All questions of Electromagnetic Waves for Grade 11 Exam

The entire electromagnetic spectrum, from the lowest to the highest frequency (longest to shortest wavelength), includes all radio waves (e.g., commercial radio and television, microwaves, radar), infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

A gamma ray or gamma radiation is a penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei. It consists of the shortest wavelength electromagnetic waves and so imparts the highest photon energy

Violet ---→greatest frequency.....
Red------→greatest wavelength...

The electromagnetic waves are not mechanical waves. There are vibrations of electric vector and magnetic vector in them. These vibrations do not need any particles present in the medium for their propagation. That's why electromagnetic waves do not require any medium for propagation.

Option(d)becoz---> infrared radiation is heat or thermal radiation, any object which has a temperature radiates in the infrared. Even objects that we think of as being very cold, such as an ice cube, emit infrared. When an object is not quite hot enough to radiate visible light, it will emit most of its energy in the infrared

It's an infrared sensor that reads temperatures.This sensor allows satellites to measure the amount of energy radiated by Earth's surface, clouds, oceans, air, and so on. Infrared sensors can be used at night—a helpful feature for forecasters, considering that the imager can only pick up data during daylight hours.

Yes. I think proof is given in the book

These are not audible because high frequency wave

Electromagnetic waves are not used to determine the global warming

Infra red rays are basically heat radiation it gives warmth which support the plant growth. It also helps in treating muscle pain in the same way we use hot water massage for muscle cramps. Hope it helps:)

These electric and magnetic waves travel perpendicular to each other and have certain characteristics, including amplitude, wavelength, and frequency. General Properties of all electromagnetic radiation: Electromagnetic radiation can travel through empty space.

According to Huygens wave theory waves travel with the speed of light in vacuum. so D is right

Explanation:

Intensity is defined as the rate at which energy is transmitted through a unit area perpendicular to the direction of propagation of the wave. The intensity of a plane electromagnetic wave is proportional to the square of the amplitude of its electric field.

I = 1/2εcE²

where I represents the intensity, ε represents the permittivity of free space, c represents the speed of light, and E represents the amplitude of the electric field.

To find the proportionality between the intensity and the electric field, we can simplify the above equation as follows:

I ∝ E²

Therefore, the correct option is D, i.e., the intensity of a plane electromagnetic wave is proportional to the square of its electric field.

Conclusion:

The intensity of a plane electromagnetic wave is proportional to the square of its electric field. This relationship is important in various applications of electromagnetic waves, such as in the design of antennas and in the measurement of electromagnetic radiation.

Bcoz x rays have less wavelength or high frequency E=hf or hc/lymda

When we get a tan, what is actually happening is that the melanocytes are producing melanin pigment in reaction to ultraviolet light in sunlight...

So, correct answer is "UV radiation"...

$$Hope it's help... $$

This has no explanation you just need to remember the answer.

The utility frequency, (power) line frequency (American English) or mains frequency (British English) is the nominal frequency of the oscillations of alternating current (AC) in an electric power grid transmitted from a power station to the end-user.An electronic oscillator is an electronic circuit that produces a periodic, oscillating electronic signal, often a sine wave or a square wave. An RF oscillator produces signals in the radio frequency (RF) range of about 100 kHz to 100 GHz.

The Importance of Displacement Current in Applying Ampere's Law in a Parallel Plate Capacitor

Introduction:
In electromagnetism, Ampere's law is an important equation that relates the magnetic field to the electric current that produces it. However, when dealing with certain situations, such as a parallel plate capacitor, the application of Ampere's law requires the concept of displacement current to be included for accurate results. In this response, we will discuss why displacement current is necessary in applying Ampere's law in a parallel plate capacitor.

Parallel Plate Capacitor:
A parallel plate capacitor is a simple device consisting of two parallel plates separated by a distance, with a potential difference applied across them. When a capacitor is charged, an electric field is created between the plates, which stores energy in the form of electrostatic potential energy.

