Chapter 25, Problem 51P

(a)

To determine

Whether the central maximum be narrower or wider if red light is used instead of blue in the single-slit experiment.

Answer to Problem 51P

The central maximum will be $\underset{\_}{\text{wider for red light}}$ than it is for the blue.

Explanation of Solution

Given that the width of the central maximum when blue light is used is $2.0\text{cm}$.

Write the condition for the diffraction minima in a single-slit diffraction.

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

Here, $a$ is the slit width, $\theta$ is the angle of diffraction, m is the order of the diffraction, $\lambda$ is the wavelength of light used.

Equation (I) 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 $m=1$,

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

From equation (II) it is clear that larger wavelengths give larger angles for the first minimum.

In this case, the width of the central bright fringe is twice the distance from the center of the screen to the first minimum so that the fringe is wider for greater wavelengths. The wavelength of the red light is greater than the wavelength of the blue light so that the central maximum is wider for the red light than it is for the blue light.

Conclusion:

Therefore, the central maximum will be $\underset{\_}{\text{wider for red light}}$ than it is for the blue.

(b)

To determine

The width of the central maximum when red light is used.

Answer to Problem 51P

The width of the central maximum when red light is used is $\underset{\_}{3.3\text{cm}}$.

Explanation of Solution

Given that the wavelength of the blue light is $0.43\text{μm}$, and the wavelength of the red light is $0.70\text{μm}$.

Let the width of the central maximum is $W$, referring to the figure 25.31 it can be seen that,

    $W=2x$                                                                                                                   (III)

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 distance $x$.

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

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

Use equation (IV) in equation (III),

    $W=2D\tan \theta$                                                                                                             (V)

The angle $\theta$ is assumed to be small so that equation (V) becomes (use the approximation $\tan \theta \approx \sin \theta$),

    $W\approx 2D\sin \theta$

Solve equation (II) for $\sin \theta$.

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

Use equation (VI) in equation (V) to find $W$.

    $W=\frac{2D\lambda }{a}$                                                                                                             (VII)

Apply equation (VII) for the red light,

    $W_{\text{R}}=\frac{2D{\lambda }_{\text{R}}}{a}$                                                                                                         (VIII)

Apply equation (VII) for the blue light,

    $W_{\text{B}}=\frac{2D{\lambda }_{\text{B}}}{a}$                                                                                                            (IX)

Divide equation (VIII) by (IX) and solve for $W_{\text{R}}$.

    $\begin{array}{c}\frac{W_{\text{R}}}{W_{\text{B}}}=\frac{\frac{2D{\lambda }_{\text{R}}}{a}}{\frac{2D{\lambda }_{\text{B}}}{a}}\\ W_{\text{R}}=W_{\text{B}}\frac{{\lambda }_{\text{R}}}{{\lambda }_{\text{B}}}\end{array}$                                                                                                            (X)

Conclusion:

Substitute $2.0\text{cm}$ for $W_{\text{B}}$, $0.70\text{μm}$ for ${\lambda }_{\text{R}}$ and $0.43\text{μm}$ for ${\lambda }_{\text{B}}$ in equation (X) to find $W_{\text{R}}$.

    $\begin{array}{c}{W}_{\text{R}}=\left(2.0\text{cm}\right)\left(\frac{0.70\text{μm}}{0.43\text{μm}}\right)\\ =3.3\text{cm}\end{array}$

Therefore, the width of the central maximum when red light is used is $\underset{\_}{3.3\text{cm}}$.