Chapter 25, Problem 91P

(a)

To determine

The width of the slits.

Answer to Problem 91P

The width of the slits is $\underset{\_}{0.10\text{mm}}$.

Explanation of Solution

Write the expression for the width of the central maximum.

    $W=2x$                                                                                                                     (I)

Here, $W$ is the width of the central maximum, $x$ is the distance between first minimum and the central maximum.

Write the expression for the single-slit diffraction minima.

  $a\sin \theta =m\lambda$                                                                                                            (II)

Here, $a$ is the slit width, $\theta$ is the angle between the central maximum and first minimum, m is the order of the diffraction, $\lambda$ is the wavelength of light used.

Equation (II) is used mainly for dark spots on the screen. The width of the central maximum can be quantified as the distance between the minima on each side with $m=1$,

    $a\sin \theta =\lambda$                                                                                                              (III)

Solve equation (III) for $a$,

    $a=\frac{\lambda }{\sin \theta }$                                                                                                                (IV)

Since $\theta$ is very small, the approximation, $\sin \theta \approx \tan \theta$ can be used in equation (IV).

    $a=\frac{\lambda }{\tan \theta }$                                                                                                                (V)

Write the expression for the distance $x$.

    $x=D\tan \theta$                                                                                                             (VI)

Here, $D$ is the distance between slit and the screen.

Use equation (VI) in (V).

  $a=\frac{\lambda D}{x}$                                                                                                                 (VII)

Conclusion:

Substitute $2.40\text{cm}$ for $W$ in equation (I) and solve for $x$.

    $\begin{array}{c}x=\frac{2.40\text{cm}}{2}\\ =1.20\text{cm}\end{array}$

Substitute $1.20\text{cm}$ for $x$, $510\text{nm}$ for $\lambda$, and $2.4\text{m}$ for $D$ in equation (VII) to find $a$.

    $\begin{array}{c}a=\frac{\left(510\text{nm}\right)\left(\text{2}\text{.4}\text{m}\right)}{1.20\text{cm}}\\ =\frac{\left(510\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}\right)\left(\text{2}\text{.4}\text{m}\right)}{1.20\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =1.02\times {10}^{-4}\text{m}\times \frac{1000\text{mm}}{\text{1}\text{m}}\\ =0.10\text{mm}\end{array}$

Therefore, the width of the slits is $\underset{\_}{0.10\text{mm}}$.

(b)

To determine

The distance between slits.

Answer to Problem 91P

The distance between slits is $\underset{\_}{0.51\text{mm}}$.

Explanation of Solution

Write the expression for the maxima in a double slit interference experiment.

    $d\sin \theta =m\lambda$                                                                                                        (VIII)

Here, $d$ is the distance between the slits.

Rewrite equation (VIII) for fifth order double slit interference experiment,

    $d\sin \theta =5\lambda$                                                                                                          (IX)

Divide equation (IX) and equation (III),

$\begin{array}{c}\frac{d\sin \theta }{a\sin \theta }=\frac{5\lambda }{\lambda }\\ \frac{d}{a}=5\\ d=5a\end{array}$                                                                                                         (X)

Conclusion:

Substitute $0.10\text{mm}$ for $a$ in equation (X) to find $d$.

    $\begin{array}{c}d=5\left(0.10\text{mm}\right)\\ =0.51\text{mm}\end{array}$

Therefore, the distance between slits is $\underset{\_}{0.51\text{mm}}$.