A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?
1 answer:
Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.
The ruling separation is d=1/(470mm-1)
Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.
Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.
we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is,find the greater integer value of m for which mλ<d.
since,d/λ
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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Complete Question
The complete question is shown on the first uploaded image
Answer:
a it is always zero
b 0
c
Explanation:ss
Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero
Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero
Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor
So We can mathematically define the charge density on the surface of the electric field as
∮
Where E is the electric field
change in unit area
is the negative charge
is the permittivity of free space
So
Where is the charge density
