D3-1. Red light with a free-space wavelength of 650 nm and blue light with a fre

Question

D3-1. Red light with a free-space wavelength of 650 nm and blue light with a frequency of 7.32x10" Hz are traveling in air. Find the frequency of the red light and the wavelength of the blue light. Which has the longer wavelength? Which has the higher frequency? [4.62x10"Hz, 410 nm; red has longer wavelength, blue has higher frequency) D3-2. A beam of light is traveling in glass with an index of 1.5. The wavelength of the light in the glass is 400 [5x10" Hz, 2x10 m/s, 600 nm] D3-3. A beam of light is traveling in air and enters a piece of glass with an index of 1.6 at an angle of incidence nm. Find the frequency of the light, the speed of the light in the glass, and its free-space wavelength. of 20°. The wavelength of the light in air is 520 nm. Find the angle that the transmitted light beam makes with the surface normal as well as the wavelength and frequency of the light inside the glass. 12.3, 325 nm, 5.77x10 Hz D3-4. Light in air hits a piece of glass with an angle of incidence of 30°. Find the angle of reflection of the reflected light and the angle of refraction for the transmitted light if the index of the glass is 1.57. [30°, 18.6

Explanation / Answer

d3-1)

We know that,

c = f lambda => f = c/lambda

lambda = c/f

f(red) = 3 x 10^8/(650 x 10^-9) = 4.62 x 10^14 Hz

lambda(blue) = 3 x 10^8/(7.32 x 10^14) = 410 x 10^-9 m = 410 nm

Hence, f(red) = 4.62 x 10^14 Hz ; lambda(blue) = 410 nm

2)We know that

n = c/v => v = c/n

v = 3 x 10^8/n = 3 x 10^8/1.5 = 2 x 10^8 m/s

lambda = n lambda' = 1.5 x 400 = 600 nm

f = 3 x 10^8/600 x 10^-9 = 5 x 10^14 Hz

Hence, f = 5 x 10^14 Hz ; v = 2 x 10^8 m/s ; lambda = 600 nm

3)lambda' = lambda/n

lambda' = 520/1.6 = 325 nm

f = 3 x 10^8/520 x 10^-9 = 5.77 x 10^14 Hz

from snell's law

n1 sin(theta1) = n2 sin(theta2)

sin(theta2) = 1 x sin20/1.6 =0.214

theta2 = sin^-1(0.214) = 12.3 deg

Hence, theta2 = 12.3 deg ; lambda = 325 nm ; f = 5.77 x 10^14 Hz

4)from snell's law

n1 sin(theta1) = n2 sin(theta2)

sin(theta2) = 1 x sin30/1.57 =0.318

theta2 = sin^-1(0.318) = 18.6 deg

again using Snells law

n2 sin(theta2) = n3 sin(theta3)

theta3 = sin^-1(1.57 x sin18.6/1) = 30 deg

Hence, theta3 = 30 deg and theta2 = 18.6 deg

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