Electromagnetic Waves Previous Year Questions — JEE Main Physics
31 past-year Electromagnetic Waves questions from JEE Main (Physics) — page 2.
Q4 (2025)
The radiation pressure exerted by a $450\mathrm{~W}$ light source on a perfectly reflecting surface placed at $2\mathrm{~m}$ away from it, is :
- $1.5 \times 10^{-8}\mathrm{~Pascals}$
- $0$
- $6 \times 10^{-8}\mathrm{~Pascals}$
- $3 \times 10^{-8}\mathrm{~Pascals}$
### Related Formula
For a perfectly reflecting surface, the radiation pressure $P_{\text{rad}}$ is given by:
$$P_{\text{rad}} = \frac{2I}{c}$$
where,
$I$ = intensity of the light source,
$c$ = speed of light $\approx 3 \times 10^8\mathrm{~m/s}$.
### Core Logic
Let's first calculate the intensity $I$ of the point source at a distance $r = 2\mathrm{~m}$:
$$I = \frac{\text{Power}}{\text{Area}} = \frac{P}{4\pi r^2}$$
Substitute the given values ($P = 450\mathrm{~W}$ and $r = 2\mathrm{~m}$):
$$I = \frac{450}{4\pi \times 2^2} = \frac{450}{16\pi}\mathrm{~W/m}^2$$
### Step 1: Calculating Radiation Pressure
Now, substitute $I$ into the radiation pressure formula:
$$P_{\text{rad}} = \frac{2 \times \left(\frac{450}{16\pi}\right)}{3 \times 10^8} = \frac{900}{16\pi \times 3 \times 10^8}$$
$$P_{\text{rad}} = \frac{300}{16\pi \times 10^8} = \frac{75}{4\pi \times 10^8}\mathrm{~N/m}^2$$
Using $\pi \approx 3.1416$:
$$P_{\text{rad}} = \frac{75}{4 \times 3.1416 \times 10^8} = \frac{75}{12.566} \times 10^{-8}$$
$$P_{\text{rad}} \approx 5.968 \times 10^{-8}\mathrm{~Pascals} \approx 6 \times 10^{-8}\mathrm{~Pascals}$$
### Pattern Recognition
Remember: Perfectly absorbing surface $\implies P = I/c$. Perfectly reflecting surface $\implies P = 2I/c$. Always pay close attention to the surface's properties mentioned in the prompt!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q23 (2025)
If an optical medium possesses a relative permeability of $\frac{10}{pi}$ and relative permittivity of $\frac{1}{0.0885}$, then the velocity of light is greater in vacuum than that in this medium by ________ times. $(\mu_0 = 4\pi \times 10^{-7}\text{ H / m}, \epsilon_0 = 8.85 \times 10^{-12}\text{ F / m}, c = 3 \times 10^8\text{ m / s})$
### Related Formula
$$v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0\mu_r \cdot \epsilon_0\epsilon_r}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$$
### Core Logic
Given parameters:
$$\mu_r = \frac{10}{\pi}$$
$$\epsilon_r = \frac{1}{0.0885}$$
Let's substitute these into the refractive index radical term $\sqrt{\mu_r \epsilon_r}$:
$$\mu_r \epsilon_r = \frac{10}{\pi} \times \frac{1}{0.0885}$$
### Step 1: Simplify Numerical Ratio
Using standard approximations $\pi \approx 3.14$:
$$\mu_r \epsilon_r = \frac{10}{3.1415 \times 0.0885} \approx \frac{10}{0.278} \approx 36$$
Taking the square root:
$$\sqrt{\mu_r \epsilon_r} = \sqrt{36} = 6$$
Therefore, $v = \frac{c}{6} \implies c = 6v$.
Velocity in a vacuum is exactly **6** times faster.