Ampere's Law:
Ampere's law states that the integral of the magnetic field around a closed loop is equal to the current passing through the loop. Mathematically, this can be expressed as:

∮B⋅dl=μ0Ienc

Where B is the magnetic field, dl is an element of the path around the loop, Ienc is the current passing through the loop, and μ0 is the permeability of free space.

Displacement Current:
Displacement current is a concept introduced by James Clerk Maxwell to account for the changing electric field in a region of space. When a capacitor is charging or discharging, the electric field between the plates is changing, which creates a displacement current. This current does not involve the flow of charge carriers, but rather the changing electric field itself.

Importance of Displacement Current in a Parallel Plate Capacitor:
When applying Ampere's law to a path parallel to the plates of a parallel plate capacitor, the displacement current must be included to obtain accurate results. This is because the electric field between the plates is changing, which creates a magnetic field that contributes to the total magnetic field around the loop. Without considering the displacement current, Ampere's law would only account for the current flow through the wires, ignoring the changing electric field between the plates.

Conclusion:
In conclusion, the concept of displacement current is necessary for accurately applying Ampere's law in a parallel plate capacitor. By including the displacement current, the changing electric field between the plates is accounted for, allowing for a more complete understanding of the magnetic field around the loop.

Radio waves are the radiation arising from two close energy levels in hydrogen.

E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation. Electromagnetic waves are transverse waves. The wave number is k = 2π/λ, where λ is the wavelength of the wave.

This answer is given in NCERT ch 16 where it is written that the average earth temp is 15 due to greenhouse effect or without it the temp would be -18.

The emission of gamma rays does not alter the number of protons or neutrons in the nucleus but instead has the effect of moving the nucleus from a higher to a lower energy state (unstable to stable). Gamma ray emission frequently follows beta decay, alpha decay, and other nuclear decay processes.

The Electromagnetic Spectrum and 1057 MHz Radiation

The electromagnetic spectrum is a range of all types of electromagnetic radiation. It includes various forms of energy such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each of these forms of energy has a specific frequency and wavelength.

1057 MHz is a specific frequency of radiation that is known as the Lamb shift. This radiation arises due to the interaction between an electron and a proton in a hydrogen atom. It is a very low-frequency radiation, and it belongs to the radio wave part of the electromagnetic spectrum.

Therefore, the correct answer to the given question is option D, i.e., radio (short wavelength end).

In summary, the Lamb shift radiation with a frequency of 1057 MHz belongs to the radio wave part of the electromagnetic spectrum.

Understanding Electromagnetic Radiation Sources
Electromagnetic waves vary in wavelength across different sources. To comprehend the correct order of wavelengths produced by various radiation sources, we can analyze each option provided.
Types of Radiation Sources:
- Radioactive Source: Emits gamma rays, which have very short wavelengths (high frequency).
- X-ray Tube: Produces X-rays with shorter wavelengths than visible light but longer than gamma rays.
- Sodium Lamp: Emits light in the visible spectrum, specifically in the yellow-orange range, with longer wavelengths than X-rays.
- Magnetron Valve: Primarily used in microwave ovens, it generates microwaves with even longer wavelengths than visible light.
Wavelength Order from Shortest to Longest:
- Gamma Rays (from Radioactive Source)
- X-rays (from X-ray Tube)
- Visible Light (from Sodium Lamp)
- Microwaves (from Magnetron Valve)
Correct Sequence Explained:
In option B, the order is:
1. Radioactive Source: Produces the shortest wavelengths (gamma rays).
2. X-ray Tube: Produces X-rays, which are longer than gamma rays but shorter than visible light.
3. Sodium Lamp: Emits visible light with longer wavelengths.
4. Magnetron Valve: Produces microwaves, which have the longest wavelengths in this grouping.
This sequence follows the increasing order of wavelengths, confirming that option B is indeed the correct answer.
Conclusion:
Understanding the properties of each radiation source helps in determining the correct sequence of electromagnetic wave wavelengths. This knowledge is essential for various applications in physics and medical imaging, especially relevant in NEET preparation.

Electromagnetic wave and its properties

An electromagnetic wave is a disturbance that travels through space, created by the motion of electric charges. It consists of two perpendicular waves - the electric field and the magnetic field, which are also perpendicular to the direction of wave propagation. Electromagnetic waves do not require a medium for their propagation and can travel through a vacuum.

