Electromagnetic Waves Class 12 MCQs Questions with Answers
MCQ On Electromagnetic Waves Class 12 Question 1.
A linearly polarized electromagnetic wave given as E = E0cos (kz-ωt) is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as
(A) Er = – E0\(\hat{i}\) cos (kz – ωt)
(B) Er = E0 \(\hat{i}\) cos (kz + ωt)
(C) Er = – E0 \(\hat{i}\) cos (kz + ωt)
(D) Er = E0 \(\hat{i}\) sin (kz – ωt)
Answer:
(B) Er = E0 \(\hat{i}\) cos (kz + ωt)
Explanation:
The phase of a wave changes by 180° or n radian after got reflected from a denser medium. But the type of waves remains identical. Therefore, for the reflected wave, we have
\(\hat{z}\) = \(\hat{-z}\), \(\hat{i}\) = \(\hat{-i}\) and additional phase of π in the incident wave. Incident electromagnetic wave.
Then,
E = E0(\(\hat{-i}\)) cos (kz – ωt)
Therefore, the reflected electromagnetic wave is given as:
Er = E0(\(\hat{i}\))cos[K(-z) – ωt + π]
[∴\(\hat{z}\) = \(\hat{-z}\) and \(\hat{i}\) = \(\hat{-i}\)]
= -E0 \(\hat{i}\) cos [π – (kz – ωt)]
= -E0 \(\hat{i}\) [-cos{(kz – ωt)}]
= E0 \(\hat{i}\) cos (kz – ωt)
Electromagnetic Waves Class 12 MCQ Chapter 8 Question 2.
Light with an energy flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered (for complete absorption) during 30 minutes is –
(A) 36 x 10-5 kg m/s.
(B) 36 x 10-4 kg m/s.
(C) 108 x 104kg m/s.
(D) 1.08 x 107 kg m/s.
Answer:
(B) 36 x 10-4 kg m/s.
Explanation:
Energy flux, ø = 20 W/cm2
Area A= 30 cm2,
time t = 30 x 60 sec
U = Total energy falling in t sec = Energy flux x Area x time = ø At
U= 20 x 30 x 30 x 60
Momentum of the incident light = \(\frac{u}{c}=\frac{20 \times 30 \times 30 \times 60}{3 \times 10^{8}}\) = 36 x 10-4kg-ms-1
As no reflection from the surface and for complete absorption, momentum of reflected radiation is zero.
Momentum delivered to surface = Change in momentum
= Pf – Pi = 0 – 36 x 10-4 kgm/s
= – 36 x 10-4 kg m/s
(-) sign shows the direction of momentum.
Electromagnetic Waves MCQ Chapter 8 Class 12 Question 3.
The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is E
(A) \(\frac {E}{2}\)
(B) 2E
(C) \(\frac{E}{\sqrt{2}}\)
(D) \(\sqrt{2} \mathrm{E}\)
Answer:
(C) \(\frac{E}{\sqrt{2}}\)
Explanation:
We know that,
Class 12 Physics Chapter 8 MCQ Question 4.
If \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along
(A) –
(B) \(\overrightarrow{\mathrm{B}}\)
(C) \(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}}\)
(D) \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)
Answer:
(D) \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)
Explanation:
The direction of propagation of electromagnetic wave is perpendicular to \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) both and by right thumb rule.
The direction of propagation of electromagnetic wave is perpendicular to The direction of propagation of electromagnetic wave is perpendicular to both and the electric field vector \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) magnetic field vector B, i.e., in the direction of \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\). Here, electromagnetic wave is along the z-direction which is given by the cross product of \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\).
MCQs On Electromagnetic Waves Class 12 Question 5.
The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is
(A) c : 1
(B) c2 : 1
(C) 1 : 1
(D) \(\sqrt{c}\) : 1
Answer:
(C) 1 : 1
ExpIanation.
Average energy by electric field E0 is Uav
MCQ On Electromagnetic Waves Class 12 Question 6.
An EM wave radiates outwards from a dipole antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as –
(A) \(\frac{1}{r^{3}}\)
(B) \(\frac{1}{r^{2}}\)
(C) \(\frac {1}{r}\)
(D) remains constant
Answer:
(B) \(\frac{1}{r^{2}}\)
Explanation:
A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector (E0) which transports significant energy from the source falls inversely as the distance (r) from the antenna. As we know that electromagnetic waves are radiated from dipole antenna and radiated energy, so E ∝ = \(\frac {1}{r}\)
Electromagnetic Waves MCQs Class 12 Question 7.
