Question

A diffraction grating having 550 lines/mm diffracts visible light at 37°. What is the light's wavelength?

Answer 1

Given:

Diffraction grating having the lines, \(n=550\) lines \(/ \mathrm{mm}\)

diffraction occurs at anangle to visible light, \(\theta=37^{\circ}\)

$$ \text { thus, } \begin{aligned} d &=\frac{1}{550 \text { linesimm }}\left(\frac{10^{6} \mathrm{~nm}}{1 \mathrm{~mm}}\right) \\ &=1818.18 \mathrm{~nm} \text { per line } \end{aligned} $$

Solution:

It is known by the formula for diffraction, \(d \sin \theta=p \lambda\)

Here , \(p\) running form \(1,2,3\)...etc.,

\((1818.18 \mathrm{~nm})\left(\sin 37^{\circ}\right)=(1)(\lambda)\)

Hence, for \(p=1\), wave lenght,

$$ \lambda=1094.20 \mathrm{~nm} $$

But this is not lies in the range of visible wavel ength now, \(\quad n=2\)

$$ \begin{aligned} (1818.18 \mathrm{~nm})\left(\sin 37^{\circ}\right) &=(2)(\lambda) \\ \lambda &=547.10 \mathrm{~nm} \end{aligned} $$

since the wavel enght range of visible is from \(400 \mathrm{~nm}\) to \(750 \mathrm{~nm}\)

Hence, Required visible wavelenght for diffraction is

$$ \lambda=547.10 \mathrm{~nm} $$

since which is in the wavel enght ragne of visible