- #1

Abu

## Homework Statement

White light containing wavelengths from 400 nm to 750 nm falls on a grating with 6000 lines/cm. How wide is the first-order spectrum on a screen 2.0 meters away

## Homework Equations

dsinθ=mλ for constructive interference

dsinθ=(m+1/2)λ for destructive interference

Δx = λL/d

## The Attempt at a Solution

I am having trouble deciding which formula to use. From my understanding of the question, we are not talking about the width of the central spectrum, but the first order, which is either the first constructive/destructive point above or below the central (lets just say above).

Different wavelengths will go to different spots on the screen, right? I think my main problem is attempting to picture this scenario.

If we were imagining that the first order spectrum was constructive interference, it would be surrounded by a destructive point above and below it, correct? So my first thought was to find the angles of each wavelength to those points but I have a feeling that is just ridiculous and hard to comprehend.

If I were to mindlessly use what I have in the equations, I would probably do:

dsinθ=mλ

(1/6000 *10^-2)sinθ = 1(400*10^-9(

θ = 13.8 degrees

dsinθ=mλ

(1/6000 *10^-2)sinθ = 1(750*10^-9)

θ = 26.7 degrees

For one, I don't even know what these angles mean, even if they are correct.

Two, I don't know if my idea to find the destructive points around the first order is correct or not

Three, I don't know how to picture this example. Is it saying that one end of the first order is the 750 wavelength and the other end of the first order is the 400 wavelength?

Sorry if my attempt seems lackluster, I'm trying my best.

Thank you.