Question

$FIGUREP33.49$shows the interference pattern on a screen $1.0m$behind an $800line/mm$diffraction grating. What is the wavelength (in $mm$) of the light?

Short answer

The wavelength's light is$\lambda =500nm.$

Step-by-step solution

Step: 1 Interference pattern:

When parallel plane waves of the same frequency meet at an angle, a simple sort of interference pattern results. Interference is fundamentally a process of energy redistribution. The energy wasted during destructive interference is recovered during positive interference.

Step: 2 Equating equation:

The angle of fringes as

$$\begin{gathered} Y_1=L\tan \left({\theta }_1\right) \\ {\theta }_1={\tan }^{-1}\left(\frac{Y_1}{L}\right) \end{gathered}$$

Equation of grating wavelength is

localid="1649148192850" $d\sin \left({\theta }_1\right)=\lambda =\frac{1\times {10}^{-3}}{800}\sin \left({\tan }^{-1}\left(\frac{Y_1}{L}\right)\right)$

Step: 3 Obtaining wavelength:

The wavelength to be

$$\begin{gathered} \lambda =\frac{1\times {10}^{-3}}{800}\sin \left({\tan }^{-1}\left(\frac{0.436}{1}\right)\right) \\ \lambda =1.25\times {10}^{-6}\sin \left({23.56}^{\circ }\right) \\ \lambda =5.00\times {10}^{-7}\mathrm{m} \\ \mathrm{\lambda }=500\mathrm{nm}. \end{gathered}$$