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Chapter 4: Problem 98

For white light \((400 \mathrm{nm}<\lambda<700 \mathrm{nm})\) falling normally on a diffraction grating, show that the second and third-order spectra overlap no matter what the grating constant \(d\) is.

Short Answer

Expert verified
To show that the 2nd and 3rd-order spectra overlap for white light on a diffraction grating, we use the diffraction grating equation and calculate the minimum and maximum angles for both orders. By comparing the angles, we find that \(\sin(\theta_{3,\min}) \le \sin(\theta_{2,\max})\), which holds true regardless of the grating constant d. Therefore, the second and third-order spectra overlap for white light on a diffraction grating, no matter what the grating constant d is.

Step by step solution

01

Understand the diffraction grating equation

The diffraction grating equation relates the diffracted angle, order of diffraction, wavelength and grating constant as follows: \(n\lambda = d\sin(\theta)\) where: - \(n\) is the order of diffraction (in this case, we're looking at the 2nd and 3rd orders), - \(\lambda\) is the wavelength of light, - \(d\) is the grating constant, and - \(\theta\) is the diffracted angle.
02

Find the range of angles for the 2nd and 3rd-order spectra

For the second-order spectrum (\(n=2\)), the diffracted angles (\(\theta_2\)) can be found as: \(2\lambda = d\sin(\theta_2)\) We can rearrange the equation to find the angle: \(\sin(\theta_2) = \frac{2\lambda}{d}\) Similarly, for the third-order spectrum (\(n=3\)), the diffracted angles (\(\theta_3\)) can be found as: \(3\lambda = d\sin(\theta_3)\) And the equation can be rearranged as: \(\sin(\theta_3) = \frac{3\lambda}{d}\) Now, we need to find the range of angles for the 2nd and 3rd-order spectra using the given wavelength range of white light, i.e., 400 nm to 700 nm.
03

Calculate the minimum and maximum angles for the 2nd-order spectrum

For the minimum angle, use the smallest wavelength (\(\lambda = 400\,\mathrm{nm}\)): \(\sin(\theta_{2,\min}) = \frac{2\cdot400\,\mathrm{nm}}{d}\) For the maximum angle, use the largest wavelength (\(\lambda = 700\,\mathrm{nm}\)): \(\sin(\theta_{2,\max}) = \frac{2\cdot700\,\mathrm{nm}}{d}\)
04

Calculate the minimum and maximum angles for the 3rd-order spectrum

Similarly, for the third-order spectrum, use the smallest and largest wavelengths to calculate the minimum and maximum angles: For the minimum angle: \(\sin(\theta_{3,\min}) = \frac{3\cdot400\,\mathrm{nm}}{d}\) For the maximum angle: \(\sin(\theta_{3,\max}) = \frac{3\cdot700\,\mathrm{nm}}{d}\)
05

Show that the 2nd and 3rd-order spectra overlap

To show that the second and third-order spectra overlap, we need to demonstrate that \(\theta_{3,\min} \le \theta_{2,\max}\). Since the sine function is monotonic, we just need to show that: \(\sin(\theta_{3,\min}) \le \sin(\theta_{2,\max})\) From the equations in steps 3 and 4, we have: \(\frac{3\cdot400\,\mathrm{nm}}{d} \le \frac{2\cdot700\,\mathrm{nm}}{d}\) Notice that the grating constant d cancels out: \(3\cdot400\,\mathrm{nm}\le2\cdot700\,\mathrm{nm}\) \(1200\,\mathrm{nm}\le1400\,\mathrm{nm}\) And since this inequality is true, this establishes that the second and third-order spectra overlap for white light on a diffraction grating, regardless of the value of the grating constant d.

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Most popular questions from this chapter

On a certain crystal, a first-order X-ray diffraction maximum is observed at an angle of \(27.1^{\circ}\) relative to its surface, using an X-ray source of unknown wavelength. Additionally, when illuminated with a different, this time of known wavelength \(0.137 \mathrm{nm},\) a second-order maximum is detected at \(37.3^{\circ} .\) Determine (a) the spacing between the reflecting planes, and (b) the unknown wavelength.

Find the wavelength of light that has its third minimum at an angle of \(48.6^{\circ}\) when it falls on a single slit of width \(3.00 \mu \mathrm{m}\)

(a) What is the minimum width of a single slit (in multiples of \(\bar{\lambda}\) ) that will produce a first minimum for a wavelength \(\lambda\) ? (b) What is its minimum width if it produces 50 minima? (c) 1000 minima?

What is the minimum diameter mirror on a telescope that would allow you to see details as small as \(5.00 \mathrm{km}\) on the moon some \(384,000 \mathrm{km}\) away? Assume an average wavelength of \(550 \mathrm{nm}\) for the light received.

Two lamps producing light of wavelength 589 nm are fixed \(1.0 \mathrm{m}\) apart on a wooden plank. What is the maximum distance an observer can be and still resolve the lamps as two separate sources of light, if the resolution is affected solely by the diffraction of light entering the eye? Assume light enters the eye through a pupil of diameter \(4.5 \mathrm{mm}\).
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