Question

Hot hydrogen gas, like that found in star formation regions, emits light only at very specific wavelengths in the visible part of the spectrum: λα = 656.3nm (red), λβ = 486.1nm (teal), λγ = 434.1nm (blue) and λδ = 410.2nm (violet). If you wanted to resolve the first order maxima of these four wavelengths into distinct spots on your screen using a double slit diffraction grating, what values for a and d would you choose? Assume that the distance from 7 the double slit to the screen is 2.00m and the spots are resolved when there is a minimum of 1mm between any two lines. Justify your answer.

Answer 1

we have a double slit diffraction grating

double slit will produce maxima for

dSin = m ( for m = 0, +/-1 , .....) - d is slit seperation

and minima for

dSin = (m+1/2) ( for m = 0, +/-1 , .....)

a is the slit width and the diffraction pattern will produce diffraction minima for

aSin = m ( for m = +/-1 , .....)

to get the lines resolved clearly first order we shall have the

interference minima coincide with the diffraction minima

being small we have Sin = Tan = y/L , y is the distance on the screen from center and L is the distance of screen from the slits

ay/L = for first order diffraction minima

dy/L = 3/2 first order interference minima

a = 2d/3 we shall have

= 410.1 nm and = 434.1 nm lines are resolved we will have all other lines resolved as they are the closest.

suppose we have the first order interference maxima line at

y = L/d = 2.0 * 410.1e-9/d = 820.1e-9 /d

we can have (434.1 nm)  line at 1 mm from the above

y +1.0 e-3 = 2.0 * 434.1e-9 / d = 868.2e-9 /d

d = 48.0 e-6 m

first order maxima of 410.1 will be at

y = 2.0 *410.1e-9 /48e-6 = 17.08 mm

and maxima of 434 will occur at

y =  L/d = 18.09 mm

Diffraction minima of 410.1 shall be at

y = L/a = 2.0 *410e-9/a = 17.58 mm

a = 46.66 e-6 m