Master all 6 problem patterns with step-by-step solutions
Every EM waves problem falls into one of these 6 categories. Learn the pattern, not just the solution. JEE tests pattern recognition.
Type 1: Direct Formula Application
These problems test if you know which formula to use. Usually single-step calculations.
Problem 1.1: Find Wavelength
Question: An EM wave has frequency 6 × 10¹⁴ Hz. Find its wavelength in vacuum.
Difficulty:Easy (CBSE/NEET)
Given:
ν = 6 × 10¹⁴ Hz
c = 3 × 10⁸ m/s
What examiner tests: Do you know c = νλ?
Concept Selection:
Wave speed relation: c = νλ
Solution:
λ = c/ν = (3×10⁸)/(6×10¹⁴)
λ = 5×10⁻⁷ m = 500 nm
Answer: 500 nm (orange-yellow light) Shortcut: For visible light, ν around 10¹⁴-10¹⁵ Hz → λ in hundreds of nm
Problem 1.2: E and B Relation
Question: In an EM wave, magnetic field amplitude is 2×10⁻⁷ T. Find electric field amplitude.
Difficulty:Easy (CBSE/JEE Main)
Given:
B₀ = 2×10⁻⁷ T
What examiner tests: Do you know E₀/B₀ = c?
Solution:
E₀ = cB₀ = (3×10⁸)(2×10⁻⁷)
E₀ = 60 V/m
Quick check: E₀ should be numerically much larger than B₀. If not, you made an error.
Problem 1.3: Photon Energy
Question: Find energy of a photon of wavelength 400 nm. (h = 6.6×10⁻³⁴ J·s)
Difficulty:Easy (NEET/JEE Main)
Given:
λ = 400 nm = 400×10⁻⁹ m
h = 6.6×10⁻³⁴ J·s
Solution:
Method 1: Using E = hc/λ
E = (6.6×10⁻³⁴ × 3×10⁸)/(400×10⁻⁹)
E = 4.95×10⁻¹⁹ J
Convert to eV:
E = 4.95×10⁻¹⁹ / 1.6×10⁻¹⁹ = 3.1 eV
Shortcut formula for visible light:
E(eV) ≈ 1240/λ(nm) = 1240/400 = 3.1 eV ✓
Type 2: Conceptual Questions
These test deep understanding. You can't just plug numbers—you must think.
Problem 2.1: E and B Phase
Question: In an EM wave propagating in vacuum, when electric field is maximum, what is the magnetic field?
Difficulty:Medium (JEE Main)
Concept Tested:
Phase relationship between E and B in EM waves
Common wrong answer: "B is zero"
Why wrong? Students confuse EM waves with LC oscillations or AC circuits.
Correct Understanding:
In EM waves, E and B oscillate in phase.
When E is maximum, B is also maximum (in its direction).
When E is zero, B is also zero.
Answer: Magnetic field is also maximum.
This is different from LC circuits where E and B oscillate 90° out of phase!
Problem 2.2: Medium Entry
Question: When light enters from air to glass, which property does NOT change?
(A) Speed (B) Wavelength (C) Frequency (D) All change
Difficulty:Easy (CBSE/NEET)
Common error: Thinking wavelength stays constant because "it looks the same color."
Analysis:
Property
Changes?
Why
Speed (v)
YES
v = c/n, n > 1
Wavelength (λ)
YES
λ decreases by factor n
Frequency (ν)
NO
Determined by source
Physical reason: Frequency is how fast the source vibrates. Medium can't change that! But wave slows down in medium, so wavelength must decrease to keep ν constant.
Answer: (C) Frequency
Color depends on frequency, which is why light doesn't change color when entering glass!
Problem 2.3: Energy Distribution
Question: In an EM wave, what fraction of total energy is in the electric field?
Difficulty:Medium (JEE Main)
Key Concept:
In EM waves, electric and magnetic fields carry equal energy.
Proof:
u_E = (1/2)ε₀E²
u_B = (1/2μ₀)B²
Since E = cB and c² = 1/(μ₀ε₀):
u_E = u_B
Answer: 50% (or 1/2)
Electric field = 50% energy
Magnetic field = 50% energy
Even though E is numerically much larger than B, they carry equal energy!
Type 3: Multi-Step Problems
Combine multiple concepts. This is where JEE separates good from great.
