Q19E - University Physics with Modern Physics

Question

Number of Fringes in a Diffraction Maximum. In Fig.36.12 c the central diffraction maximum contains exactly seven interference fringes, and in this case d =a =4. (a) What must the ratio d /a be if the central maximum contains exactly five fringes? (b) In the case considered in part (a), how many fringes are contained within the first diffraction maximum on one side of the central maximum?

Step-by-Step Solution

Answer

a) The ratio d/a must be 3.0.

b) There are 3.0 fringes contained with the first diffraction maximum on one side of the central maximum.

Step 1: Formula used to solve the question

Equation for the first minimum of the dark fringes:

$\mathbf{a}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{\theta }\mathbf{=}\mathbf{m\lambda }$ 

Intensity is given by

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

Where 

$\mathbf{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi a}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}$

Step 2: Calculate the ratio

Since the bright fringe contains 5 fringes, so the minimum fringe that makes an interference of total dark is for $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{3}$.

The equation is given by

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

Hence, 

$\mathit{a}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathbf{=}\mathit{\lambda }$                                                            (1)

While it is the third bright fringe in the double-slit pattern,

$\mathit{d}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathbf{=}\mathbf{3}\mathit{\lambda }$                                                       (2)

Now, the intensity is given by

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

Where 

$\mathit{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi a}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}$

Thus, 

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

Now, for $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{3}$, the intensity is zero.

$$\begin{gathered} \mathbf{0}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2 \frac{\varphi }{2}\right){\left(\frac{sin \frac{\pi asin \theta }{\lambda }}{\frac{\pi asin \theta }{\lambda }}\right)}^{\mathbf{2}} \\ \mathbf{\Rightarrow }\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\frac{\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}}{\frac{\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}}\mathbf{=}\mathbf{0} \\ \mathbf{\Rightarrow }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\frac{\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{\lambda }}\mathbf{=}\mathbf{0} \end{gathered}$$ 

 

Plug the value,

$\begin{array}{c}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\left(\frac{3\pi asin \theta }{dsin \theta }\right)\mathbf{=}\mathbf{0}\\ \mathbf{\Rightarrow }\frac{\mathbf{3}\mathbf{a}}{\mathbf{d}}\mathbf{=}\mathbf{1}\\ \mathbf{\Rightarrow }\mathbf{d}\mathbf{/}\mathbf{a}\mathbf{=}\mathbf{3}\mathbf{.0}\end{array}$

Step 3: Calculate the number of fringes

Now, the fringe for the central maximum fringe are for

$\mathit{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}$ 

So, 

$\mathit{a}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathbf{=}\mathbf{2}\mathit{\lambda }$                                                            (3)

And 

$\mathit{d}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$                                                       (4)

Divide (4) by (3) and plug d /a,

$$\begin{gathered} \frac{\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}{\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{\theta }}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{2}\mathbf{\lambda }} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\Rightarrow }\frac{\mathbf{m}}{\mathbf{2}}\mathbf{=}\mathbf{3} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\Rightarrow }\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{6} \end{gathered}$$ 

 

 

This means that the second bright fringe envelope is contained between $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{3}$ and $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{6}$.

Implies that this envelope only contains two bright fringes which are for $\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{\pm }\mathbf{5}$.

Thus, the ratio d /a must be 3.0. There are 3.0 fringes contained with the first diffraction maximum on one side of the central maximum.