Question

If it takes light 4 ns to travel 0.960 m in an optical cable, what is the index of refraction of the cable?

What is the wavelength of the radio signal emitted by an AM station broadcasting at 1300 kHz? Radio waves travel at the speed of light.

A transparent material is known to have an index of refraction equal to 2.4. What is the wavelength of light in this material if it has a wavelength of 620 nm in a vacuum?

Answer 1

1) v = d/t = 0.96/4*10^-9 = 2.4*10^8 m/s

n = c/v = 3*10^8/2.4*10^8 = 1.25

2) c = lamda*f

==> lamda = c/f = 3*10^8/1300*10^3 = 230.8 m

3) lamda' = lamdao/n = 620/2.4 = 258.33 nm

Answer 2

The velocity of the light is the distance divided by the time taken to travel the distance;

v = d/t
v = 0.960 / 4x10^-9
v = 2.4x10^8 metres per second

The refractive index of air x the speed of light in air = refractive index of cable x speed of light in cable

The refractive index of air = 1.000, speed of light in air is 3x10^8 metres per second

(1.000 x 3x10^8) = refractive index of cable x2.4x10^8

Refractive index of cable = (3x10^8) / (2.4x10^8)

Refractive index of cable = 1.25

2.wavelenth=c/f=(3*10^8)/(1.3*10^6)=230.76 m

3.lambda' will be lambda/n

so wavelength=620 nm/2.4=258.33 nm