A light ray with wavelength lambda = 500 nm is incident on the air/diamond inter

Question

A light ray with wavelength lambda = 500 nm is incident on the air/diamond interfaces shown below. Calculate the angle of the reflected, and refracted, light rays R_1, R_2. theta_r = 40 degree, theta_R = 38 degree theta_r = 45 degree, theta_R = 23.7 degree theta_r = 45 degree, theta_R = 18 degree theta_r = 40 degree, theta_R = 27 degree For the light ray in problem #3, calculate the following properties: 1) wavelength in diamond, lambda_0; 2) speed of light ray in diamond; and 3) the energy of the light ray in the diamond, in electron volts. Lambda_0 = 207 nm; v = 1.24 times 10^8 m/s; epsilon = 6 eV Lambda_0 = 250 nm; v = 1.8 times 10^8 m/s; epsilon = 4 eV Lambda_0 = 207 nm; v = 2.2 times 10^8 m/s; epsilon = 2.5 eV Lambda_0 = 203 nm; v = 1.9 times 10^8 m/s; epsilon = 3 eV

Explanation / Answer

(3)

Using Snell's law we have,

sini/sinR = n2/n1 , where i is the angle of incidence

or, sin450/sinR = 2.42

or, R = 170

Since, angle of incidence is equal to angle of refraction so,

r = 450

correct option would be b)

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4)

Speed of the light ray in air is,

v = 3X108 m/s

so frequency of the light is f = v/ = (3X108 m/s)/(500X10-9m) = 6X1014 Hz

So speed in diamond is,

vd = v/n2  = (3X108 m/s)/(2.42)

or, vd = 1.24X108m/s

So wavelength in diamond is,

d = vd/f = (1.24X108m/s)/(6X1014 Hz)

or, d = 207 nm

So, the correct option is a).

P.S. - Amplitude of wave is necessary to calculate energy of the wave.

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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