Q49P - University Physics with Modern Physics

Question

Red light with a wavelength of 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, in the center of the third bright fringe 1m = 32 of the red light appears pure red, with none of the other colors. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

Step-by-Step Solution

Answer

The wavelength is 467nm and 600nm.

Step 1: Interference

At m = 3 there is no contribution of the second light so the interference is destructive. The red light has however the constructive interference and we can write

                                       $d\sin \theta =m{\lambda }_{red}=3\times 700nm=2100nm$

In destructive interference of second light we have 

                                       $\mathrm{dsin\theta }=\left(\mathrm{m}+\frac{1}{2}\right){\mathrm{\lambda }}_2$

The condition for it to be in the visible light part of the EM spectrum is 

                                        $$\begin{gathered} {\mathrm{\lambda }}_2=\frac{2100\mathrm{nm}}{\left(\mathrm{m}+{\displaystyle \frac{1}{2}}\right)} \\ 400\mathrm{nm}<\frac{2100\mathrm{nm}}{\left(\mathrm{m}+{\displaystyle \frac{1}{2}}\right)}<700\mathrm{nm} \end{gathered}$$

 

And this holds for m = 3,4 with ${\mathrm{\lambda }}_2$ = 467nm, 600nm.

Step 2: Conclusion

Hence, the wavelength is 467nm and 600nm.