Ok. The explanation for why we don't see the n=2 beam.
This is the beam that we would expect to lie between the 2 that we have in the question. The n=1 and the n=3.
So far we have the
n=1 beam at 14
o 55' and the
n=3 beam at 50
o 40'
So where should the n=2 beam be. At what angle? Somewhere between those two angle values, of course.
If you apply the formula for the grating again we can find out exactly where it should be.
d sin θ = nλ will do it, putting
n=2 in the formula
we know
λ is 430nm from the question
we know
d from the calculation we did before with the same formula for n=1 and angle 14.92deg
So, work out what the angle is for n=2
Answer:________
Now the theory for why it's missing.In a
single slit diffraction, as opposed to this multi slit diffraction grating diffraction, there are no sharp beams, but rather a series of so called fringes. A light and dark alternating pattern. The theory for single slits (I will leave you to look this up) says that the dark parts (where there is no light) of the light/dark fringe pattern are separated by a distance on the screen such that they form an angle where
sin θ = λ/a where
a is the actual width of the single slit. (We are given this in the question. This is why. )
When you do a multi-slit diffraction grating experiment, the bright beams you see, are actually contained in an 'envelope' determined by the single slit pattern of the individual slits. The diagram here shows this:
The dotted line shows the diffraction pattern due to a single slit, and the vertical bars show the bright beams you get with the grating.
The bright beams have to fall inside the dotted curve. The vertical height of the graph shows the brightness of the light at any angle.
In
this diagram, the n=1 and n=2 beams fall inside the curve and would be visible. In our question it is similar but we need to do a calculation first to see where these points are in our example.
Now we know where the n=2 beam
should be from the calculation above.
We need to look at where that 'minimum' is on the dotted line, to the right of centre, at the position given by that other formula, sin θ = λ/a
At that point, if there is a bright beam expected, it will
not appear, as
it occurs at a minimum of the envelope curve, due to the diffraction pattern of the single slits that make up the grating.
So sub the values we know into this second formula to find t
he angle theta at which this minimum occurs.
λ is 430nm and a = 0.83 μm as given in the question.
What do we find? I think you may already have guessed.
If the two coincide to within a degree or 2, we have our explanation.