### Pattern Recognition
The expression $\sqrt{\mu_r \epsilon_r}$ is identical to the definition of refractive index $n$. Simplifying indices down to a perfect square ($36 \implies 6$) clarifies the ratio immediately.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q3 (2025)
The unit of $\sqrt{\frac{2I}{\epsilon_{0}c}}$ is:
($I$ = intensity of an electromagnetic wave, $c$: speed of light) [cite: 10]
- $\mathrm{Vm}$ [cite: 29]
- $\mathrm{NC}$ [cite: 31]
- $\mathrm{Nm}$ [cite: 30]
- $\mathrm{NC}^{-1}$ [cite: 31]
### Related Formula
$$I = \frac{1}{2}\epsilon_0 E_0^2 c$$ [cite: 670]
### Core Logic
Rearranging the equation for intensity $I$ to express the peak electric field amplitude $E_0$: [cite: 670]
$$E_0^2 = \frac{2I}{\epsilon_0 c}$$
$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$ [cite: 670]
Thus, the given quantity is simply the magnitude of the peak electric field $E_0$[cite: 672]. The standard SI unit of an electric field is Newtons per Coulomb ($\mathrm{N\cdot C^{-1}}$) or Volts per meter ($\mathrm{V\cdot m^{-1}}$)[cite: 673]. Matching with the given structural choices, $\mathrm{NC}^{-1}$ is the correct unit[cite: 31, 669].
### Pattern Recognition
Recognize the standard configuration for energy flux density (intensity) $I = u_{\text{avg}}c$[cite: 670]. Identifying that $\sqrt{\frac{2I}{\epsilon_0 c}}$ resolves to the electric field dimension directly yields the solution unit $\mathrm{NC}^{-1}$ or $\mathrm{V/m}$[cite: 670, 672, 673].
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q4 (2025)
The dimension of $\frac{\mu_{0}}{\epsilon_{0}}$ is equal to that of:
($\mu_{0}=$ Vacuum permeability and $\epsilon_{0}=$ Vacuum permittivity) [cite: 33, 34]
- Voltage [cite: 35]
- Capacitance [cite: 36]
- Inductance [cite: 37]
- Resistance [cite: 38]
### Related Formula
$$L = \frac{\mu_0 N^2 A}{l} \implies \mu_0 \propto L$$ [cite: 675]
$$C = \frac{\epsilon_0 A}{d} \implies \epsilon_0 \propto C$$ [cite: 677]
### Core Logic
From the basic formulas of inductance and capacitance, we can note the proportional parameters: [cite: 675, 677]
$$\frac{\mu_0}{\epsilon_0} \propto \frac{L}{C}$$ [cite: 678]
We know that the time constant for an $LR$ circuit is $\tau = \frac{L}{R}$ and for a $RC$ circuit is $\tau = RC$[cite: 679]. Equating these time dimensions: [cite: 679]
$$\frac{L}{R} = RC \implies \frac{L}{C} = R^2$$ [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 $\sqrt{\frac{\mu_0}{\epsilon_0}}$ represents the intrinsic impedance of free space, which has the value $\approx 377\ \Omega$[cite: 679]. Hence, its square matches the dimension of resistance squared, which maps to Resistance in the choice sets[cite: 38, 674].
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q2 (2025)
Arrange the following in the ascending order of wavelength $(\lambda)$ :
(A) Microwaves $(\lambda_{1})$
(B) Ultraviolet rays $(\lambda_{2})$
(C) Infrared rays $(\lambda_{3})$
(D) X-rays $(\lambda_{4})$
Choose the most appropriate answer from the options given below :
- $\lambda_{4} < \lambda_{3} < \lambda_{2} < \lambda_{1}$
- $\lambda_{3} < \lambda_{4} < \lambda_{2} < \lambda_{1}$
- $\lambda_{4} < \lambda_{2} < \lambda_{3} < \lambda_{1}$
- $\lambda_{4} < \lambda_{3} < \lambda_{1} < \lambda_{2}$
### Core Logic
The order of components in the electromagnetic spectrum in increasing order of wavelength ($\lambda$) is:
$$\gamma\text{-rays} < \text{X-rays} < \text{U.V. rays} < \text{Visible rays} < \text{IR rays} < \text{Microwaves} < \text{Radio waves}$$
Given components:
- Microwaves: $\lambda_1$
- Ultraviolet rays: $\lambda_2$
- Infrared rays: $\lambda_3$
- X-rays: $\lambda_4$
Comparing these yields:
$$\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$$
### 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).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Waves
