Question

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are $656\mathrm{nm}$ (red) and $486\mathrm{nm}$ (blue). Light from a hydrogen lamp illuminates a diffraction grating with $500\mathrm{lines}/\mathrm{mm}$, and the light is observed on a screen $1.50\mathrm{m}$ behind the grating. What is the distance between the first-order red and blue fringes?

Short answer

The distance between red and blue fringes is$14.5\mathrm{cm}$.

Step-by-step solution

Formula for distance

The center of the$\mathrm{m}$-th fringe is given by,

${\mathrm{Y}}_{\mathrm{m}}={\mathrm{Ltan\theta }}_{\mathrm{m}}$

$=\mathrm{Ltan}\left({\sin }^{-1}\left(\frac{\mathrm{m\lambda }}{\mathrm{d}}\right)\right)$

localid="1649218383524" $\mathrm{d}$is the distance between gratings

localid="1649218434130" $\mathrm{\lambda }$is wavelength

localid="1649218442071" ${\mathrm{\theta }}_{\mathrm{m}}$is diffraction angle

Calculation for distance

The grating constant is$500\mathrm{lines}/\mathrm{mm}$.

There will be a space between two succeeding gratings.

$\mathrm{d}=\frac{1\times {10}^{-3}}{500}$

$=\underset{¯}{2\times {10}^{-6}\mathrm{m}}$

So,

$\mathrm{\Delta Y}={\mathrm{Ltan\theta }}_{1,\text{red}}-{\mathrm{Ltan\theta }}_{1,\text{blue}}$

$=\mathrm{L}\left[\tan \left({\sin }^{-1}\left(\frac{{\mathrm{\lambda }}_{\text{red}}}{\mathrm{d}}\right)\right)-\tan \left({\sin }^{-1}\left(\frac{{\mathrm{\lambda }}_{\text{blue}}}{\mathrm{d}}\right)\right)\right]$

In terms of numbers,

$\mathrm{\Delta Y}=1.5\left[\tan \left({\sin }^{-1}\left(\frac{0.656}{2}\right)\right)-\tan \left({\sin }^{-1}\left(\frac{0.486}{2}\right)\right)\right]$

$=14.5\mathrm{cm}$