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Chapter 25: Problem 37

A red line (wavelength \(630 \mathrm{nm}\) ) in the third order overlaps with a blue line in the fourth order for a particular grating. What is the wavelength of the blue line?

Short Answer

Expert verified
Answer: The wavelength of the blue line is 472.5 nm.

Step by step solution

01

Write down the given information

We are given: - Wavelength of red line: \(\lambda_{red} = 630 \, \mathrm{nm}\) - Order of red line: \(m_{red} = 3\) - Order of blue line: \(m_{blue} = 4\) We need to find the wavelength of the blue line (\(\lambda_{blue}\)).
02

Use the grating equation for overlapping lines

Diffraction grating equation is given by $$d\sin\theta = m\lambda$$ Since the red and blue lines are overlapping, they are at the same angle (\(\theta\)). Therefore, we can write down two equations: $$d\sin\theta = m_{red}\lambda_{red}$$ $$d\sin\theta = m_{blue}\lambda_{blue}$$
03

Form an equation with only one variable

From the previous step, we can create a relation between the red and blue wavelengths by dividing the first equation by the second equation: $$\frac{m_{red}\lambda_{red}}{m_{blue}\lambda_{blue}} = 1$$ Now, we can solve for the variable we need, \(\lambda_{blue}\).
04

Solve for the blue wavelength

Rearrange the equation from Step 3 to solve for \(\lambda_{blue}\): $$\lambda_{blue} = \frac{m_{red}\lambda_{red}}{m_{blue}}$$ Now plug in the given values: $$\lambda_{blue} = \frac{3 \times 630 \, \mathrm{nm}}{4}$$ $$\lambda_{blue} = \frac{1890 \, \mathrm{nm}}{4}$$ $$\lambda_{blue} = 472.5 \, \mathrm{nm}$$ The wavelength of the blue line is \(472.5 \, \mathrm{nm}\).

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