Lternating quarter-wave layers of refractive indices 1.6 and 1.7 are used to reduce the reflectivity from a substrate ns = 3.5 at 1000 nm light wavelength and normal incidence in air. determine the number of pairs (integer) and the order of the layers that will result in the minimum reflectivity. what is the minimum reflectivity?

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared. 

The sunlight shines uv radiation on earth,a microwave oven radiates waves on food