A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The fifth-order (m=5) maximum of the unknown wavelength exactly overlaps the third-order (m=3) maximum of red light.

What is the unknown wavelength?

402 nm .

Explanation:

Let the wave length of unknown light be λ .

For fifth order maxima the condition is

path difference at the point of fringe = 5 λ

For red light of 670 nm , for 3 rd order maxima

path difference at the point of fringe = 3  x wave length

= 3 x 670 nm

From these two equations

5 λ = 3 x 670 nm

λ = 3 x 670 / 5 nm

= 402 nm .

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .