Question

White light$\left(400-700nm\right)$incident on a $600line/mm$diffraction grating produces rainbows of diffracted light. What is the width of the first-order rainbow on a screen $2.0m$behind the grating?

Short answer

On a screen behind the grating, the breadth of the rainbow is$0.43m.$

Step-by-step solution

Step: 1 Diffraction grating:

A optical device is an device that splits multicoloured light into its beam is split (disperses colors). The polychromatic light received on the grating is scattered, ends up in each wavelength becoming reflected at a replacement angle from the grating.

Step: 2 Finding position:

The fringe position by

$Y_m=L\tan \left({\theta }_m\right)$

The angle by

localid="1649097338452" $$\begin{gathered} d\sin \left({\theta }_m\right)=m\lambda \\ \Rightarrow {\theta }_m={\sin }^{-1}\left(\frac{m\lambda }{d}\right) \\ {Y}_m=L\tan \left({\sin }^{-1}\left(\frac{m\lambda }{d}\right)\right) \end{gathered}$$

Step: 3 Finding the width:

Two variable wavelength by

$\mathrm{\Delta }{Y}_m=L\left[\tan \left({\sin }^{-1}\left(\frac{m{\lambda }_2}{d}\right)\right)-\tan \left({\sin }^{-1}\left(\frac{m{\lambda }_1}{d}\right)\right)\right]$

where,

localid="1649148081063" $$\begin{gathered} d=1\mathrm{mm}/600 \\ d=5/3\times {10}^{-6}\mathrm{m}. \end{gathered}$$

The rainbow width by

localid="1649156772212" $$\begin{gathered} \mathrm{\Delta }{Y}_1=2\times \left[\tan \left({\sin }^{-1}\left(\frac{7\times {10}^{-7}}{5/3\times {10}^{-6}}\right)\right)-\tan \left({\sin }^{-1}\left(\frac{4\times {10}^{-7}}{5/3\times {10}^{-6}}\right)\right)\right] \\ \mathrm{\Delta }{Y}_1=0.43\mathrm{m}. \end{gathered}$$