Chapter 25, Problem 41P

(a)

To determine

The number of maxima seen on the screen.

Answer to Problem 41P

The number of maxima seen on the screen is five.

Explanation of Solution

Find the equation to find the maximum possible value of $m$.

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

Here, $d$ is the slit width, $\theta$ is the angle of incidence, $\lambda$ is the wavelength and $m$ is the order that has the values $0,\pm 1,\pm \mathrm{2...}$.

Conclusion:

Substitute $0.600\text{μm}$ for $\lambda$, $1.50\text{μm}$ for $d$ and $90°$ for $\theta$ to find the maximum value of $m$.

    $\begin{array}{c}m=\frac{\left(1.50\text{μm}\right)\sin 90°}{0.600\text{μm}}\\ =\frac{1.50\text{μm}\left(1\right)}{0.600\text{μm}}\\ =2.50\end{array}$

 As the order has the values $0,\pm 1,\pm \mathrm{2...}$, the possible values are $-2,-1,0,1,2$.

Thus, the number of maxima seen on the screen is five.

(a)

To determine

Sketch the pattern for the distance $3.0\text{m}$ from the grating.

Answer to Problem 41P

The pattern is drawn in figure 2.

Explanation of Solution

Sketch the diagram showing the situation.

The central maxima is seen at an angle ${\theta }_0=0°$.

Find the equation to find the distance of the first maxima from the central maxima.

    $x_1=D\tan {\theta }_1$                                                                                                       (II)

Here, $x_1$ is the distance of the first maxima from the central maxima, $D$ is the distance between the slit and the screen and ${\theta }_1$ is the angle for first maxima.

Find the equation for the angle for first maxima.

    $\begin{array}{c}d\sin {\theta }_1=\lambda \\ {\theta }_1={\sin }^{-1}\frac{\lambda }{d}\end{array}$                                                                                             (III)

Substitute equation (III) in equation (II) to find $x_1$.

    $x_1=D\tan \left({\sin }^{-1}\frac{\lambda }{d}\right)$                                                                                       (IV)

Find the equation to find the distance of the second maxima from the central maxima.

    $x_2=D\tan {\theta }_2$                                                                                                    (V)

Here, $x_2$ is the distance of the second maxima from the central maxima and ${\theta }_2$ is the angle for second maxima.

Find the equation for the angle for second maxima.

    $\begin{array}{c}d\sin {\theta }_2=2\lambda \\ {\theta }_2={\sin }^{-1}\frac{2\lambda }{d}\end{array}$                                                                                         (VI)

Substitute equation (VI) in equation (V) to find $x_2$.

    $x_2=D\tan \left({\sin }^{-1}\frac{2\lambda }{d}\right)$                                                                                     (VII)

Conclusion:

Substitute $0.600\text{μm}$ for $\lambda$, $1.50\text{μm}$ for $d$ and $3.0\text{m}$ for $D$ in equation (IV) to find $x_1$.

    $\begin{array}{c}{x}_1=\left(3.0\text{m}\right)\tan \left({\sin }^{-1}\frac{0.600\text{μm}}{1.50\text{μm}}\right)\\ =\left(3.0\text{m}\right)\tan 23.58°\\ =1.3\text{m}\end{array}$                                                               (VIII)

Substitute $0.600\text{μm}$ for $\lambda$, $1.50\text{μm}$ for $d$ and $3.0\text{m}$ for $D$ in equation (VII) to find $x_2$.

    $\begin{array}{c}{x}_2=\left(3.0\text{m}\right)\tan \left({\sin }^{-1}\frac{2\left(0.600\text{μm}\right)}{1.50\text{μm}}\right)\\ =\left(3.0\text{m}\right)\tan 53.13°\\ =4.0\text{m}\end{array}$                                                             (IX)

Sketch the pattern using the results in equation (VIII) and (IX).