Problem 3.1: Intensity and Fields
Question: An EM wave in vacuum has intensity 12.5 W/m². Find amplitudes of E and B fields.
Difficulty:Medium (JEE Main)
Given:
I = 12.5 W/m²
ε₀ = 8.85×10⁻¹² C²/N·m²
c = 3×10⁸ m/s
Strategy: I → E₀ → B₀
Can't directly get B₀ from I without finding E₀ first.
Solution:
Step 1: Find E₀ from intensity
I = (1/2)cε₀E₀²
E₀² = 2I/(cε₀)
E₀² = 2(12.5)/[(3×10⁸)(8.85×10⁻¹²)]
E₀² = 9.4×10³
E₀ = 97 V/m ≈ 100 V/m
Step 2: Find B₀ from E₀
B₀ = E₀/c = 100/(3×10⁸)
B₀ = 3.3×10⁻⁷ T
Shortcut: Remember (1/2)cε₀ ≈ 1.3×10⁻³
So E₀ ≈ √(I/1.3×10⁻³)
Problem 3.2: Radiation Pressure
Question: Light of intensity 18 W/m² falls normally on a perfectly reflecting surface. Find radiation pressure.
Difficulty:Medium (JEE Main)
Given:
I = 18 W/m²
Surface is perfectly reflecting
Key decision: Absorption or reflection?
Surface reflects → use P = 2I/c
Solution:
For perfect reflection:
P = 2I/c
P = 2(18)/(3×10⁸)
P = 1.2×10⁻⁷ N/m²
Common error: Using P = I/c (this is for absorption only!)
Why double? When wave reflects, momentum change is 2p (like elastic collision). Initial momentum +p, final momentum -p, change = 2p.
Type 4: Graph-Based Problems
Visual interpretation. JEE Advanced loves these.
Problem 4.1: E vs t Graph
Question: Electric field of an EM wave is E = 100 sin(ωt). At t = π/(2ω), magnetic field is:
(A) Maximum (B) Zero (C) Half maximum (D) √2 times maximum
Difficulty:Medium (JEE Advanced)
Analysis:
At t = π/(2ω):
E = 100 sin(ω · π/(2ω)) = 100 sin(π/2)
E = 100 (maximum)
Key concept: E and B are in phase in EM waves.
When E is maximum, B is also maximum.
Answer: (A) Maximum
Both E and B reach their peaks at the same time and are zero at the same time.
Problem 4.2: EM Spectrum Identification
Question: Which part of EM spectrum has wavelength around 1 mm?
(A) Radio (B) Microwave (C) Infrared (D) Visible
Difficulty:Easy (NEET)
Spectrum Ranges:
Radio: > 0.1 m (> 100 mm)
Microwave: 1 mm - 0.1 m
Infrared: 700 nm - 1 mm
Visible: 400-700 nm
1 mm is at the boundary between microwave and infrared.
Typically classified as microwave.
Answer: (B) Microwave
Memory trick: Microwaves are measured in millimeters to centimeters.
Type 5: Assertion-Reason Questions
CBSE and NEET favorite format. Test logical connection between statements.
Problem 5.1
Assertion (A): EM waves do not require a medium for propagation.
Reason (R): EM waves are transverse in nature.
Options:
(a) Both A and R are true, and R is correct explanation of A
(b) Both A and R are true, but R is NOT correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true
Difficulty:Medium (CBSE)
Analysis:
Is Assertion true? YES
EM waves can travel through vacuum (sunlight reaches Earth through space).
Is Reason true? YES
EM waves are transverse (E ⊥ B ⊥ direction).
Is R the correct explanation of A?
NO! Being transverse doesn't explain why medium isn't needed.
Correct explanation: Self-propagating nature of E and B fields (changing E creates B, changing B creates E).
Answer: (b) Both true, but R is NOT correct explanation.
Note: Water waves are also transverse but DO need a medium!
Mobile phones use microwaves for communication. A mobile tower transmits signals at frequency 900 MHz with average power 50 W equally in all directions.
Question (i)
Wavelength of the transmitted signal is:
(A) 0.33 m (B) 3.3 m (C) 33 m (D) 0.033 m
λ = c/ν = (3×10⁸)/(900×10⁶)
λ = 3×10⁸ / 9×10⁸ = 1/3 = 0.33 m
Answer: (A) 0.33 m
This is in microwave range (mm to meters).
Question (ii)
At distance 100 m from tower, intensity of signal is approximately: