Apiece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. it is illuminated with light of wavelength 680 nm. determine the minimum thickness of the coating that will result in no reflection

123.18 nm

Explanation:

Given:

Let t be the thickness of the film. The relation between the wavelength of light  in film and thickness of the film  in order to have no reflection is given  by

Now According to question we have

where n is the refractive index of the film.

Now thickness t is given by

Hence the thickness of the film is calculated in order to have no reflection.

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

f=ma newton's second law 7x8=56n