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Chapter 32: Problem 42

Find the total number of lines in a 2.5 -cm-wide diffraction grating whose third-order spectrum puts the 656 -nm hydrogen- \(\alpha\) spectral line \(37^{\circ}\) from the central maximum.

Short Answer

Expert verified
To find the final answer, perform the calculations described in the solution steps. This will yield the total number of lines in the 2.5 cm-wide diffraction grating.

Step by step solution

01

Convert Wavelength to meters

Given wavelength of the hydrogen-alpha spectral line is 656 nm, convert it to meters. 1 nm = \(1 \times 10^{-9}\) meters. Hence, the wavelength in meters is \(656 \times 10^{-9}\) m.
02

Convert Angle to radians

Convert the given angle from degrees to radians. To do this, multiply the given value by \(\frac{\pi}{180}\). So, the angle in radians is \(37 \times \frac{\pi}{180}\) rad.
03

Calculate the line Spacing

Plug the values into the grating equation \(d = \frac{m \lambda}{\sin(\theta)}\) where m=3 (third order spectrum), \(\lambda = 656 \times 10^{-9}\) m, and \(\theta = 37 \times \frac{\pi}{180}\) rad. Solve for \(d\) (spacing between grating lines) to get \(d=\frac{3 \times (656 \times 10^{-9})}{\sin(37 \times \frac{\pi}{180})} \) meters.
04

Calculate Number of Lines per Meter

The reciprocal of the line spacing \(d\) gives the number of lines per meter. So the number of lines per meter, often represented as \(N\), is \(\frac{1}{d}\) lines per meter.
05

Calculate Total Number of Lines

The total number of lines in the diffraction grating is the product of the number of lines per meter and the width of the grating in meters. The grating is given to have a width of 2.5 cm, which converts to 0.025 m. Hence, the total number of lines is \(N \times 0.025\).

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Most popular questions from this chapter

Light with wavelength \(633 \mathrm{nm}\) is incident on a \(2.50-\mu \mathrm{m}\) -wide slit. Find the angular width of the central peak in the diffraction pattern, taken as the angular separation between the first minima.

Light is incident on a diffraction grating at angle \(\alpha\) to the normal. Show that the condition for maximum light intensity becomes \(d(\sin \theta \pm \sin \alpha)=m \lambda\).

In deriving the intensity in double-slit interference, why can't you simply add the intensities from the two slits?

A double-slit experiment has slit spacing \(0.035 \mathrm{mm},\) slit-toscreen distance \(1.5 \mathrm{m},\) and wavelength \(490 \mathrm{nm} .\) What's the phase difference between two waves arriving at a point \(0.56 \mathrm{cm}\) from the center line of the screen?

A five-slit system with 7.5 - \(\mu\) m slit spacing is illuminated with 633-nm light. Find the angular positions of (a) the first two maxima and (b) the third and sixth minima.
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