Exams › JEE Advanced › Physics

The speed of light in free space is 3 * 10⁸ m/s. An electromagnetic wave has frequency 2 * 10⁶ Hz. Its wavelength equals 25x metres. Find the value of x.

1. 6
2. 12
3. 15
4. 60

Correct answer: 6

Solution

The wavelength of an EM wave is lambda = c/f = (3*10⁸)/(2*10⁶) = 150 m. Given lambda = 25x, we get x = 150/25 = 6.

Related JEE Advanced Physics questions

• A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 × 10⁸ m/s. The final momentum of object is
• A floodlight filtered to pass only red light produces a sinusoidal plane electromagnetic wave whose electric field component is given by Eₓ = 36 sin(1.20 * 10⁷ * z - 3.6 * 10¹⁵ * t) V/m. The average intensity of this beam is closest to sqrt(n) W/m², where n is an integer. Find n.
• The electric field of an electromagnetic wave is E = 20 sin(w*(t - x/c)) j V/m, where w is the angular frequency and c is the speed of light. Calculate the total electromagnetic energy stored in a volume of 5 * 10^(-4) m³. (Given: epsilon₀ = 8.85 * 10^(-12) C²/(N*m²))
• A plane electromagnetic wave of frequency 25 GHz propagates in vacuum along the +z direction. At a particular point in space and time, the magnetic field is B = 5 * 10^(-8) T in the +x direction. What is the corresponding electric field vector E? (Speed of light c = 3 * 10⁸ m/s)
• The electric field component of an electromagnetic wave in vacuum is given by E = (j-hat + b*k-hat) * 10⁻³ * sin[10⁷(x + 2y + 3z - beta*t)] in SI units. Using c = 3 * 10⁸ m/s, which of the following statements are correct?
• The magnetic field of a plane electromagnetic wave is given by B = 3 * 10^(-8) * sin[200*pi*(y + c*t)] * i_hat T, where c = 3 * 10⁸ m/s. What is the corresponding electric field?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →