Light containing two different wavelengths passes through a diffraction grating
ID: 1433487 • Letter: L
Question
Light containing two different wavelengths passes through a diffraction grating with 1000 slits/cm. On a screen 17.0 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first order maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light? (Hint: Use the small angle approximation.)
_______nm (longer wavelength) <------- I got 82.66666nm but it's off by 10%
_______nm (shorter wavelength) <--------I got 496.2nm but it's 100% off
Explanation / Answer
This is an exercise or little complicated and as you have already tried to do we will do it step by step.
We seek the separation distance between the slits
d = 1/ 1000 = 1 10-3cm
We look for each maximum angles that give us suggestion and use the small angles which means
Sin =
The consecutive maximum length of the longest wavelength (L) are spaced 0.844 cm, seek the angle trigonometry
Tan = 0.844/17 = 0.04964
= 2.842 deg deg symbol for degrees
as they are consecutive they must correspond to the diffraction orders n y n+1
By analyzing the expression of diffraction
d Sin = m
d = m
With calculated data can be found
Data
n= 2
= 2.842
d sin = n
L= d Sin / n
L= 1 10-3Sin 2.842 /2
L= 0.02479 10-3cm
1m = 102cm = 109 nm
1 cm = 107 nm
L= 247.9 nm
As we expression to a linear form, the first maximum of long wavelength is at an angle that is half the value found, remember that all these expressions the angle must be in radians, (very important)
2 = 2.842 deg ( rad/180deg)
2 = 0.0496 rad
1 = 0.0248 rad
Now he tells us that the maximum short wavelength (S ) is in the middle
3 = 0.0124006
n=3
d = n
= d / n
S= 1 10-30.0124006 / 3
S= 0.00413352 10-3cm (107nm/1 cm)
S= 41.3352 nm
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