Q6E - University Physics with Modern Physics

Question

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, $\mathbf{2}\mathbf{.}\mathbf{04}\mathbf{\hspace{0.33em}}\mathbf{\mu m}$ apart, and in line with an observer, so that one source is $\mathbf{2}\mathbf{.}\mathbf{04}\mathbf{\hspace{0.33em}}\mathbf{\mu m}$ farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is $\mathbf{2}\mathbf{.}\mathbf{04}\mathbf{\hspace{0.33em}}\mathbf{\mu m}$ farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

Step-by-Step Solution

Answer

a)  The wavelengths for which the brightest light is witnessed is 680 nm, 510 nm and 408 nm.

                                                                                                                                                       

b)  The wavelength for this case is unchanged when compared to part (a).

c)  The destructive interference at the location of the observer is observed for ${\mathrm{\lambda }}_3=583 \mathrm{nm} \mathrm{and}\hspace{0.33em}{\mathrm{\lambda }}_4=453 \mathrm{nm}$

Step 1: Given Data

Distance between two sources: $2.04 \mathrm{\mu m}$ 

Visible wavelength range: $380 \mathrm{nm} \mathrm{to} 750 \mathrm{nm}$ 

Step 2: (a) Determination of the wavelengths for which the brightest light is witnessed.

For constructive interference, the condition for path difference $\left(d\right)$  is that it should be  an integral multiple of the wavelength of light $\mathbf{\left(}\mathbf{\lambda }\mathbf{\right)}$.

$\mathbf{d}\mathbf{ }\mathbf{=}\mathbf{\hspace{0.33em}}\mathbf{m\lambda }\mathbf{,}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$                             ...(i)

 

For destructive interference, the condition for path difference  $\mathbf{\left(}\mathbf{d}\mathbf{\right)}$ is that it should be a half- integral multiple of the wavelength of light $\mathbf{\left(}\mathbf{\lambda }\mathbf{\right)}$.

$\mathbf{d}\mathbf{ }\mathbf{=}\mathbf{\hspace{0.33em}}\left(m+\frac{1}{2}\right)\mathbf{\lambda }\mathbf{,}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$                    ...(ii)

               

The path difference given is d = 2040 nm.

 

Brightest light is witnessed when the waves undergo constructive interference.

So, using equation (i) and substituting the values, we get-

$\mathrm{d}={\mathrm{m\lambda }}_{\mathrm{m}}$

${\mathrm{\lambda }}_{\mathrm{m}}=\frac{\mathrm{d}}{\mathrm{m}}$

$\begin{array}{l}\mathrm{for} \mathrm{m}=3,\\ {\mathrm{\lambda }}_3=\frac{2040 \mathrm{nm}}{3}\\ =680 \mathrm{nm}\\ \begin{array}{l}\mathrm{for} \mathrm{m}=4,\\ {\mathrm{\lambda }}_3=\frac{2040 \mathrm{nm}}{4}\\ =510 \mathrm{nm}\end{array}\\ \begin{array}{l}\mathrm{for} \mathrm{m}=5,\\ {\mathrm{\lambda }}_3=\frac{2040 \mathrm{nm}}{5}\\ =408 \mathrm{mm}\end{array}\end{array}$

  Thus, the above mentioned wavelengths are the required values.

Step 3: (b) Determination of the wavelength when sources are not inline with the observer.

The path difference between the sources remains the same and so the wavelengths and the nature of the interference remain unchanged.

Step 4: (c) Determination of the wavelengths when the nature of interference is destructive.

From equation (ii).

$$\begin{gathered} d=\left(m+\frac{1}{2}\right){\lambda }_m \\ {\lambda }_m=\frac{d}{m+{\displaystyle \frac{1}{2}}} \\ =\frac{2040 \mathrm{nm}}{\mathrm{m}+{\displaystyle \frac{1}{2}}} \end{gathered}$$

For $m=3$ and $m=4$, we get-

${\lambda }_{\mathrm{m}}=\frac{2040 \mathrm{nm}}{3+{\displaystyle \frac{1}{2}}}$ 

And

$$\begin{gathered} {\lambda }_m =\frac{2040 \mathrm{nm}}{4+{\displaystyle \frac{1}{2}}} \\ =2040 \mathrm{nm}\times \frac{2}{9} \\ =453 \mathrm{nm} \end{gathered}$$

Calculating wavelengths in the visible range by putting $m=3$and $m=4$ in above equation gives${\lambda }_3=583 \mathrm{nm}$ and ${\lambda }_4=453 \mathrm{nm}$