Question

Helium atoms emit light at several wavelengths. Light from a helium lamp illuminates a diffraction grating and is observed on a screen$50.00cm$behind the grating. The emission at wavelength $501.5nm$creates a first-order bright fringe $21.90cm$from the central maximum. What is the wavelength of the bright fringe that is $31.60cm$from the central maximum?

Short answer

The luminous fringe from the central maxima has a wavelength,$\lambda =667.8\mathrm{nm}.$

Step-by-step solution

Step: 1 Finding the space grating:

Equation of grating by

$d\sin \left({\theta }_m\right)=m\lambda$

Grating space $d$is constant as

$\frac{m_a{\lambda }_a}{\sin \left({\theta }_{m,a}\right)}=d=\frac{m_b{\lambda }_b}{\sin \left({\theta }_{m,b}\right)}.$

Step: 2 Equating equation:

We have to find $\sin \left({\theta }_{m,a}\right),\sin \left({\theta }_{m,b}\right)$as easy as follows by

role="math" localid="1649090960925" $$\begin{gathered} {\lambda }_{\mathrm{b}}=\frac{{\mathrm{m}}_{\mathrm{a}}{\mathrm{\lambda }}_{\mathrm{a}}\sin \left({\mathrm{\theta }}_{\mathrm{m},\mathrm{b}}\right)}{{\mathrm{m}}_{\mathrm{b}}\sin \left({\mathrm{\theta }}_{\mathrm{m},\mathrm{a}}\right)} \\ {Y}_{\mathrm{x}}=L\tan \left({\mathrm{\theta }}_{\mathrm{m},\mathrm{x}}\right) \\ \Rightarrow {\theta }_{\mathrm{m},\mathrm{x}}={\tan }^{-1}\left(\frac{{\mathrm{Y}}_{\mathrm{x}}}{\mathrm{L}}\right). \end{gathered}$$

Step: 3 Calculating wavellength:

We may proceed to identify the unknown wavelength for our numerical example after being told explicitly that $m_a=1$and assuming that $m_b=1$as well (since this issue would have an infinite number of solutions if m b were not determined):

localid="1649156207047" $\begin{array}{l}{\lambda }_b=\frac{1\times 5.015\times {10}^{-7}\times \sin \left({\tan }^{-1}\left(\frac{0.316}{0.5}\right)\right)}{1\times \sin \left({\tan }^{-1}\left(\frac{0.219}{0.5}\right)\right)}\\ \begin{array}{l}{\lambda }_b==\frac{5.015\times {10}^{-7}\times \sin \left({32.29}^{\circ }\right)}{\sin \left({23.65}^{\circ }\right)}\\ \lambda =667.8\mathrm{nm}.\end{array}\end{array}$