 \times 10^{10} \mathrm{~Hz}\)

Step-by-Step Solution

Answer

The frequency is approximately \(1.5 \times 10^{10} \mathrm{~Hz}\).

Step 1: Understanding the Problem

When microwaves strike a plane reflector, interference patterns are formed. The distance traveled between 14 successive maxima will equal the wavelength as each successive maxima corresponds to a half-wavelength shift.

Step 2: Determining the Total Distance Between Successive Maxima

The distance between 14 successive maxima is given as 0.14 m. Each peak to next peak corresponds to half a wavelength, hence we have the ridge-to-ridge distance being half-wavelengths. Since 14 maxima involve 13 half-full wavelength cycles, this gives us 13/2 wavelengths = 0.14 m.

Step 3: Calculating the Wavelength

The total distance covered by 14 maxima is \(0.14\,\text{m}\). Since there are 13 segments that form half-wavelength per segment between these maxima, solve for \( \lambda \), the wavelength: \[\frac{13}{2}\lambda = 0.14\] So, \[\lambda = \frac{0.14}{6.5} = 0.0215\, \text{m}\].

Step 4: Using Speed of Light to Find Frequency

The speed of light \(c\) is approximately \(3 \times 10^8\, \text{m/s}\). The formula connecting speed, wavelength, and frequency \(u\) is \( c = u \cdot \lambda \). Solving for frequency gives: \[u = \frac{c}{\lambda} = \frac{3 \times 10^8}{0.0215} = 1.395 \times 10^{10}\, \text{Hz}\].

Step 5: Selecting the Closest Option

Comparing \(1.395 \times 10^{10} \mathrm{~Hz}\) to the given options, \(1.5 \times 10^{10} \mathrm{~Hz}\) is the closest frequency.

Key Concepts

Wavelength Calculation

When trying to determine the wavelength of microwaves, a key method involves observing interference patterns. When a detector is moved along the normal to a reflecting surface, it observes points, or maxima, where the waves constructively interfere due to being in phase. Each maximum occurs at a position where the path difference from the two source points is an integer multiple of the wavelength, causing constructive interference.

In the exercise example, you are given that the distance between 14 successive maxima is 0.14 meters. Since the maxima require a full wavelength path difference, and given there are 13 half-wavelength intervals (each interval consisting of one-half cycle), you can write the relationship for wavelength \( \frac{13}{2} \lambda = 0.14 \). Solving for \( \lambda \), the equation simplifies to: \[ \lambda = \frac{0.14}{6.5} \], resulting in a wavelength, \( \lambda \), of approximately 0.0215 m. Fund understanding of this calculation is crucial for analyzing wave behavior in other contexts.

Frequency Determination

Frequency determination of electromagnetic waves, such as microwaves, directly relates to their wavelength and speed. The basic formula to connect these elements is:\( c = u \cdot \lambda \), where \(c\) is the speed of light (approximately \(3 \times 10^8\, \text{m/s}\)), \(u\) is the frequency, and \(\lambda\) is the wavelength.

To find the frequency, rearrange the formula to solve for \(u\):\[ u = \frac{c}{\lambda} \]. Substituting in the known speed of light and the calculated wavelength \(0.0215\,\text{m}\), you have:\[ u = \frac{3 \times 10^8}{0.0215} \], which equals approximately \(1.395 \times 10^{10}\,\text{Hz}\). This calculated frequency is then compared with given options to select the closest match, which is \(1.5 \times 10^{10} \mathrm{~Hz}\). Understanding such calculations helps predict how waves behave in different mediums and situations.

Interference Patterns

Interference patterns occur when waves overlap, leading to areas of constructive interference (maxima) and destructive interference (minima). Constructive interference produces brighter or louder spots as waves align peaks to peaks, while destructive interference leads to cancellation.

In the scenario of microwaves striking a plane reflector, the incident and reflected waves interact to form a distinct pattern of maxima and minima. The detector detects these changes as it moves, effectively mapping the wave behavior. Eight maxima between the starting point and the 0.14 m mark tell us much about the wave's characteristics.

These patterns are not only fundamental to understanding standing waves but also underpin other applications, like noise-canceling technologies and antenna design. Recognizing how interference patterns work will be beneficial for exploring wave-related technologies and physics phenomena in greater depth.