Question

Light with a frequency of 6.8×10¹⁴ Hz is travelling through glass and strikes a glass/air interface as shown above. Angle X₁ is 29.0^o to the normal. Some of the light emerges from the glass and some is reflected back into the glass.

air glass X1 X2

a.) If the index of refraction for this glass is 1.36, what is the value of angle X₃?

b.) What is the value of angle X₂?

c.) What is the speed of light in this glass?

d.) What is the wavelength of this light in the glass?

e.) Finally, what is the largest angle that the light in this glass could make with the normal and still produce an emergent ray into air?

Answer 1

Given:
frequency = 6.8×1014 Hz
Angle X1 = 29 degrees
refraction for this glass = 1.36

sin(x3) / sin(x1) = n(glass)/n(air) = 1.36/1 (law of Snellius)
sinx(3) = sin(x1)*1.36
sinx3 = sin(29)*1.36
sinx3 = 0.7847
x3 = 41.24°

sin(x2) / sin(x3) = n(glass)/n(air) = 1.36/1 (law of Snellius)
sinx(2) = sin(x3)*1.36
sinx2 = sin(41.24)*1.36
x2 = 63.70°

speed in air/speed in glass = 1.36/1
3*10^8 m/s / 1.36 = speed in glass = 2.205*10^8 m/s

wavelength = c/f = 2.205*10^8/(6.8*10^14) = 3.24*10^-7 m

x3 is 90° in the border case

sin(x1)/sin(x3) = 1/1.36
sinx1/1 = 1/1.36
sinx1 = 47.33° --- maximum angle