Please give full answer for best answer. Thank you. It is a little hard to see b

Question

Please give full answer for best answer. Thank you. It is a little hard to see but s = .1mm.

A monochromator consider an incident beam on a reflection diffraction grating as in Figure 1.60. Each incident wavelength will result in a diffracted wave with a different angle. We can place a small slit and allow only one diffracted wave lambda m to pass through to the photodetector. The diffracted beam would consist of wavelengths in the incident beam separated (or fanned) out after diffraction. Only one wavelength lambda m will be diffracted favorably to pass through the slit and reach the photodetector. Suppose that the slit is placed so that it is at right angles to the incident beam: theta i + theta m = pi / 2. The grating has a corrugation periodically of 1 mu m. What is the range of wavelengths that can be captured by the photodetector when we rotate the grating from theta i = 1 degree to 40 degree? Suppose that theta i = 15 degree. What is the wavelength that will be detected? What is the resolution, that is, the range of wavelengths that will pass through the slit? How can you improve the resolution? What would be the advantage and disadvantage in decreasing the slit width s?

Explanation / Answer

a)

For a transmission diffraction gratting we have the relation

d*( sin( t(i) ) +sin*( t(m)) ) = k*lambda

(http://en.wikipedia.org/wiki/Diffraction_grating)

For a reflection gratting the relation is the same .

Here d is the grating prtiodicity d =10^-6 m

We take k=1 it means the central diffracted maximum.

ti =1 deg means tm = 89 deg

lambda (min)= 10^-6(sin(1) +sin(89)) =1.19*10^-6 m =1.19 micron

ti =40 deg means tm =50 deg

lambda (max)=10^-6(sin(40)+sin(50)) =1.53*10^-6m =1.53 micron

b) ti =15 deg means tm =75 deg

lambda = 10^-6*(sin(15) +sin(75)) =1.2247 micron

The difference in path between a wave through the center of the slit and at one end of the slit is

(see http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction)

L = s*sin(alpha)

here the angle alpha is coming from

tan(alpha) = (s/2)/R =s/(2R)

Because alpha is small we can write

tan(alpha) = sin(alpha) = s/(2R)

Therefore

the path difference betwen wave arriving at the slit is

L =s*sin(alpha) = s^2/(2R)

The condition for maxima on the detector is

L= m*delta(lambda)

s^2/(2R) =m*delta(lambda) (1)

for m =1 we have

delta(lambda) = s^2/(2R) = (10^-4)^2/(2*0.05) =10^-7 m =0.1 micron

The resolution of the slit in wavelength is 0.1 microns. The range of the wavelengths detected is

lambda(min) =1.225-0.05 =1.175 micron

lambda (max) =1.225+0.05 =1.275 micron

To improve the resolution delta(lambda) one need to decrease the slit width s. (See equation (1)). Decreasing the slit s will have the disatvantege of decreasing the amplitude of the wave that need to be detected but it will improve the uncertainity in wavelength.

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