Chapter 37, Problem 1P

To determine

The width of the central maximum on the screen.

Answer to Problem 1P

The width of the central maximum on the screen is $4.22\text{mm}$.

Explanation of Solution

Given info: The wavelength of the light is $632.8\text{nm}$, the width of the slit is $0.300\text{mm}$. And the distance between the slit and the screen is $1.00\text{m}$.

The condition for the case of destructive interference in the diffraction is,

$\sin \theta =m\frac{\lambda }{a}$

Here,

$\lambda$ is the wavelength of the light.

$\theta$ is the angel at which first minima is found.

$m$ is the order of diffraction.

$a$ is the width of the sit.

The order of diffraction for central maxima is $1$.

Substitute $632.8\text{nm}$ for $\lambda$ and $0.300\text{mm}$ for $a$ and $1$ for $m$ in the above equation.

$\begin{array}{c}\sin \theta =\left(1\right)\left[\frac{632.8\text{nm}}{0.300\text{mm}}\right]\\ =\left[\frac{632.8\text{nm}\times \left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{0.300\text{mm}\times \left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\right]\\ =2.11\times {10}^{-3}\end{array}$

The formula to calculate the width of the central maximum for small values of $\theta$ is,

$\frac{y}{L}=\sin \theta$

Here,

$y$ is the half width of the central maxima.

$L$ is the distance between the screen and the slit.

Substitute $1.00\text{m}$ for $L$ and $2.11\times {10}^{-3}$ for $\sin \theta$ in the above equation.

$\begin{array}{c}y=L\sin \theta \\ =\left(1.00\text{m}\right)\left(2.11\times {10}^{-3}\right)\\ =2.11\text{mm}\end{array}$

Thus, half width of central maximum is $2.11\text{mm}$.

The width of the central bright maxima is,

$\text{width}=2y$

Substitute $2.11\text{mm}$ for $y$ in above equation.

$\begin{array}{c}\text{width}=2\times 2.11\text{mm}\\ =\text{4}\text{.22}\text{mm}\end{array}$

Conclusion:

Therefore, width of the central maxima is $4.22\text{mm}$.