Chapter 33: Q. 47 (page 957)

$(a)$ 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 $\tan θ \approx \sin θ \approx θ$ to be valid. Your expression should be in terms of $d, L$ and $λ$ .
$(b)$ . 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 $∆ λ$ .
$(c)$ 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 $333 l i n e s / m m$ grating and a detector with $100 p i x e l s / m m$ located $12 c m$ behind the grating. The resolution of a spectrometer is the smallest wavelength separation $∆ λ_{m i n}$ that can be measured reliably. What is the resolution of this spectrometer for wavelengths near localid="1649156925210" $550 n m$ , 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

Expert verified

Part $(a)$

$(a)$ The position of expression is $y_{1} = \frac{L λ}{d} .$

Part $(b)$

$(b)$ The seperation expression is $Δ y_{1} = \frac{L}{d} Δ λ .$

Part $(c)$

$(c)$ The seperation of smallest wavelength is $Δ λ_{min} = 0.25 nm .$

Step by step solution

Step: 1 Position expression: (part a)

In diffraction,the bright fringe has

$\sin (θ_{m}) = \frac{m λ}{d}$

Applying small angle as

$θ_{1} = \frac{λ}{d}$

The first spectral line position is

role="math" localid="1649130878882" $\begin{array}{l} y_{1} = L \tan (θ_{1}) \\ y_{1} = L θ_{1} = \frac{L λ}{d} . \end{array}$

Step: 2 Seperation expression: (part b)

The total differentiation gives and smallelements are replaced as

$d y_{1} = \frac{L}{d} d λ Δ y_{1} = \frac{L}{d} Δ λ .$

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, $∆ y_{1, m i n}$ equals one pixel width, which may be computed as follows:

$Δ y_{1, min} = \frac{1 mm}{100} = 0.01 mm$

and the width between the grating lines is

$d = \frac{1 mm}{333} d = 0.003 mm .$

Finally, we could reassemble the equation from section $(b)$ and apply it to get $∆ λ_{m i n}$ .

$\begin{array}{l} Δ λ_{min} = \frac{d Δ y_{1, min}}{L} \\ Δ λ_{min} = \frac{(0.003 mm) \times (0.01 mm)}{120 mm} \\ Δ λ_{min} = 0.25 \times 10^{- 6} mm \\ Δ λ_{min} = 0.25 nm . \end{array}$