You deposit a thin film of magnesium difluoride on a glass lens (n > 1.60), reducing the reflection of yellow light, at normal incidence, to a minimum. you find that the thinnest coating that accomplishes this is 106 nm thick. the index of refraction for mgf2 for yellow light (λ = 585 nm) is
a. 1.50.
b. 1.38.
c. 1.15
d. 1.00.
e. 0.707.

To solve this problem it is necessary to apply the concepts related to destructive interference.

The concept refers to an overlap of two or more waves of identical or similar frequency that, when interfering, creates a new pattern of waves of lower intensity

By definition destructive interference is given by

Where,

n=integer (1,2,3,4,5,6...etc)

t = thickness

M= Index of refrqaction

For minimum thickness to satisfy this condition n will be minimum there,

n=0

Therefore

Solving to find M,

Therefore the correct answer is B. 1.38

answer: (431,2)

step-by-step explanation:

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.