Chapter 36, Problem 1P

Two slits are separated by 0.320 mm. A beam of 500-nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range −30.0° < θ < 30.0°.

To determine

The number of maxima observed in angular range $-30.0°<\theta <30.0°$.

Answer to Problem 1P

The number of maxima observed in the angular range $-30.0°<\theta <30.0°$ is $641$.

Explanation of Solution

Given info: The two slits are separated by a distance of $0.320\text{mm}$ and the wavelength of beam is $500\text{nm}$.

The diagram is shown as

Figure 1

Consider there are $n$ fringes (bright and dark) of width $x$ on one side of the central bright.

The formula to calculate the interference is as,

$d\sin \theta =n\lambda$ (1)

Here,

$n$ is the order.

$\lambda$ is the wavelength.

$d$ is the separation between the slits.

Rearrange the above formula to find $n$,

$\begin{array}{l}d\sin \theta =n\lambda \\ n=\frac{d\sin \theta }{\lambda }\end{array}$

Substitute $0.320\text{}\text{nm}$ for $d$, $30°$ for $\theta$, $500\text{nm}$ for $\lambda$ in the above formula as,

$\begin{array}{c}n=\frac{d\sin \theta }{\lambda }\\ =\frac{\left(0.320\text{nm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\sin 30°}{\left(500\text{nm}\times {\frac{{10}^{-9}\text{m}}{1\text{nm}}}^{}\right)}\\ =\frac{\left(0.320\text{nm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{2\left(500\text{nm}\times {\frac{{10}^{-9}\text{m}}{1\text{nm}}}^{}\right)}\\ =320\end{array}$

The approximate value of $n$ is $320.5$.

The maxima are used for bright fringe and minima for dark fringe. Thus, the total number of fringes (bright and dark) in the angular separation from $-30°$ to $30°$ is $2\times 320.5$ that is $641$.

Conclusion:

Therefore, the number of maxima observed in the angular range $-30.0°<\theta <30.0°$ is $641$