In electromagnetic waves, the phase difference between magnetic and electric field vectors is –
(A) zero
(B) π
(C) π/2
(D) π/4
Answer:
(A) zero
Explanation:
Peaks of magnetic and electric waves of electromagnetic wave form at the same time. Hence, there is no phase difference between these two waves.
Electromagnetic Waves Class 12 MCQ Pdf Question 8.
From Maxwell’s hypothesis, a changing electric field gives rise to –
(A) an electric field
(B) an induced emf
(C) a magnetic field
(D) a magnetic torque.
Answer:
(C) a magnetic field
Explanation:
A changing magnetic field induces an electromotive force (emf) and, hence, an electric field. The direction of the emf opposes the change. The third Maxwell’s equations is Faraday’s law of induction, and includes Lenz’s law.
MCQ Electromagnetic Waves Class 12 Chapter 8 Question 9.
One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in
(A) visible region
(B) infrared region
(C) ultraviolet region
(D) microwave region
Answer:
(C) ultraviolet region
Explanation:
E = 11 eV = 11 x 1.6 x 10-19
h = 6.62 x 10-34 Js
E = hv
v = \(\frac {E}{h}\) = \(\frac{11 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}\) = \(\frac{8.8 \times 10^{-19+34}}{3.31}\)
= \(\frac {880}{331}\) x 10-15 Hz=2.65 x 1015 Hz
This frequency radiation belongs to the ultraviolet region.
MCQ Of Electromagnetic Waves Chapter 8 Class 12 Question 10.
The phenomenon which shows quantum nature of electromagnetic radiation is
(A) Piezoelectric effect
(B) Photoelectric effect
(C) Hall effect
(D) Tyndall effect
Answer:
(B) Photoelectric effect
Explanation:
Photoelectric effect allows us to perceive the quantum nature of light and I ultimately electromagnetic radiation.
MCQ Of Electromagnetic Waves Class 12 Question 11.
The electromagnetic radiations used for water purification and eye surgery is
(A) Infrared
(B) Microwave
(C) X-rays
(D) None of the above
Answer:
(D) None of the above
Explanation:
Ultraviolet rays are used for water purification and eye surgery.
Electromagnetic Waves MCQ Class 12 Question 12.
Electromagnetic wave having frequency 5 x 1011 Hz is
(A) Ultraviolet wave
(B) Radio wave
(C) Microwave
(D) X-rays
Answer:
(C) Microwave
Explanation:
Microwave frequency ranges from 1014 to 109 Hz.
Electromagnetic Waves MCQ Questions Chapter 8 Class 12 Question 13.
Proper arrangement of Gamma rays, Microwave, IR wave and UV rays in ascending order of frequency is
(A) Gamma rays > UV rays > IR rays > Microwave
(B) Microwave > IR rays > UV rays > Gamma rays
(C) UV rays > Gamma rays > Microwave > IR rays
(D) IR rays > UV rays > Microwave > Gamma rays
Answer:
Option (A) is correct
Explanation:
Frequency range of Gamma rays: 1022 – 1019 Hz Frequency range of UV rays: 1017 – 1015 Hz Frequency range of IR rays: 1014 -1012Hz Frequency range of Microwave: 1013-109 Hz
Electromagnetic Waves MCQs Chapter 8 Class 12 Question 14.
In vacuum, the physical property which remains same for microwave of wavelength 1 mm and UV radiation 1600 A is
(A) Wavelength
(B) Frequency
(C) Speed
(D) None of the above
Answer:
(C) Speed
Explanation:
All types of electromagnetic waves travel with speed of light in vacuum.
Electromagnetic Waves MCQ Pdf Chapter 8 Question 15.
In vacuum, the wavelength of the electromagnetic wave of frequency 5 x 1019Hz is
(A) 6 x 10-12m
(B) 3 x 10-8 m
(C) 1.6 x 1011 m
(D) 15 x 1027 m
Answer:
(A) 6 x 10-12m
Explanation:
λ = v/c = \(\frac{3 \times 10^{8}}{5 \times 10^{19}}\)= 6 x 10-12m
Electromagnetic Wave MCQ Class 12 Chapter 8 Question 16.
Which one of the following statements are correct?
(A) X -rays are suitable for radar system and aircraft navigation.
(B) Water molecules readily absorb infrared radiation and their thermal motion increases.
