Question

$\left(a\right)$Find an expression for the positions $y_1$of the first-order fringes of a diffraction grating if the line spacing is large enough for the small-angle approximation $\tan \theta \approx \sin \theta \approx \theta$to be valid. Your expression should be in terms of $d,L$and$\lambda$.
$\left(b\right)$. Use your expression from part a to find an expression for the separation$∆y$on the screen of two fringes that differ in wavelength by$∆\lambda$.
$\left(c\right)$Rather than a viewing screen, modern spectrometers use detectors-similar to the one in your digital camera-that are divided into pixels. Consider a spectrometer with a $333lines/mm$grating and a detector with $100pixels/mm$located $12cm$behind the grating. The resolution of a spectrometer is the smallest wavelength separation $∆{\lambda }_{min}$that can be measured reliably. What is the resolution of this spectrometer for wavelengths near localid="1649156925210" $550nm$, in the center of the visible spectrum? You can assume that the fringe due to one specific wavelength is narrow enough to illuminate only one column of pixels.

Short answer

Part $\left(a\right)$

$\left(a\right)$The position of expression is$y_1=\frac{L\lambda }{d}.$

Part $\left(b\right)$

$\left(b\right)$The seperation expression is$\mathrm{\Delta }{y}_1=\frac{L}{d}\mathrm{\Delta }\lambda .$

Part $\left(c\right)$

$\left(c\right)$The seperation of smallest wavelength is$\mathrm{\Delta }{\lambda }_{min}=0.25\mathrm{nm}.$

Step-by-step solution

Step: 1 Position expression: (part a)

In diffraction,the bright fringe has

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

Applying small angle as

${\theta }_1=\frac{\lambda }{d}$

The first spectral line position is

role="math" localid="1649130878882" $\begin{array}{l}{y}_1=L\tan \left({\theta }_1\right)\\ y_1=L{\theta }_1=\frac{L\lambda }{d}.\end{array}$

Step: 2 Seperation expression: (part b)

The total differentiation gives and smallelements are replaced as

$$\begin{gathered} \mathrm{d}{y}_1=\frac{L}{d}\mathrm{d}\lambda \\ \mathrm{\Delta }{y}_1=\frac{L}{d}\mathrm{\Delta }\lambda . \end{gathered}$$

Step: 3 The smallest wavelength seperation: (part c)

The lowest wavelength separation occurs when $∆y_1$is the smallest, and the smallest separation between$y_1$of two fringes produced by two specific wavelengths is localid="1649156999554" $1$pixel. This is because one fringe can only illuminate one column of pixels, so we're talking about two bright fringes, each of which illuminates one column of pixels, and we're measuring $∆y$from the centre of each column. As a result, $∆y_{1,min}$equals one pixel width, which may be computed as follows:

$\mathrm{\Delta }{y}_{1,min}=\frac{1\mathrm{mm}}{100}=0.01\mathrm{mm}$

and the width between the grating lines is

$$\begin{gathered} d=\frac{1\mathrm{mm}}{333} \\ d=0.003\mathrm{mm}. \end{gathered}$$

Finally, we could reassemble the equation from section $\left(b\right)$and apply it to get $∆{\lambda }_{min}$.

$\begin{array}{l}\mathrm{\Delta }{\lambda }_{min}=\frac{d\mathrm{\Delta }{y}_{1,min}}{L}\\ \mathrm{\Delta }{\lambda }_{min}=\frac{\left(0.003\mathrm{mm}\right)\times \left(0.01\mathrm{mm}\right)}{120\mathrm{mm}}\\ \mathrm{\Delta }{\lambda }_{min}=0.25\times {10}^{-6}\mathrm{mm}\\ \mathrm{\Delta }{\lambda }_{min}=0.25\mathrm{nm}.\end{array}$