Q.
Following diffraction pattern was obtained using a diffraction grating using two different wavelengths $\lambda_{1}$ and $\lambda_{2}.$ With the help of the figure identify which is the longer wavelength and their ratios.
A) $\lambda_{2}$ is longer than $\lambda_{1}$ and the ratio of the longer to the shorter wavelength is 1.5
B) $\lambda_{1}$ is longer than $\lambda_{2}$ and the ratio of the longer to the shorter wavelength is 1.5
C) $\lambda_{1}$ and $\lambda_{2}$ are equal and their ratio is 1.0
D) $\lambda_{2}$ is longer than $\lambda_{1}$ and the ratio of the longer to the shorter wavelength is 2.5
Explanation
(c) The equation of $n$th principal maxima for wavelength $\lambda$ is given by
$(a + b)\sin\theta = n\lambda$
Where $a\ $is the width of transparent portion and $b$ is that of opaque portion. The width $(a + b)$ is called the grating element.
The spectral lines will overlap, $ie$, they will have the same angle of diffraction of
$\lambda_{1} = \lambda_{2}$
When a line of wavelength $\lambda_{1}$ in order $n_{1}$ coincides with a line of unknown wavelength $\lambda_{2}$ in order $n_{2},\ $then
$n_{2}\lambda_{2} = n_{1}\lambda_{1}$
Or $\frac{\lambda_{1}}{\lambda_{2}} = \frac{n_{2}}{n_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.