Question

The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating having 750 lines/mm, over what range of angles does the visible m = 1 spectrum extend?

Answer 1

Answer:

The range of angles is from 17.50° to 31.76°

Explanation:

The diffraction grid equation is as follows:

$dsen\theta=m\lambda$

Clearing for $\theta$

$sen\theta=\frac{m\lambda}{d}$

$\theta=sen^{-1}(\frac{m\lambda}{d})$

where $\theta$ is the angle, $m$ is the order, in this case $m=1$, $\lambda$ is the wavelength, and $d$ is defined as follows:

$d=\frac{1}{resolution}$

and since the resolution is 750 lines/mm wich is the same as $750lines/1x10^{-3}m$

$d$ will be:

$d=\frac{1}{750lines/1x10^{-3}m}=\frac{1x10^{-3}m}{750lines}=1.33x10^{-6}m$

wich is the distance between each line of the diffraction grating.

substituting the values for $m$ and $d$:

$\theta=sen^{-1}(\frac{(1)\lambda}{(1.33x10^{-6}m)})$

And we need to find two angle values: one for when the wavelength is 400nm and one for when it is 700 nm. So we will get the angle range

$\theta=sen^{-1}(\frac{(400x10^{-9})}{(1.33x10^{-6}m)})=17.50$

and

$\theta=sen^{-1}(\frac{(700x10^{-9})}{(1.33x10^{-6}m)})=31.76$

The range of angles is from 17.50° to 31.76°