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Chapter 25: Problem 43

White light containing wavelengths from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) is shone through a grating. Assuming that at least part of the third-order spectrum is present, show that the second-and third-order spectra always overlap, regardless of the slit separation of the grating.

Short Answer

Expert verified
Answer: Yes, the second-order and third-order spectra always overlap for a wavelength range of 400 nm to 700 nm, regardless of the slit separation.

Step by step solution

01

Write down the grating equation

For a diffraction grating, the grating equation is given by: \( m \lambda = d \sin{\theta} \) where \(m\) is the order of the spectrum, \(\lambda\) is the wavelength of the light, \(d\) is the slit separation, and \(\theta\) is the angle of diffraction.
02

Set up the inequalities for the second and third-order spectrums

Given the range of wavelengths from 400 nm to 700 nm, we need to show that there is always an overlap between the second and third order spectra despite the slit separation of the grating. For the second-order spectrum, m = 2, and for the third-order spectrum, m = 3. We can now write down two inequalities for the spectral lines: Second-order spectrum: \(2 \lambda_{\text{min}} \leq d \sin{\theta} \leq 2 \lambda_{\text{max}}\) Third-order spectrum: \(3 \lambda_{\text{min}} \leq d \sin{\theta} \leq 3 \lambda_{\text{max}}\) We will now analyze if there is a range of overlap by comparing the right-hand side of the first inequality with the left-hand side of the second inequality.
03

Compare the right-hand side of the second-order inequality to the left-hand side of the third-order inequality

If the right-hand side of the second-order inequality is equal or smaller than the left-hand side of the third-order inequality, it means the second and third-orders spectra will overlap. Comparing: \( 2 \lambda_{\text{max}} \leq 3 \lambda_{\text{min}} \) Plugging in the given values for the wavelengths, we get: \( 2 (700 \mathrm{nm}) \leq 3 (400 \mathrm{nm}) \) which simplifies to: \( 1400 \mathrm{nm} \leq 1200 \mathrm{nm} \)
04

Analyzing the result

Since the inequality is false (1400 nm > 1200 nm), we can conclude that second-order and third-order spectra would always overlap, regardless of the slit separation of the grating.

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