Question

5. (a) A radio station broadcasts at a frequency of 91.5 MHz (1 MHz = 10 Hz). At a receiver some distance from the antenna, the maximum magnetic field of the electromagnetic wave detected is 2.10 x 10-11 T. ) (2 pts) What is the maximum electric field of the electromagnetic wave at this receiver? (ii) (1 pt) What is the wavelength of the electromagnetic wave? (b) Sunlight strikes a piece of crown glass at an angle of incidence of 40.00 (0) (1 pts) What is the angle of reflection? (ii) (3 pts) Calculate the difference in the angle of refraction between red (660 nm) and blue (470 nm) rays within the glass. The index of refraction is n = 1.512 for red and n = 1.524 for blue light (ii) (3 pts) The light now travels inside the glass. What is the minimum angle of incidence at which the red ray can hit the surface of the glass and become totally internally reflected?

Answer 1

5. (a)

Frequency of radio waves, f = 91.5 x 10⁶ Hz

Speed of light, c = 3 x 10⁸ m/s

Magnetic field of the electromagnetic wave at the receiver, B = 2.10 x 10^-11 m

(i)

Maximum electric field at the receiver, E = Bc = 2.10 x 10^-11 x 3 x 10⁸ = 6.30 x 10^-3 N/C

(ii)

The wavelength of the electromagnetic wave, λ = c/f = 3 x 10⁸ /91.5 x 10⁶ = 3.28 m

(b)

(i)

Angle of incidence on crown glass, i=40⁰

Angle of reflection, r = angle of incidence,i

Angle of reflection, r = 40⁰

(ii)

Index of refraction for red, n_R = 1.512

Index of refraction for blue, n_B = 1.524

Using Snell’s law, n = Sin i/Sin r

For red:

Sin r_R = Sin 40/1.512

Angle of refraction for red, r_R = 25.2⁰

For blue:

Sin r_B = Sin 40/1.524

Angle of refraction for blue, r_B = 26.2⁰

The difference in angle of refraction = 26.2⁰ = 25.2⁰ = 1.0⁰

(iii)

The minimum angle of incidence on a denser medium to get the light totally internally reflected is the critical angle C.

Sin C = 1/n = 1/ 1.518 = 0.6587 ( the average value of n_B and n_R is taken as the index of refraction of glass)

Critical angle, C = Sin ^-1 0.6587 = 41.2⁰