In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is : [Use 0 = 8.85 x 10-12 , C2 / Nm2 ]

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is : [Use 0 = 8.85 x 10-12 , C2 / Nm2 ]

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50…

Year	2026	2025	2024	Total
Questions	10	12	9	31

Q4 jee_main_2025_03_april_morning Radiation Pressure

The radiation pressure exerted by a

450~W

light source on a perfectly reflecting surface placed at

2~m

away from it, is :

A. $1.5 \times 10⁻⁸~Pascals$ $1.5 \times 10^{-8}\mathrm{~Pascals}$
B. $0$ $0$
C. $6 \times 10⁻⁸~Pascals$ $6 \times 10^{-8}\mathrm{~Pascals}$
D. $3 \times 10⁻⁸~Pascals$ $3 \times 10^{-8}\mathrm{~Pascals}$

Solution

Related Formula

For a perfectly reflecting surface, the radiation pressure $Prad$ $P_{\text{rad}}$ is given by:

Prad = (2I)/(c)

where, $I$ $I$ = intensity of the light source, $c$ $c$ = speed of light $\approx 3 \times 10⁸~m/s$ $\approx 3 \times 10^8\mathrm{~m/s}$.

Core Logic

Let's first calculate the intensity $I$ $I$ of the point source at a distance $r = 2~m$ $r = 2\mathrm{~m}$:

I = PowerArea = (P)/(4π r²)

Substitute the given values ( $P = 450~W$ $P = 450\mathrm{~W}$ and $r = 2~m$ $r = 2\mathrm{~m}$):

I = (450)/(4π \times 2²) = (450)/(16π)~W/m²

Step 1: Calculating Radiation Pressure

Now, substitute $I$ $I$ into the radiation pressure formula:

Prad = (2 \times ((450)/(16π)))/(3 \times 10⁸) = (900)/(16π \times 3 \times 10⁸)

Prad = (300)/(16π \times 10⁸) = (75)/(4π \times 10⁸)~N/m²

Using $π \approx 3.1416$ $\pi \approx 3.1416$:

Prad = (75)/(4 \times 3.1416 \times 10⁸) = (75)/(12.566) \times 10⁻⁸

Prad \approx 5.968 \times 10⁻⁸~Pascals \approx 6 \times 10⁻⁸~Pascals

Pattern Recognition

Remember: Perfectly absorbing surface $P = I/c$ $\implies P = I/c$. Perfectly reflecting surface $P = 2I/c$ $\implies P = 2I/c$. Always pay close attention to the surface's properties mentioned in the prompt!

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q23 jee_main_2025_04_april_evening Velocity of EM Waves

If an optical medium possesses a relative permeability of

(10)/(pi)

and relative permittivity of

(1)/(0.0885)

, then the velocity of light is greater in vacuum than that in this medium by ________ times.

(μ₀ = 4π \times 10⁻⁷ H / m, ε₀ = 8.85 \times 10⁻¹² F / m, c = 3 \times 10⁸ m / s)

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

v = 1\sqrt(μ ε) = 1\sqrt(μ₀μᵣ \cdot ε₀εᵣ) = c\sqrt(μᵣ εᵣ)

Core Logic

Given parameters:

μᵣ = (10)/(π)

εᵣ = (1)/(0.0885)

Let's substitute these into the refractive index radical term $\sqrt(μᵣ εᵣ)$ $\sqrt{\mu_r \epsilon_r}$:

μᵣ εᵣ = (10)/(π) \times (1)/(0.0885)

Step 1: Simplify Numerical Ratio

Using standard approximations $π \approx 3.14$ $\pi \approx 3.14$:

μᵣ εᵣ = (10)/(3.1415 \times 0.0885) \approx (10)/(0.278) \approx 36

Taking the square root:

\sqrt(μᵣ εᵣ) = \sqrt(36) = 6

Therefore, $v = (c)/(6) c = 6v$ $v = \frac{c}{6} \implies c = 6v$. Velocity in a vacuum is exactly 6 times faster.

Pattern Recognition

The expression $\sqrt(μᵣ εᵣ)$ $\sqrt{\mu_r \epsilon_r}$ is identical to the definition of refractive index $n$ $n$. Simplifying indices down to a perfect square ( $36 6$ $36 \implies 6$) clarifies the ratio immediately.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q3 jee_main_2025_07_april_evening Intensity of EM Waves

The unit of

2Iε₀c

is: (

I

= intensity of an electromagnetic wave,

c

: speed of light) [cite: 10]

A. $Vm$ $\mathrm{Vm}$ [cite: 29]
B. $NC$ $\mathrm{NC}$ [cite: 31]
C. $Nm$ $\mathrm{Nm}$ [cite: 30]
D. $NC⁻¹$ $\mathrm{NC}^{-1}$ [cite: 31]

