Q21E - University Physics with Modern Physics

Question

An interference pattern is produced by light of wavelength from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320mm. In terms of the intensity ${\mathit{I}}_{\mathbf{0}}$ at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Step-by-Step Solution

Answer

(a) The angular positions of the first-order and second-order, two-slit interference maxima are $\mathbf{0}\mathbf{.}{\mathbf{0627}}^{\mathbf{\circ }}$ and $\mathbf{0}\mathbf{.}{\mathbf{1254}}^{\mathbf{\circ }}$ respectively.

(b) The intensity at each of the angular positions in part (a) is $\mathbf{0}\mathbf{.}\mathbf{249}{\mathit{I}}_{\mathbf{0}}$ and $\mathbf{0}\mathbf{.}\mathbf{0256}{\mathit{I}}_{\mathbf{0}}$.

Step 1: Formula used to solve the question

Intensity is given by

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2 \frac{\varphi }{2}\right){\left(\frac{sin \frac{\beta }{2}}{\frac{\beta }{2}}\right)}^{\mathbf{2}}$ 

Where 

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

And

$\mathit{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}$ 

Step 2: Determine the angular positions of the first-order and second-order

Given:

  $$\begin{gathered} \mathit{\lambda }\mathbf{=}\mathbf{580}\mathit{n}\mathit{m}\mathbf{=}\mathbf{580}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathit{m} \\ \mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{530}\mathit{m}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{530}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m} \\ \mathit{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{320}\mathit{m}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{320}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{m} \end{gathered}$$         

             

             

When the two slits are very narrow, the interference pattern will be affected only by the two-slits effect. And hence, there is no diffraction pattern due to the single-slit diffraction. 

So, the angles of bright fringes of the double-slit interference is given by

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

Hence, the angular position is given by

$\mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right)}$                                                            (1)

For first order and second order maximum, $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}$.

Plug these values in (1),

$\begin{array}{r}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{ }\left(\frac{1.0\ast 580\ast {10}^{-9}}{0.530\ast {10}^{-3}}\right)\mathbf{=}\mathbf{0}{\mathbf{.0627}}^{\mathbf{\circ }}\\ {\mathbf{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{ }\left(\frac{2.0\ast 580\ast {10}^{-9}}{0.530\ast {10}^{-3}}\right)\mathbf{=}\mathbf{0}{\mathbf{.1254}}^{\mathbf{\circ }}\end{array}$ 

 

Step 3: Determine the intensity at each of the angular positions

The intensity of four fringes including the diffraction effect is given by,

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\left({\cos }^2 \frac{\varphi }{2}\right){\left(\frac{\sin \frac{\beta }{2}}{\frac{\beta }{2}}\right)}^{\mathbf{2}}$ 

Where 

 $\mathbf{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi dsin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}$

And

$\mathit{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}$ 

Therefore,

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\mathbf{(}{\mathbf{cos}}^{\mathbf{2}}\mathbf{ }\frac{\mathbf{\pi dsin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}\mathbf{)}{\mathbf{\left(}\frac{\mathbf{sin}\frac{\mathbf{\pi asin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}}{\frac{\mathbf{\pi asin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}}\mathbf{\right)}}^{\mathbf{2}}$ 

Plug the given values for first-order,

$$\begin{gathered} {\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2 \frac{\pi d\lambda }{\lambda d}\right){\left(\frac{sin \frac{\pi a\lambda }{\lambda d}}{\frac{\pi a\lambda }{\lambda d}}\right)}^{\mathbf{2}} \\ \mathbf{\Rightarrow }{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2 \pi \right){\left(\frac{sin\frac{\pi a}{d}}{\frac{\pi a}{d}}\right)}^{\mathbf{2}} \\ {\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2 \pi \right){\left(\frac{sin\frac{\pi {\pi }^{+}0{.320}^{+}+{10}^{-3}}{0{.50}^{\ast }+0^{-3}}}{\frac{{\pi }^{\ast }0{.320}^{\ast }{10}^{-3}}{0{.530}^{\ast }{0}^{-3}}}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.249}{\mathit{I}}_{\mathbf{0}} \end{gathered}$$ 

 

 

Similarly for second-order,

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0256}{\mathit{I}}_{\mathbf{0}}$ 

Thus, the angular positions of the first-order and second-order, two-slit interference maxima are $\mathbf{0}\mathbf{.}{\mathbf{0627}}^{\mathbf{\circ }}$ and $\mathbf{0}\mathbf{.}{\mathbf{1254}}^{\mathbf{\circ }}$respectively. The intensity at each of the angular positions in part (a) is $\mathbf{0}\mathbf{.}\mathbf{249}{\mathit{I}}_{\mathbf{0}}$ and $\mathbf{0}\mathbf{.}\mathbf{0256}{\mathit{I}}_{\mathbf{0}}$.