
A plane electromagnetic wave of wave intensity $6W{m^{ - 2}}$ strikes a small mirror area $40c{m^2}$, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be:
(A) $6.4 \times {10^{ - 7}}kgm{s^{ - 2}}$
(B) $4.8 \times {10^{ - 8}}kgm{s^{ - 2}}$
(C) $3.2 \times {10^{ - 9}}kgm{s^{ - 2}}$
(D) $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
Answer
161.1k+ views
Hint: In order to solve this question, we will use the general relation between intensity, and momentum and are to calculate the net transferred momentum by the electromagnetic wave to the mirror each second.
Formula used:
If I is the intensity of an electromagnetic wave, A is the area of the object on which its falling then net momentum transferred by the wave to the object is given by:
$p = 2IAc$
where, p is the momentum transferred $c = 3 \times {10^8}m{s^{ - 1}}$ is the speed of the light
Complete answer:
According to the question, we have given that,
The intensity of an electromagnetic wave is $I = 6W{m^{ - 2}}$
Area of the small mirror is $A = 40c{m^2} = 4 \times {10^{ - 3}}{m^2}$ and $c = 3 \times {10^8}m{s^{ - 1}}$
In order to calculate momentum transferred, put the values in the formula $p = 2IAc$ we get,
$p = 2(6)(4 \times {10^{ - 3}})(3 \times {10^8})$
on solving for p we get the value,
$p = 1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
So, the net momentum transferred by the electromagnetic wave to the small mirror has the magnitude of $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
Hence, the correct answer is option (D) $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
Note: It should be remembered that electromagnetic waves are a form of energy but as light energy carries photon and photons has a mass, due to this particle nature of light, it carries a momentum along with it and the value of the speed of light is taken in free space as the speed of light varies in a different medium but it’s maximum in free space.
Formula used:
If I is the intensity of an electromagnetic wave, A is the area of the object on which its falling then net momentum transferred by the wave to the object is given by:
$p = 2IAc$
where, p is the momentum transferred $c = 3 \times {10^8}m{s^{ - 1}}$ is the speed of the light
Complete answer:
According to the question, we have given that,
The intensity of an electromagnetic wave is $I = 6W{m^{ - 2}}$
Area of the small mirror is $A = 40c{m^2} = 4 \times {10^{ - 3}}{m^2}$ and $c = 3 \times {10^8}m{s^{ - 1}}$
In order to calculate momentum transferred, put the values in the formula $p = 2IAc$ we get,
$p = 2(6)(4 \times {10^{ - 3}})(3 \times {10^8})$
on solving for p we get the value,
$p = 1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
So, the net momentum transferred by the electromagnetic wave to the small mirror has the magnitude of $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
Hence, the correct answer is option (D) $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$
Note: It should be remembered that electromagnetic waves are a form of energy but as light energy carries photon and photons has a mass, due to this particle nature of light, it carries a momentum along with it and the value of the speed of light is taken in free space as the speed of light varies in a different medium but it’s maximum in free space.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

If a wire of resistance R is stretched to double of class 12 physics JEE_Main
