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course PHY 202
`aGOOD STUDENT SOLUTION
We are given that the spectrum is from 400-750 nm. We are also given that the screen is 2.3 meters away and that the grating is 7500 lines/cm. To find this I will find where 400 nm wavelength falls on the screen and also where 750 nm wavelength falls onto the screen. Everything in between them will be the spectrum. I will use the formula...
sin of theta = m * wavelength / d
since these are first order angles m will be 1.
since the grating is 7500 lines/cm, d will be 1/7500 cm or 1/750000 m.
Sin of theta(400nm) =
1 * (4.0 * 10^-7)/1/750000
sin of theta (400nm) = 0.300
theta (400nm) = 17.46 degrees
This is the angle that the 1st order 400nm ray will make.
sin of theta (750nm) = 0.563
theta (750nm) = 34.24 degrees
This is the angle that the 1st order 750 nm ray will make.
We were given that the screen is 2.3 meters away. If we draw an imaginary ray from the grating to to the screen and this ray begins at the focal point for the rays of the spectrum and is perpendicular to the screen (I will call this point A), this ray will make two triangles, one with the screen and the 400nm angle ray and one with the screen and the 750 nm angle ray. Using the trigonomic function; tangent, we can solve for the sides of the triangles which the screen makes up.
Tan of theta = opposite / adjacent
tan of 34.24 degrees = opposite / 2.3 meters
0.6806 = opposite / 2.3 meters
opposite = 1.57 meters
tan of 17.46 degrees = opposite / 2.3 meters
opposite = 0.72 meters
So from point A to where the angle(400nm) hits the screen is 0.72 meters.
And from point A to where the angle(750nm) hits the screen is 1.57 meters.
If you subtract the one segment from the other one you will get the length of the spectrum on the screen.
1.57 m - 0.72 m = 0.85 meters is the width of the spectrum on the screen.
CORRECTION ON LAST STEP:
spectrum width = 2.3m * tan (31.33)) - 2.3m * tan (17.45) = 0.68m
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