(C) Microwaves are produced in Coolidge tube
(D) Gamma radiations generate due to electron transitions between upper and lower energy levels of heavy element when excited by electron bombardment
Answer:
(B) Water molecules readily absorb infrared radiation and their thermal motion increases.
Explanation:
Water molecules readily absorb infrared radiation and their thermal motion increases and therefore, they heat their surroundings.
Assertion And Reason Based MCQs (1 Mark each)
Directions:
In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true
Class 12 Physics Electromagnetic Waves MCQ Question 1.
Assertion (A): Electromagnetic radiation exerts pressure.
Reason (R): Electromagnetic waves carry momentum and energy.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Electromagnetic radiation is composed of photons which has momentum (X/h) and energy (hv). When photon is incident on a surface, its momentum changes which gives rise to radiation pressure. So, the assertion and reason both are true but the reason does not explain the assertion.
Question 2.
Assertion (A): Electromagnetic wave does not require any medium to travel.
Reason (R): Electromagnetic wave cannot travel through any medium.
Answer:
(C) A is true but R is false
Explanation:
Electromagnetic waves are not mechanical waves. Hence, they do not require any medium to travel. It does not mean that electromagnetic wave cannot travel through a medium. Hence the assertion is true. But the reason is false.
Question 3.
Assertion (A): Dipole oscillation produce electromagnetic waves.
Reason (R): Accelerated charge produce electromagnetic waves.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
A source of electromagnetic waves is accelerating charge, since accelerated charge produces both electric and magnetic field. Again according to Maxwells’ classical theory, dipole oscillation produces electromagnetic wave, since charges are accelerated. So, assertion and reason both are true and the reason explains the assertion.
Question 4.
Assertion (A): X-ray travels with the speed of light.
Reason (R): X-ray is an e.m. wave.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Velocity of all electromagnetic I wave is 3 x 108 m/s which is the velocity of I light. X-ray is an electromagnetic wave. So, it travels with the velocity 3 x 108 m/s which is the velocity of light. So, assertion and reason both are correct and reason properly explains the assertion.
Question 5.
Assertion (A): Microwaves are considered suitable for radar system.
Reason (R): Microwaves are of shorter wavelength.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Wavelength of microwaves ranges from 10-3 to 0.1 m. Hence, it can be bounced from any small object. Hence, it is suitable for radar system. So, the assertion and reason both are true and the assertion is properly explained by the reason.
Question 6.
Assertion(A): The Ozone layer present at the top of stratosphere is very crucial for human survival.
Reason (R): Ozone layer prevents IR radiation.
Answer:
(C) A is true but R is false
Explanation:
There is a layer present at the top of stratosphere which is known as Ozone layer. This layer prevents UV radiations, mainly coming from the Sun, to reach Earth. UV radiation is harmful for human beings. So, the assertion is true. But the reason is false.
Question 7.
Assertion (A): Gamma rays are electromagnetic waves having the smallest wavelength.
Reason (R): Gamma rays are having the lowest frequency.
Answer:
(C) A is true but R is false
Explanation:
Gamma rays are electromagnetic waves having the smallest wavelength. So, the assertion is true. Relation between wavelength and frequency is v = c/λ. c is the velocity of light and same for all electromagnetic waves. For Gamma rays wavelength being smallest, frequency will be highest. So, the reason is false.
Case-Based MCQs
Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
Microwave oven:
The spectrum of electromagnetic radiation contains a part known as microwaves. These waves have frequency and energy smaller than visible light and wavelength larger than it. What is the principle of a microwave oven and how does it work ? Our objective is to cook food or warm it up. All food items such as fruit, vegetables, meat,
cereals, etc., contain water as a constituent. Now, what does it mean when we say that a certain object has become warmer? When the temperature of a body rises, the energy of the random motion of atoms and molecules increases and the molecules travel or vibrate or rotate with higher energies. The frequency of rotation of water molecules is about 2.45 gigahertz (GHz). If water receives microwaves of this frequency, its molecules absorb this radiation, which is equivalent to heating up water.
These molecules share this energy with neighbouring food molecules, heating up the food. One should use porcelain vessels and non metal containers in a microwave oven because of the danger of getting a shock from accumulated electric charges. Metals may also melt from heating. The porcelain container remains unaffected and cool, because its large molecules vibrate and rotate with much smaller frequencies, and thus cannot absorb microwaves.
