CBSE Class 12 Physics Chapter 15: Electromagnetic Waves NCERT Solutions

The given electric field vector is \vec{E} = \{(3.1 \text{ N/C})\} \cos [(1.8 \text{ rad/m}) \text{ y} + (5.4 \times 10^8 \text{ rad/s})t])^{\hat{i}}.

(a) The general form of a plane electromagnetic wave propagating along the y-axis is \vec{E} = E_0 \cos(ky + \omega t) \hat{i} or \vec{E} = E_0 \cos(ky - \omega t) \hat{i}. Comparing the given equation with the general form \vec{E} = E_0 \cos(ky + \omega t) \hat{i}, we see that the wave is propagating along the negative y-direction.

(b) The wave number k is given as k = 1.8 \text{ rad/m}. The wavelength \lambda is related to the wave number by the formula \lambda = \frac{2\pi}{k}.

Substituting the value of k:

\lambda = \frac{2\pi}{1.8} \approx 3.49 \text{ m}

(c) The angular frequency \omega is given as \omega = 5.4 \times 10^8 \text{ rad/s}. The frequency \nu is related to the angular frequency by \nu = \frac{\omega}{2\pi}.

Substituting the value of \omega:

\nu = \frac{5.4 \times 10^8}{2\pi} \approx 8.6 \times 10^7 \text{ Hz}

(d) The amplitude of the electric field is E_0 = 3.1 \text{ N/C}. The amplitude of the magnetic field B_0 is related to the amplitude of the electric field by B_0 = \frac{E_0}{c}, where c is the speed of light in vacuum (c \approx 3 \times 10^8 \text{ m/s}).

Calculating B_0:

B_0 = \frac{3.1 \text{ N/C}}{3 \times 10^8 \text{ m/s}} \approx 1.03 \times 10^{-8} \text{ T}

(e) The electric field is propagating along the negative y-direction and oscillating along the x-direction. Therefore, the magnetic field must oscillate along the z-direction. The expression for the magnetic field part of the wave is given by \vec{B} = B_0 \cos(ky + \omega t) \hat{k}.

Substituting the values:

\vec{B} = \{(1.03 \times 10^{-8} \text{ T})\} \cos [(1.8 \text{ rad/m}) \text{ y} + (5.4 \times 10^8 \text{ rad/s})t] \hat{k}

The power rating of the bulb is P_{total} = 100 \text{ W}.

It is given that 5% of this power is converted into visible radiation.

Therefore, the power of the visible radiation emitted by the bulb is:

P_{visible} = \frac{5}{100} \times 100 \text{ W} = 5 \text{ W}

Since the radiation is emitted isotropically, it spreads out uniformly in all directions. The intensity I of the radiation at a distance r from the source is given by the power divided by the surface area of the sphere of radius r:

I = \frac{P_{visible}}{4\pi r^2}

(a) At a distance of r = 1 \text{ m} from the bulb:

I_1 = \frac{5 \text{ W}}{4\pi (1 \text{ m})^2} = \frac{5}{4\pi} \text{ W/m}^2 \approx 0.398 \text{ W/m}^2

(b) At a distance of r = 10 \text{ m} from the bulb:

I_{10} = \frac{5 \text{ W}}{4\pi (10 \text{ m})^2} = \frac{5}{400\pi} \text{ W/m}^2 \approx 0.00398 \text{ W/m}^2