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Why is red light deviated more than violet light in diffraction grating? (1 Viewer)

Momentazeus

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So I know that the angle of diffraction is greater but how would you explain it in terms of the equation dsin0 = m(lambda). The issue I am having is that d is constant, red wavelength is greater but then when I rearrange for 0, I know that the angle domain is 0-90 so I can explain it through that but the m is the confusing part. The respective values of m would be different for red and violet right? If so, red's m would be greater and so would I just assume that it is an integer greater than the value of violet's m?
 

ssoans

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in the double-slit equation, m refers to the number of maxima from the central maximum

so an m value of 2 would refer to the second maxima from the middle

1663819707925.png
* second maximum in this diagram

an m value of 3 for a red wavelength (third maximum) would be much further away than an m value of 1 (first maximum) for the violet wavelength

thus, when measuring the deviance, m should be the same value for both red and violet wavelengths ~ otherwise the measurements would be quite unfair

now, knowing that both m and d are constants, the only variable is the different wavelengths, implying that the greater deviance of a red wavelength is due to its larger wavelength



(an example)

rearranging the formula,

m λ / d = sin θ

let m = 1 ~ (the first maxima from the central maximum)

and d = 0.05


for the red wavelength ~~ sin θ = (620*10^-9)/0.05

whilst

for a violet wavlength ~~ sin θ = (380*10^-9)/0.05


thus, θ for the red wavelength (as calculated) is larger than the violet wavelength,
telling us that the red light diffracted more than the violet light

hope this helps!!
 

bruhwth

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dsin@ = mλ

Tan@ = y/L , where y is the distance between two successive bright or dark bands and L is the distance between the slits and the screen

At small angles of @ Tan@ = Sin@ roughly

Hence, dy/L = mλ

d, L and m are constant (let m=1 for both)

Therefore you get the following relationship between wavelength and y: y proportional to λ.

Another way to show the relationship, but definitely ssoans' method is perfect for more quantitative responses.
 

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