Class 12/NEET/JEE/Dual nature of radiation and matter
Introduction
Light is a form of electromagnetic radiation that carries energy and momentum. Although photons have zero rest mass, they possess momentum and can exert a force when they are absorbed or reflected by a surface.
The force exerted by electromagnetic radiation on a surface is called radiation force. The force exerted per unit area is known as radiation pressure.
This topic is important for CBSE Class 12 Physics and is useful for NEET and JEE preparation, particularly in modern physics and photon-related numerical problems.
1. Momentum of a Photon
A photon is a quantum of electromagnetic radiation. The energy of a photon is given by:
E = hν = hc/λ
Where:
- E = energy of the photon
- h = Planck’s constant = 6.626 × 10⁻³⁴ J s
- ν = frequency of radiation
- c = speed of light in vacuum = 3 × 10⁸ m/s
- λ = wavelength of radiation
The momentum of a photon is calculated using:
p = E/c
Substituting the expression for photon energy:
p = hν/c = h/λ
Therefore, the momentum of a photon is inversely proportional to its wavelength.
Important concept: A photon has zero rest mass, but it carries momentum because it possesses energy.
2. Force Due to Radiation
Force is defined as the rate of change of momentum.
Force = Change in momentum / Time interval
When photons strike a surface, their momentum changes. This change in momentum produces a force on the surface.
The magnitude of the force depends on whether the photons are absorbed or reflected.
Case 1: Complete Absorption of Radiation
When light falls normally on a perfectly absorbing surface, the photons transfer all their momentum to the surface.
If the total energy of the incident radiation during time t is E, the momentum transferred is E/c.
Therefore,
F = E/(ct)
Since power is energy transferred per unit time,
P = E/t
Hence, the radiation force on a perfectly absorbing surface is:
F = P/c
Thus, the force equals the incident power divided by the speed of light.
Case 2: Complete Reflection of Radiation
When photons strike a perfectly reflecting surface normally, their momentum reverses direction.
If the initial momentum of a photon is p, its final momentum is −p.
The magnitude of the change in momentum is:
Change in momentum = 2p
Therefore, the force exerted by the radiation is:
F = 2E/(ct)
Since P = E/t,
F = 2P/c
Conclusion: For the same incident power, a perfectly reflecting surface experiences twice the radiation force of a perfectly absorbing surface, assuming normal incidence and ideal reflection.
3. Radiation Pressure
Radiation pressure is the force exerted by radiation per unit area.
Radiation pressure = Force / Area
The SI unit of radiation pressure is the pascal (Pa), equivalent to newton per square metre.
Radiation Pressure on a Perfectly Absorbing Surface
The intensity of radiation is the power incident per unit area.
I = P/A
For a perfectly absorbing surface:
F = P/c
Dividing both sides by the area:
Radiation pressure = I/c
Therefore, the radiation pressure on a perfectly absorbing surface is:
pᵣ = I/c
Radiation Pressure on a Perfectly Reflecting Surface
For a perfectly reflecting surface:
F = 2P/c
Dividing by the area:
Radiation pressure = 2I/c
Therefore,
pᵣ = 2I/c
For the same intensity, the radiation pressure on a perfectly reflecting surface is twice that on a perfectly absorbing surface.
4. General Formula for Radiation Force
For radiation falling normally on a surface with reflectivity R, assuming the remaining fraction is absorbed and no radiation is transmitted, the force is:
F = (1 + R)P/c
Where:
- F = radiation force
- R = fraction of incident power reflected
- P = incident power
- c = speed of light
Special cases:
- For complete absorption, R = 0; therefore, F = P/c.
- For complete reflection, R = 1; therefore, F = 2P/c.
- For 40% reflection, R = 0.40; therefore, F = 1.40P/c.
For oblique incidence, the normal component of the force on a perfectly absorbing surface is:
Fₙ = (P/c) cos²θ
Here, θ is the angle between the incident beam and the surface normal, and P is the total power in the incident beam.
For a perfectly reflecting surface, the normal force is twice this value.
5. Important Formula Summary
| Physical quantity | Formula |
|---|---|
| Energy of a photon | E = hν = hc/λ |
| Momentum of a photon | p = h/λ |
| Radiation force on an absorbing surface | F = P/c |
| Radiation force on a reflecting surface | F = 2P/c |
| Radiation pressure on an absorbing surface | pᵣ = I/c |
| Radiation pressure on a reflecting surface | pᵣ = 2I/c |
| Intensity of radiation | I = P/A |
| Force for partial reflection | F = (1 + R)P/c |
6. Solved Numerical Problems
Numerical 1: Force on a Perfectly Absorbing Surface
Question: A beam of light of power 60 W falls normally on a perfectly absorbing surface. Calculate the force exerted by the radiation.
Given:
- Power, P = 60 W
- Speed of light, c = 3 × 10⁸ m/s
Formula:
F = P/c
Solution:
F = 60/(3 × 10⁸)
Answer: F = 2 × 10⁻⁷ N
Numerical 2: Force on a Perfectly Reflecting Surface
Question: A laser beam of power 150 W falls normally on a perfectly reflecting mirror. Find the radiation force.
