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

Answer 1

(3)

Using Snell's law we have,

sinθ_i/sinθ_R = n₂/n₁ , where θ_i is the angle of incidence

or, sin45⁰/sinθ_R = 2.42

or, θ_R = 17⁰

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

θ_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/λ = (3X10⁸ m/s)/(500X10^-9m) = 6X10¹⁴ Hz

So speed in diamond is,

v_d = v/n₂  = (3X10⁸ m/s)/(2.42)

or, v_d = 1.24X10⁸m/s

So wavelength in diamond is,

λ_d = v_d/f = (1.24X10⁸m/s)/(6X10¹⁴ Hz)

or, λ_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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