91Ó°ÊÓ

Distance between two sources:$2.04\mathrm{μ³¾}$

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

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

$\mathbf{d}\mathbf{}\mathbf{=}\mathbf{ }\mathbf{³¾Î»}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{Â\pm }\mathbf{1}\mathbf{,}\mathbf{Â\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$ ...(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$\mathbf{\left(}\mathbf{λ}\mathbf{\right)}$.

$\mathbf{d}\mathbf{}\mathbf{=}\mathbf{ }\left(m+\frac{1}{2}\right)\mathbf{λ}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{Â\pm }\mathbf{1}\mathbf{,}\mathbf{Â\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$ ...(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-

$\mathrm{d}={\mathrm{³¾Î»}}_{\mathrm{m}}$

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

localid="1663930359789" $\begin{array}{l}\mathrm{for}\mathrm{m}=3,\\ {\mathrm{λ}}_3=\frac{2040\mathrm{nm}}{3}\\ =680\mathrm{nm}\\ \begin{array}{l}\mathrm{for}\mathrm{m}=4,\\ {\mathrm{λ}}_3=\frac{2040\mathrm{nm}}{4}\\ =510\mathrm{nm}\end{array}\\ \begin{array}{l}\mathrm{for}\mathrm{m}=5,\\ {\mathrm{λ}}_3=\frac{2040\mathrm{nm}}{5}\\ =408\mathrm{mm}\end{array}\end{array}$

Thus, the above mentioned wavelengths are the required values.

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

From equation (ii).

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

For$m=3$and$m=4$, we get-

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

And

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

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