Given:
- P = 150 W
- c = 3 × 10⁸ m/s
Formula:
F = 2P/c
Solution:
F = (2 × 150)/(3 × 10⁸)
Answer: F = 1 × 10⁻⁶ N = 1 μN
Numerical 3: Radiation Pressure on an Absorbing Surface
Question: Light of intensity 900 W/m² falls normally on a perfectly absorbing surface. Calculate the radiation pressure.
Formula:
Radiation pressure = I/c
Solution:
Radiation pressure = 900/(3 × 10⁸)
Answer: Radiation pressure = 3 × 10⁻⁶ Pa
Numerical 4: Momentum of a Photon
Question: Calculate the momentum of a photon having a wavelength of 500 nm.
Given:
Wavelength = 500 nm = 500 × 10⁻⁹ m
Planck’s constant = 6.626 × 10⁻³⁴ J s
Formula:
p = h/λ
Solution:
p = (6.626 × 10⁻³⁴)/(500 × 10⁻⁹)
Answer: p ≈ 1.33 × 10⁻²⁷ kg m/s
Numerical 5: Number of Photons Absorbed Per Second
Question: Monochromatic light of wavelength 600 nm and power 12 W is completely absorbed by a surface. Calculate the number of photons absorbed per second.
Step 1: Calculate the energy of one photon.
Energy = hc/λ
Energy = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(600 × 10⁻⁹)
Energy = 3.313 × 10⁻¹⁹ J
Step 2: Calculate the number of photons per second.
Power = Number of photons per second × Energy of one photon
Therefore,
Number of photons per second = Power / Energy of one photon
N = 12/(3.313 × 10⁻¹⁹)
Answer: N ≈ 3.62 × 10¹⁹ photons/s
Numerical 6: Force When Only a Fraction of Light Is Reflected
Question: A beam of power 90 W falls normally on a surface that reflects 40% of the incident power and absorbs the remaining 60%. Calculate the radiation force, assuming no transmission.
Given:
- P = 90 W
- Reflectivity, R = 0.40
- c = 3 × 10⁸ m/s
Formula:
F = (1 + R)P/c
Solution:
F = (1 + 0.40) × 90/(3 × 10⁸)
F = 126/(3 × 10⁸)
Answer: F = 4.2 × 10⁻⁷ N
7. Practice Questions
Try these questions before checking the answers.
Question 1: A 300 W light beam falls normally on a perfectly absorbing surface. What is the radiation force?
A. 1 × 10⁻⁶ N
B. 2 × 10⁻⁶ N
C. 3 × 10⁻⁸ N
D. 6 × 10⁻⁷ N
Correct answer: A. 1 × 10⁻⁶ N
Explanation: F = P/c = 300/(3 × 10⁸) = 1 × 10⁻⁶ N.
Question 2: The radiation pressure on a perfectly reflecting surface is:
A. I/c
B. I/2c
C. 2I/c
D. Ic
Correct answer: C. 2I/c
Explanation: For normal incidence on a perfect reflector, the photon momentum reverses, producing twice the momentum change.
Question 3: If the wavelength of a photon is doubled, its momentum becomes:
A. Twice
B. Half
C. Four times
D. Unchanged
Correct answer: B. Half
Explanation: Photon momentum is p = h/λ. Momentum is inversely proportional to wavelength.
Question 4: A perfectly absorbing surface receives radiation of constant wavelength. If the incident power doubles, the radiation force:
A. Becomes half
B. Remains unchanged
C. Doubles
D. Becomes four times
Correct answer: C. Doubles
Explanation: For complete absorption, F = P/c. Hence, force is directly proportional to incident power.
8. Applications of Radiation Force
Radiation force is extremely small under ordinary conditions, but it has important applications in science and technology.
1. Solar Sails
Spacecraft can use momentum transferred by sunlight for propulsion. Solar sails can operate without conventional rocket fuel for generating thrust from the incident light.
2. Optical Tweezers
Highly focused laser beams can trap and manipulate microscopic particles using optical forces.
3. Astrophysics
Radiation from stars can exert pressure on dust and gas in space, influencing the motion of matter in astronomical environments.
4. Precision Experiments
Radiation pressure can affect extremely sensitive measurements and must be considered in some precision optical experiments.
9. Important NEET and JEE Exam Tips
- Remember that photon momentum is p = h/λ, even though a photon has zero rest mass.
- For normal incidence, use F = P/c for complete absorption and F = 2P/c for complete reflection.
- Do not confuse radiation force with radiation pressure. Force is measured in newtons, whereas pressure is measured in pascals.
- If intensity and area are given, calculate power using P = IA.
- For partial reflection with no transmission, use F = (1 + R)P/c.
- Convert wavelength from nanometres to metres before calculating photon momentum.
- Check whether the question specifies absorption, reflection, or partial reflection before selecting the formula.
Conclusion
Radiation force demonstrates that electromagnetic radiation carries momentum. When photons are absorbed or reflected, their momentum changes and a force acts on the surface.
By understanding the relationship between photon energy, momentum, incident power, intensity and radiation pressure, students can solve a wide range of Class 12 Physics, NEET and JEE numerical problems.
For effective revision, practise problems involving absorption, reflection, photon momentum and radiation pressure. Always identify the physical situation correctly and check the units of your final answer.
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