Hence, they do not get eaten up. Thus, the basic principle of a microwave oven is to generate microwave radiation of appropriate frequency in the working space of the oven where we keep food. This way energy is not wasted in heating up the vessel. In the conventional heating method, the vessel on the burner gets heated first and then the food inside gets heated because of transfer of energy from the vessel. In the microwave oven, on the other hand, energy is directly delivered to water molecules which is shared by the entire food.
Question 1.
As compared to visible light microwave has frequency and energy:
(A) more than visible light.
(B) less than visible light.
(C) equal to visible light.
(D) Frequency is less but energy is more
Answer:
(B) less than visible light.
Explanation:
Microwaves have frequency and energy smaller than visible light and I wavelength larger than it.
Question 2.
When the temperature of a body rises:
(A) the energy of the random motion of atoms and molecules increases.
(B) the energy of the random motion of atoms and molecules decreases.
(C) the energy of the random motion of atoms and molecules remains same.
(D) the random motion of atoms and molecules becomes streamlined.
Answer:
(A) the energy of the random motion of atoms and molecules increases.
Explanation:
When the energy of the random motion of atoms and molecules of a substance increases and the molecules travel or vibrate or rotate with higher energies, the substance becomes hot.
Question 3.
The frequency of rotation of water molecules is about:
(A) 2.45 MHz.
(B) 2.45 kHz.
(C) 2.45 GHz.
(D) 2.45 THz.
Answer:
(C) 2.45 GHz.
Explanation:
The frequency of rotation of water molecules is about 2.45 gigahertz.
Question 4.
Why should one use porcelain vessels and non¬metal containers in a microwave oven ?
(A) Because it will get too much hot.
(B) Because it may crack due to high frequency.
(C) Because it will prevent the food items to become hot.
(D) Because of the danger of getting a shock from accumulated electric charges.
Answer:
(D) Because of the danger of getting a shock from accumulated electric charges.
Explanation:
One should use porcelain vessels and non-metal containers in a microwave oven because of the danger of getting a shock from accumulated electric charges. Metals may also melt from heating. The porcelain container remains unaffected and cool, because its large molecules vibrate and rotate with much smaller frequencies and thus cannot absorb microwaves. Hence, they do not get heated up.
Question 5.
In the microwave oven,
(A) energy is directly delivered to water molecules which is shared by the entire food.
(B) the vessel gets heated first, and then the food grains inside.
(C) the vessel gets heated first and then the water molecules collect heat from the body of the vessel.
(D) energy is directly delivered to the food grains.
Answer:
(A) energy is directly delivered to water molecules which is shared by the entire food.
Explanation:
In the conventional heating method, the vessel on the burner gets heated first and then the food inside gets heated because of transfer of energy from the vessel. In the microwave oven, on the other hand, energy is directly delivered to water molecules which is shared by the entire food.
I. Read the following text and answer the following questions on the basis of the same:
Laser:
Electromagnetic radiation is a natural phenomenon found in almost all areas of daily life, from radio waves to sunlight to x-rays. Laser radiation – like all light – is also a form of electromagnetic radiation. Electromagnetic radiation that has a wavelength between 380 nm and 780 nm is visible to the human eye and is commonly referred to as light. At wavelengths longer than 780 nm, optical radiation is termed infrared (IR) and is invisible to the eye.
At wavelengths shorter than 380 nm, optical radiation is termed ultraviolet (UV) and is also invisible to the eye. The term “laser light” refers to a much broader range of the electromagnetic spectrum that just the visible spectrum, anything between 150 nm up to 11000 nm (i.e. from the UV up to the far IR). The term laser is an acronym which stands for “light amplification by stimulated emission of radiation”. Einstein explained the stimulated emission.
In an atom, electron may move to higher energy level by absorbing a photon. When the electron comes back to the lower energy level it releases the same photon. This is called spontaneous emission. This may also so happen that the excited electron absorbs another photon, releases two photons and returns to the lower energy state. This is known as stimulated emission.
Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and ablating the stone inside human body organ. Apart from medical usage, laser is used for optical disk drive, printer, barcode reader etc.
Question 1.
What is the full form of LASER ?
(A) Light amplified by stimulated emission of radiation
(B) Light amplification by stimulated emission of radiation
(C) Light amplification by simultaneous emission of radiation
(D) Light amplified by synchronous emission of radiation
Answer:
(B) Light amplification by stimulated emission of radiation
Explanation:
The term laser is an acronym which stands for “light amplification by stimulated emission of radiation”.
Question 2.
The “stimulated emission” is the process of:
(A) release of a photon when electron comes back from higher to lower energy level.
