Chapter 36: Q56P (page 1113)

(a) How many rulings must a 4.00-cm-wide diffraction grating have to resolve the wavelengths 415.496 and 415.487 nm in the second order? (b) At what angle are the second-order maxima found?

Short Answer

Expert verified

The number of rulings is 23100.
The second-order maxima can be found at $θ = 28.7 °$ .

Step by step solution

The resolving power

It is known the the resolving power of a grating is given by $R = Nm = \frac{λ_{avg}}{∆ λ}$ , where $N$ is the number of rulings in the grating and $m$ is the order of the lines, $λ_{avg}$ is the average of wavelengths and $∆ λ$ is the separation.

The number of rulings

It is known that the $N m = \frac{λ_{a v g}}{Δ λ}$ . From the given values in the question, we have

$N (2) = \frac{\frac{415.496 + 415.487}{2}}{415.496 - 415.487} N (2) = \frac{\frac{830.983}{2}}{0.009} N = 23100$

Thus, the number of rulings is 23100.

The second-order maxima

The width of the single grating, $d$ , is given by $d = \frac{L}{N}$ . So, the width can be calculated as follows:

$d = \frac{L}{N} = \frac{0.04}{23100} = 1.732 \times 10^{- 6} m$

The angle of maxima can be calculated as follows:

$\sin θ = \frac{m λ}{d} \sin θ = \frac{2 (415.496 \times 10^{- 9})}{1.73 . \times 10^{- 6}} θ = \sin^{- 1} (\frac{830.992}{1730}) θ = 28.7 °$

Thus, the second-order maxima can be found at $θ = 28.7 °$

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(a) How many rulings must a 4.00-cm-wide diffraction grating have to resolve the wavelengths 415.496 and 415.487 nm in the second order? (b) At what angle are the second-order maxima found?

Short Answer

Step by step solution

The resolving power

The number of rulings

The second-order maxima

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