Chapter 35, Problem 68PQ

(a)

To determine

The distance from the slit to the viewing screen.

Answer to Problem 68PQ

The distance from the slit to the viewing screen is $2.1\text{m}$.

Explanation of Solution

Write the expression for the minima of single slit diffraction pattern to find $\text{sin}{\theta }_{\text{m}}$.

  $\begin{array}{c}w\text{sin}{\theta }_{\text{m}}=m\lambda \\ \text{sin}{\theta }_{\text{m}}=\frac{m\lambda }{w}\end{array}$                          (I)

Here, $w$ is width of the slit and $m$ is an integer $\left(m=\pm 1,\pm 2,\pm 3,\mathrm{.........}\right)$, $\theta$ is the angular position and $\lambda$ is the wavelength of light.

Substitute $\text{m}=\text{1}$ for the first minima in above equation to find $\sin {\theta }_1$.

  $\text{sin}{\theta }_1=\frac{\lambda }{w}$                                (II)

Write the equation for the width of the central maximum which is surrounded by the first minima.

  $\begin{array}{c}B=2y_1\\ =2x\tan {\theta }_1\end{array}$                            (III)

Here, $B$ is the width of central maximum and $x$ is the distance from the slit to the screen.

For small angle of ${\theta }_1$, $\tan {\theta }_1\approx \sin {\theta }_1$.

Substitute $\frac{\lambda }{w}$ for $\tan {\theta }_1$ in above equation to find $x$.

  $\begin{array}{c}B=2x\left(\frac{\lambda }{w}\right)\\ x=\frac{Bw}{2\lambda }\end{array}$                            (IV)

Conclusion:

Substitute $6.60\text{mm}$ for $B$, $0.314\text{mm}$ for $w$ and $490.0\text{nm}$ for $\lambda$ in equation (IV) to find $x$.

  $\begin{array}{c}x=\frac{\left(\text{6}\text{.60}\text{mm}\times \frac{{\text{10}}^{\text{-3}}\text{m}}{\text{1}\text{mm}}\right)\left(\text{0}\text{.314}\text{mm}\times \frac{{\text{10}}^{\text{-3}}\text{m}}{\text{1}\text{mm}}\right)}{\text{2}\left(\text{490}\text{.0}\text{nm}\times \frac{{\text{10}}^{\text{-9}}\text{m}}{\text{1}\text{nm}}\right)}\\ =2.1\text{m}\end{array}$

Therefore, the distance from the slit to the viewing screen is $2.1\text{m}$.

(b)

To determine

The width of the first side maximum on the viewing screen.

Answer to Problem 68PQ

The width of the first side maximum on the viewing screen is $3.32\text{mm}$.

Explanation of Solution

Substitute $\text{m}=\text{2}$ for the second minima in equation (I) to find $\sin {\theta }_2$.

  $\text{sin}{\theta }_2=\frac{2\lambda }{w}$                                (V)

Write the equation for the distance from the center of the central maximum to the second minima.

  $y_2=x\tan {\theta }_2$                                  (VI)

For small angle of ${\theta }_2$, $\tan {\theta }_2\approx \sin {\theta }_2$.

Substitute $\frac{2\lambda }{w}$ for $\tan {\theta }_2$ in above equation to find $y_2$.

  $y_2=\frac{2\lambda x}{w}$                                     (VII)

Substitute $m=\text{1}$ for the second minima in equation (I) to find $\sin {\theta }_2$.

  $y_1=\frac{\lambda x}{w}$                                     (VIII)

Write $\Delta y$ from Equations (VII) and (VIII).

$\begin{array}{c}\Delta y=y_2-y_1\\ =\frac{\lambda x}{w}\end{array}$

Conclusion:

Substitute $490.0\text{nm}$ for $\lambda$, $2.1\text{m}$ for $x$ and $0.314\text{mm}$ for $w$ in equation (VII) to find $y_2$.

  $\begin{array}{c}\Delta y=\frac{\left(\text{490}\text{.0}\text{nm}\times \frac{{\text{10}}^{\text{-9}}\text{m}}{\text{1}\text{nm}}\right)\times 2.1\text{m}}{\left(\text{0}\text{.314}\text{mm}\times \frac{{\text{10}}^{\text{-3}}\text{m}}{\text{1}\text{mm}}\right)}\\ =3.32\times {10}^{-3}\text{m}\times \frac{\text{1}\text{mm}}{{\text{10}}^{\text{-3}}\text{m}}\\ =3.32\text{mm}\end{array}$

Therefore, the width of the first side maximum on the viewing screen is $3.32\text{mm}$.