Chapter 4, Problem 4.1CYU

Check Your Understanding Suppose the slit width in Example 4.1 is increased to $1.8\times {10}^{-6}$ m. What are the new angular positions for the first, second, and third minima? Would a fourth minimum exist?

To determine

The angualr position for the $1^{\text{st}},2^{\text{nd}}\text{and}{\text{3}}^{\text{rd}}$ minima.

Answer to Problem 4.1CYU

Angular spread for $1^{\text{st}},2^{\text{nd}}\text{and}{\text{3}}^{\text{rd}}$ minimum are $\text{17}.{\text{8}}^0\text{,37}.{\text{66}}^0\text{and}\text{66}.{\text{44}}^0$ respectively, and the $4^{\text{th}}$ order of minima does not exists.

Explanation of Solution

Given:

Light wavelength, $\lambda =550\text{nm}$

Order of minimum, $m=2$

Angular spread ${\theta }_2={45}^0$

Unit conversion,

  $\begin{array}{l}1\text{nm}\text{=}{\text{10}}^{-9}\text{m}\\ \text{1μm}\text{=}{\text{10}}^{-6}\text{m}\end{array}$

Formula used:

Formula for angular separation for minimum condition,

  $D\sin \theta =m\lambda$  ........(1)

Here D indicates the slit width, $\theta$ indicates the angular separation, m indicates the order of minimum and $\lambda$ indicates the light wavelength.

Calculation:

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, ${45}^0$ for $\theta$ and $2$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ D=\left[\frac{m\lambda }{\sin \theta }\right]\\ =\left[\frac{2\times 550\times {10}^{-9}\text{m}}{\sin {45}^0}\right]\\ =\left(\frac{1100\times {10}^{-9}}{0.707}\right)\end{array}$

  $=1.56\times {10}^{-6}\text{m}$

Hence, the original slit thickness is $\boxed{1.56\times {10}^{-6}\text{m}}$.

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, $1.56\times {10}^{-6}\text{m}$ for D and $1$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ {\theta }_1={\sin }^{-1}\left[\frac{m\lambda }{D}\right]\\ =\left[\frac{1\times 550\times {10}^{-9}\text{m}}{1.56\times {10}^{-6}\text{m}}\right]\\ =\left(\frac{550\times {10}^{-3}}{1.56}\right)\\ ={\sin }^{-1}\left(0.352\right)\end{array}$

  $={20.60}^0$

Hence, the angular spread at $1^{\text{st}}$ order minima is $\boxed{{20.60}^0}$.

According to question slit width is increased to a value of $1.8\times {10}^{-6}\text{m}$, now for $1^{\text{st}}$ order minima,

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, $1.8\times {10}^{-6}\text{m}$ for D and $1$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ {\theta }_1={\sin }^{-1}\left[\frac{m\lambda }{D}\right]\\ ={\sin }^{-1}\left[\frac{1\times 550\times {10}^{-9}\text{m}}{1.8\times {10}^{-6}\text{m}}\right]\\ ={\sin }^{-1}\left(305.55\times {10}^{-3}\right)\\ ={\sin }^{-1}\left(0.305\right)\end{array}$

  $={17.8}^0$

Hence, the angular spread at $1^{\text{st}}$ order minima for a slit width of $1.8\times {10}^{-6}\text{m}$ is $\boxed{{17.8}^0}$.

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, $1.8\times {10}^{-6}\text{m}$ for D and $2$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ {\theta }_1={\sin }^{-1}\left[\frac{m\lambda }{D}\right]\\ ={\sin }^{-1}\left[\frac{2\times 550\times {10}^{-9}\text{m}}{1.8\times {10}^{-6}\text{m}}\right]\\ ={\sin }^{-1}\left(611.11\times {10}^{-3}\right)\\ ={\sin }^{-1}\left(0.611\right)\end{array}$

  $={37.66}^0$

Hence, the angular spread at $2^{\text{nd}}$ order minima for a slit width of $1.8\times {10}^{-6}\text{m}$ is $\boxed{{37.66}^0}$.

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, $1.8\times {10}^{-6}\text{m}$ for D and $3$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ {\theta }_1={\sin }^{-1}\left[\frac{m\lambda }{D}\right]\\ ={\sin }^{-1}\left[\frac{3\times 550\times {10}^{-9}\text{m}}{1.8\times {10}^{-6}\text{m}}\right]\\ ={\sin }^{-1}\left(916.66\times {10}^{-3}\right)\\ ={\sin }^{-1}\left(0.916\right)\end{array}$

  $={66.44}^0$

Hence, the angular spread at $2^{\text{nd}}$ order minima for a slit width of $1.8\times {10}^{-6}\text{m}$ is $\boxed{{66.44}^0}$.

Reduce $\text{550}\times {\text{10}}^{-9}\text{m}$ for $\lambda$, $1.8\times {10}^{-6}\text{m}$ for D and $3$ for m in formula (1),

  $\begin{array}{c}D\sin \theta =m\lambda \\ {\theta }_1={\sin }^{-1}\left[\frac{m\lambda }{D}\right]\\ ={\sin }^{-1}\left[\frac{4\times 550\times {10}^{-9}\text{m}}{1.8\times {10}^{-6}\text{m}}\right]\\ ={\sin }^{-1}\left(1222.22\times {10}^{-3}\right)\\ ={\sin }^{-1}\left(1.22\right)\end{array}$

As the inverse sine angle cannot be more than 1, thus $4^{\text{th}}$ order of minima does not exist.

Conclusion:

Angular spread for $1^{\text{st}},2^{\text{nd}}\text{and}{\text{3}}^{\text{rd}}$ minimum are $\text{17}.{\text{8}}^0\text{,37}.{\text{66}}^0\text{and}\text{66}.{\text{44}}^0$ respectively, and the $4^{\text{th}}$ order of minima does not exists.