(B) release of two photons by absorbing one photon when electron comes back from higher to lower energy level.
(C) absorption of a photon when electron moves from lower to higher energy level.
(D) None of the above
Answer:
(B) release of two photons by absorbing one photon when electron comes back from higher to lower energy level.
Explanation:
Einstein explained the stimulated emission. In an atom, electron may move to higher energy level by absorbing a photon. When the electron comes back to the lower energy level, it releases the same photon. This is called spontaneous emission. This may also so happen that the excited electron absorbs another photon, releases two photons and returns to the lower energy state. This is known as stimulated emission.
Question 3.
What is the range of amplitude of LASER?
(A) 150 nm – 400 nm
(B) 700 nm – 11000 nm
(C) Both the above
(D) None of the above
Answer:
(C) Both the above
Explanation:
The term “laser light” refers to a much broader range of the electromagnetic spectrum that just the visible spectrum, anything between 150 nm up to 11000 nm (i.e. from the UV up to the far IR).
Question 4.
Lithotripsy is:
(A) an industrial application.
(B) a medical application.
(C) laboratory application.
(D) process control application.
Answer:
(B) a medical application.
Explanation:
Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and ablating the stone inside human body organ.
Question 5.
LASER is used in:
(A) optical disk drive.
(B) transmitting satellite signal.
(C) radio communication.
(D) ionization.
Answer:
(A) optical disk drive.
Explanation:
An optical disc drive (ODD) is a disc drive that uses laser light or electromagnetic waves within or near the visible light spectrum as part of the process of reading or writing data to form optical discs.
II. Read the following text and answer the following questions on the basis of the same:
Ozone layer depletion:
We are all exposed to UV radiation from the sun. The sun is by far the strongest source of ultraviolet radiation. UV radiation spectrum is divided into three regions called UVA, UVB and UVC. As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide.
UVA is not filtered as significantly by the atmosphere. The three types of UV radiation are classified according to their wavelength. They differ in their biological activity and the extent to which they can penetrate the skin. The shorter the wavelength, the more harmful the UV radiation.
The UV region covers the wavelength range 100-400 nm and is divided into three bands:
- UVA (315-400 nm)
- UVB (280-315 nm)
- UVC (100-280 nm).
Short-wavelength UVC is the most damaging type of UV radiation. However, it is completely filtered by the atmosphere and does not reach the earth’s surface.UV level reaching the earth changes with latitude and altitude. UV levels are higher closer to the equator. Closer to the equator the sun’s rays have a shorter distance to travel through the atmosphere and therefore harmful UV radiation absorption is less.
With increasing altitude less atmosphere is available to absorb UV radiation. With every 1000 m in altitude, UV levels increase by approximately 10%. Ozone is a particularly effective absorber of UV radiation. As the ozone layer gets thinner, the protective filter activity is progressively reduced. Consequently, the people and the environment are exposed to higher levels of UV radiation, especially UVB. Ozone depletion is caused by human-made chemicals released into the atmosphere.
Question 1.
How many bands are there in UV radiation spectrum?
(A) 2
(B) 5
(C) 3
(D) 4
Answer:
(C) 3
Explanation:
There are 3 bands in UV spectrum – UVA, UVB, UVC.
UVA (315-400 run)
UVB (280-315 run)
UVC (100-280 nm).
Question 2.
Most harmful UV radiation band is
(A) UVA
(B) UVB
(C) UVC
(D) all of them
Answer:
(C) UVC
Explanation:
The shorter the wavelength, the more harmful the UV radiation. UVC has the shortest wavelength.
Question 3.
Which UV band is not absorbed by the atmosphere?
(A) UVA
(B) UVB
(C) UVC
(D) none of them
Answer:
(A) UVA
Explanation:
UVA is not filtered significantly by the atmosphere.
Question 4.
Ozone layer depletion is caused by
(A) cosmic rays.
(B) human-made chemicals released into the atmosphere.
(C) electrical spark in the atmosphere.
(D) None of the above
Answer:
(B) human-made chemicals released into the atmosphere.
Explanation:
Ozone depletion is caused by human-made chemicals released into the atmosphere.
Question 5.
UV level is –
(A) low at equator, high at poles.
(B) low at poles, high at equator.
(C) same at pole and equator.
(D) None of the above
Answer:
(B) low at poles, high at equator.
Explanation:
UV level is higher at the equator compared to that at poles. Closer to the equator, the Sun’s rays have a shorter distance to travel through the atmosphere and therefore, harmful UV radiation absorption is less.