Solution

Related Formula

$I = (1)/(2)ε₀ E₀² c$ $$I = \frac{1}{2}\epsilon_0 E_0^2 c$$ [cite: 670]

Core Logic

Rearranging the equation for intensity $I$ $I$ to express the peak electric field amplitude $E₀$ $E_0$: [cite: 670]

E₀² = (2I)/(ε₀ c)

$E₀ = \sqrt((2I)/(ε₀ c))$ [cite: 670]

Thus, the given quantity is simply the magnitude of the peak electric field $E₀$ $E_0$[cite: 672]. The standard SI unit of an electric field is Newtons per Coulomb ( $N\cdot C⁻¹$ $\mathrm{N\cdot C^{-1}}$) or Volts per meter ( $V\cdot m⁻¹$ $\mathrm{V\cdot m^{-1}}$)[cite: 673]. Matching with the given structural choices, $NC⁻¹$ $\mathrm{NC}^{-1}$ is the correct unit[cite: 31, 669].

Pattern Recognition

Recognize the standard configuration for energy flux density (intensity) $I = uavgc$ $I = u_{\text{avg}}c$[cite: 670]. Identifying that $\sqrt((2I)/(ε₀ c))$ $\sqrt{\frac{2I}{\epsilon_0 c}}$ resolves to the electric field dimension directly yields the solution unit $NC⁻¹$ $\mathrm{NC}^{-1}$ or $V/m$ $\mathrm{V/m}$[cite: 670, 672, 673].

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q4 jee_main_2025_07_april_evening Impedance of Free Space

The dimension of

μ₀ε₀

is equal to that of: (

μ₀=

Vacuum permeability and

ε₀=

Vacuum permittivity) [cite: 33, 34]

A. Voltage [cite: 35]
B. Capacitance [cite: 36]
C. Inductance [cite: 37]
D. Resistance [cite: 38]

Solution

Related Formula

$L = (μ₀ N² A)/(l) μ₀ \propto L$ [cite: 675]

$C = (ε₀ A)/(d) ε₀ \propto C$ [cite: 677]

Core Logic

From the basic formulas of inductance and capacitance, we can note the proportional parameters: [cite: 675, 677]

$(μ₀)/(ε₀) \propto (L)/(C)$ [cite: 678]

We know that the time constant for an $LR$ $LR$ circuit is $τ = (L)/(R)$ $\tau = \frac{L}{R}$ and for a $RC$ $RC$ circuit is $τ = RC$ $\tau = RC$[cite: 679]. Equating these time dimensions: [cite: 679]

$(L)/(R) = RC (L)/(C) = R²$ [cite: 679]

Taking the square root or matching parameters from the text solution layout yields the characteristic dimension of resistance[cite: 679].

Pattern Recognition

The quantity $\sqrt((μ₀)/(ε₀))$ $\sqrt{\frac{\mu_0}{\epsilon_0}}$ represents the intrinsic impedance of free space, which has the value $\approx 377 Ω$ $\approx 377\ \Omega$[cite: 679]. Hence, its square matches the dimension of resistance squared, which maps to Resistance in the choice sets[cite: 38, 674].

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q2 jee_main_2025_24_jan_evening Electromagnetic Spectrum

Arrange the following in the ascending order of wavelength

(λ)

: (A) Microwaves

(λ₁)

(B) Ultraviolet rays

(λ₂)

(C) Infrared rays

(λ₃)

(D) X-rays

(λ₄)

Choose the most appropriate answer from the options given below :

A. $λ₄ < λ₃ < λ₂ < λ₁$
B. $λ₃ < λ₄ < λ₂ < λ₁$
C. $λ₄ < λ₂ < λ₃ < λ₁$
D. $λ₄ < λ₃ < λ₁ < λ₂$

Solution

Core Logic

The order of components in the electromagnetic spectrum in increasing order of wavelength ( $λ$ $\lambda$) is:

γ-rays < X-rays < U.V. rays < Visible rays < IR rays < Microwaves < Radio waves

Given components:

Microwaves: $λ₁$ $\lambda_1$
Ultraviolet rays: $λ₂$ $\lambda_2$
Infrared rays: $λ₃$ $\lambda_3$
X-rays: $λ₄$ $\lambda_4$

Comparing these yields:

λ₄ < λ₂ < λ₃ < λ₁

Pattern Recognition

Remember the mnemonic for the EM spectrum in increasing wavelength: "Good Xylophones Use Very Interesting Micro Radios" (Gamma, X-ray, UV, Visible, IR, Microwave, Radio).

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Solution & Explanation

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 4

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Related Formula

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Pattern Recognition

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Related Formula

Core Logic

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Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Core Logic

Pattern Recognition

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