Two slits spaced 0.260mm apart are 0.990m from a screen and illuminated by coherent light of wavelength . The intensity at the center of the central maximum $\left(\theta =0°\right)$is ${\mathit{I}}_{\mathbf{0}}$. What is the distance on the screen from the center of the central maximum (a) to the first minimum; (b) to the point where the intensity has fallen to $\frac{{\mathbf{I}}_{\mathbf{0}}}{\mathbf{2}}$?

Angle first dark fringe after the central maxima is for m = 0 and is given by

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{(}\mathbf{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathit{\lambda }$ (1)

Distance on the screen between the central maxima and the first dark fringe is given by

${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{R}\mathbf{tan}\mathit{\theta }$ (2)

Intensity is given as:

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}{\mathbf{cos}}^{\mathbf{2}}\left(\frac{\pi dy}{\lambda R}\right)$ (3)

Given: $\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{26}\mathit{m}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{26}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}$

R = 0.90m

$\mathit{\lambda }\mathbf{=}\mathbf{660}\mathit{n}\mathit{m}\mathbf{=}\mathbf{660}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}$

$\mathit{I}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{0}}}{\mathbf{2}}$

From the equation (1), plug the given

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{\lambda }$

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\displaystyle \frac{1}{2}*660*{10}^{-9}}}{\mathbf{0}\mathbf{.}\mathbf{26}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0727}\mathbf{°} \end{gathered}$$

In equation (2), plug the given

${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{90}\mathbf{*}\mathit{t}\mathit{a}\mathit{n}\mathbf{0}\mathbf{.}\mathbf{0727}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}$

From equation (3), plug from given

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

localid="1663994413719" $$\begin{gathered} \mathbf{\Rightarrow }\frac{\mathbf{y}\mathbf{\pi }\mathbf{d}}{\mathbf{\lambda }\mathbf{R}}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\sqrt{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\frac{\mathbf{y\pi d}}{\mathbf{\lambda R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathit{\pi } \\ \mathbf{\Rightarrow }\frac{\mathbf{y\pi d}}{\mathbf{\lambda R}}\mathbf{=}\frac{\mathbf{660}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{90}}{\mathbf{4}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{26}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{71}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{m} \end{gathered}$$

Thus, the distance on the screen from the center of the central maximum to the first minimum is 1.14mm. The distance on the screen from the center of the central maximum to the point where the intensity has fallen to$\frac{{\mathbf{I}}_{\mathbf{°}}}{\mathbf{2}}$ is 0.571mm.