Formula for wavelength

The wavelength (λ) of an electromagnetic wave is the distance between two consecutive points on the wave that are in phase. It is related to the frequency (f) of the wave and the speed of light (c) in vacuum by the formula:

λ = c/f

Where λ is the wavelength in meters, c is the speed of light in meters per second, and f is the frequency in Hertz.

Calculation

Given that the frequency of the electromagnetic wave is 30 MHz, we can convert it to units of Hertz by multiplying it by 10^6. Therefore, f = 30 × 10^6 Hz.

The speed of light in vacuum is a constant value of 3 × 10^8 m/s.

Substituting the values in the formula for wavelength, we get:

λ = c/f = 3 × 10^8 / (30 × 10^6) = 10 m

Therefore, the wavelength of the electromagnetic wave is 10 meters.

Conclusion

The correct answer is option D, i.e. the wavelength of the electromagnetic wave with a frequency of 30 MHz is 10 meters.

Microwaves are electromagnetic waves with wavelengths longer than those of terahertz (THz) wavelengths, but relatively short for radio waves. Microwaves have wavelengths approximately in the range of 30 cm (frequency = 1 GHz) to 1 mm (300 GHz).

Electromagnetic waves are produced by the acceleration of charged particles. When a charged particle accelerates, it creates a changing electric field, which in turn creates a changing magnetic field. These two fields then interact with each other and propagate away from the source as an electromagnetic wave.

Explanation of options:

a) Point charge oscillating in simple harmonic motion: When a point charge oscillates back and forth in a simple harmonic motion, it accelerates back and forth, creating a changing electric field. This changing electric field then creates a changing magnetic field, which in turn creates an electromagnetic wave.

b) Point charge travelling in a circle at constant velocity: When a point charge travels in a circle at a constant velocity, it is not accelerating. Therefore, it does not create a changing electric field, and hence does not produce electromagnetic waves.

c) Point charge travelling in a straight line with constant velocity: When a point charge travels in a straight line at a constant velocity, it is also not accelerating. Therefore, it does not create a changing electric field, and hence does not produce electromagnetic waves.

d) Point charge drifting slowly: When a point charge drifts slowly, it is not accelerating significantly. Therefore, it does not create a changing electric field, and hence does not produce electromagnetic waves.

Conclusion: Thus, the correct answer is option 'A' as only a point charge oscillating in simple harmonic motion can create a changing electric field and hence produce electromagnetic waves.

Speed is inversely proportional to the refractive index. As vacume has the least refractive index the speed of em waves will be max in it. So, option D is corect

See, when white light enters the prism it has actually entered out atmosphere and so the parts of EM spectrum like UV don't appear. After the light comes out of the prism all the parts seprate but we can only see the visible part i.e. 400-700nm range. So we look at only a part of the EM spectrum. Hence option B is correct.

Understanding the Formula
The formula used here is λmT = 0.29 cm·K, where:
- λm represents the wavelength of peak emission,
- T is the temperature in Kelvin.
This relationship helps us find the characteristic temperature when the wavelength is known.
Given Data
- λm = 5 × 10^-7 m (which is equivalent to 500 nm, or 0.5 μm)
Calculating Temperature
To find the temperature (T), rearranging the formula gives:
T = 0.29 cm·K / λm
First, convert λm from meters to centimeters:
- 5 × 10^-7 m = 5 × 10^-5 cm
Now, substitute λm into the equation:
T = 0.29 cm·K / (5 × 10^-5 cm)
Performing the Calculation
- T = 0.29 / 5 × 10^-5
- T = 0.29 / 0.00005
- T = 5800 K
Comparing with Options
The calculated temperature (5800 K) is closest to option 'c' which is 6000 K.
Conclusion
The correct answer is option 'C' (6000 K) as it aligns closely with the derived temperature from the given wavelength, confirming its suitability in the context of blackbody radiation.
This method effectively demonstrates how to determine characteristic temperatures using peak wavelength values in thermal physics.