CBSE · Class 12 · Physics · Electromagnetic WavesLight of wavelength $500\text{ nm}$ falls on a plane reflecting surface. What is the wavelength and frequency of the reflected light? Also, if the total energy of the electromagnetic wave is $E = 15\text{ J}$, find the peak value of the electric field if the amplitude of the magnetic field is $B0 = 2 \times 10^{-7}\text{ T}$ (assume appropriate theoretical context if required, and solve the numerical step-by-step).

Step 1: Wavelength and Frequency of Reflected Light\nGiven the wavelength of incident light, $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m} = 5 \times 10^{-7}\text{ m}$.\nWhen light is reflected from a surface, its speed and wavelength remain unchanged in the same medium. Therefore, the wavelength of the reflected light is:

$$\lambda' = 500\text{ nm} = 5 \times 10^{-7}\text{ m}$$ \nThe frequency $\nu$ of the light wave is calculated using the relation: $$\nu = \frac{c}{\lambda}$$\nwhere $c = 3 \times 10^8\text{ m/s}$ is the speed of light. $$\nu = \frac{3 \times 10^8\text{ m/s}}{5 \times 10^{-7}\text{ m}}$$ $$\nu = 0.6 \times 10^{15}\text{ Hz} = 6 \times 10^{14}\text{ Hz}$$\nThus, the frequency of the reflected light is $6 \times 10^{14}\text{ Hz}$ (frequency also remains unchanged during reflection).

Step 2: Peak Value of Electric Field ($E_0$)\nThe relation between the peak value of the electric field ($E_0$) and the peak value of the magnetic field ($B_0$) in an electromagnetic wave is given by:

$$\frac{E_0}{B_0} = c$$ \nGiven: $B_0 = 2 \times 10^{-7}\text{ T}$ $c = 3 \times 10^8\text{ m/s}$ \nSubstitute the values into the formula: $$E_0 = c \times B_0$$ $$E_0 = (3 \times 10^8\text{ m/s}) \times (2 \times 10^{-7}\text{ T})$$ $$E_0 = 60\text{ V/m}$$

Final Answer:

• Wavelength of reflected light = $500\text{ nm}$
• Frequency of reflected light = $6 \times 10^{14}\text{ Hz}$
• Peak value of electric field ($E_0$) = $60\text{ V/m}$