Question

In a diffraction experiment, light of 600.nm wavelength produces a first-order maximum 0.350 mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source?

Answer 1

The eqn for a diffraction grating is    n λ = d sin θ.

Our method here is to use the first set of given information to calculate d, the distance between adjacent rulings on the grating.   Then after knowing d, we can find the requested λ in the second part of the given information.

STEP 1:        sin θ = y / L ,   where y is distance from central max out to the n= 1 max , and L is distance from the grating to the screen.  

n λ = d y / L        so     (1) (600 nm) = d ( 0.350 mm) / L   both d and L are unknown, so we can find the ratio of d/L.    The only tricky part in this eqn is that nm and mm are different units, so we must make them the same:             (1) ( 0.0006 mm) =   (0.350 mm) d/L  

Thus,   d/L = .0006 / .350 = 0.0017143   there are no units on this ratio.

STEP 2:      n λ = d y / L = y (d/L)

                (3) λ = (0.870 mm) (0.0017143)     solve for wavelength to get

                     λ = 0.00049714 mm   or 497.14 nm   final answer.