Question

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?

Short answer

The intensity at point on the screen that are (a) 2.00 mm is $\mathbf{0}\mathbf{.}\mathbf{150}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$ and (b) 1.50mm is$\mathbf{0}\mathbf{.}\mathbf{300}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$ from the center of the central maximum.

Step-by-step solution

Formulas used to solve the question

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)

Calculate the wavelength

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,

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

Determine the equation for intensity

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)

Calculate the intensity

1. Plug the given,

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

1. Plug the given,

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

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.