White light containing wavelengths from 400 nm to 750 nm falls on a grating with 8500 lines/cm. how wide is the first-order spectrum on a screen 2.50 m away?

Width = 1.165m

Explanation:

Let us define terms before we start the problem:

Diffraction grating: An array of large number of parallel slits with the same width "a" and spaced equal distance "d."

From the definition, we can use the equation for multiple slits to find the maxima, or where they constructively interfere.

d is the spacing, θ is angle, m is any integer and λ is wavelength

d*sinθ = m*λ

The problem ask us the use the first-order lines, thus m = ±1. We are also told that the grating is 8500 lines/cm. To get the distance d, we have to invert 8500lines/cm. This gives us our grating spacing of 1.176*10^-6 m.

The next step is to solve for θ for both wavelengths. θr will denote the angle for 750nm because it is red light and θv will denote the angle 400nm for violet light

θr = arcsin ((750*10^-9) / (1.176*10^-6)) = 39.6°

θv = arcsin ((400*10^-9) / (1.176*10^-6)) = 19.88°

Next we can use this angle to solve the distance from the center to the first order line. This is done using trig. Let x be the distance from the slit to the screen and y be the distance from the center of the screen to the maxima. This forms a right triangle. That means we can use tangent to solve for y for both wavelengths. x is define in the problem as 2.5 meters

So from above we get:

tanθ = y/x ⇒ xtanθ = y

Substitute both angles in to get yr (red) and yv (violet)

yr = 2.5m*tan (39.6°) = 2.06859m

yv = 2.5m * tan (19.88°) = 0.9038m

Finally to get the width, subtract yr and yv

Width = 2.07 - 0.904 = 1.165m