Two slits spaced 0.0720mm apart are 0.800m from a screen. Coherent light of wavelength $\mathit{\lambda }$passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is 3.00mm . If the intensity at the peak of the central maximum is $\mathbf{0}\mathbf{.}\mathbf{600}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$, what is the intensity at points on the screen that are (a) 2.00mm and (b) 1.50mm from the center of the central maximum?

Intensity is given by

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\varphi }}{\mathbf{2}}\mathbf{\right)}$ (1)

Location of first dark fringe is given by

$\mathit{y}\mathbf{=}\frac{\mathbf{(}\mathbf{m}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{\lambda }\mathbf{R}}{\mathbf{d}}$ (2)

Given: R = 0.800m

$$\begin{gathered} {\mathit{y}}_{\left(minimum=0\right)}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{m}\mathit{m}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathbf{\backslash } \\ {\mathit{I}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0600}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}} \\ {\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{m}\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m} \\ {\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathit{m}\mathit{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m} \end{gathered}$$

From equation (2), at

${\mathit{y}}_{\left(m=0\right)}\mathbf{=}\frac{\mathbf{\lambda }\mathbf{R}}{\mathbf{2}\mathbf{d}}$

Solve for and plug the values,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{2}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{072}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{*}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{0}\mathbf{.}\mathbf{800}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{40}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathit{m}$

From equation (1),

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{d}\mathbf{sin}\mathbf{\theta }}{\mathbf{\lambda }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

It is obvious that the distance between the first dark fringe and the central maxima is too small compared to the distance between the screen and the double-slit. This means that the angles are too small, so small scale approximations can be taken.

So,

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{d}\mathbf{y}}{\mathbf{\lambda }\mathbf{R}}$

Plug the equation into equation (1),

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{d}\mathbf{y}}{\mathbf{\lambda }\mathbf{R}}\mathbf{\right)}\mathbf{}$ (4)

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0600}\mathbf{*}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{072}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{*}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{5}\mathbf{.}\mathbf{40}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{80}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0150}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0600}\mathbf{*}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{072}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{*}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{5}\mathbf{.}\mathbf{40}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{80}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0300}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$

Thus, the intensity at point on the screen that are (a) 2.00 mm is0.0150W/${\mathit{m}}^{\mathbf{2}}$and

(b) 1.50 mm is 0.0300W/${\mathit{m}}^{\mathbf{2}}$ from the center of the central maximum.