Question

A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm. What is the width of the central maximum on a screen 2.0 m behind the slit?

Answer 1

The width of the central maximum of the single-slit diffraction pattern is given by,

$w=\frac{2 \lambda L}{a}$

Here, $w$ is the width of the central maximum, $\lambda$ is the wavelength of the light, $a$ is slit width and $L$ is

the separation between slit and screen.

Convert wavelength from Nano-meter to meter

$$\begin{aligned} \lambda &=500 \mathrm{nm} \\ &=(500 \mathrm{nm})\left(\frac{1.0 \times 10^{-9} \mathrm{~m}}{1.0 \mathrm{nm}}\right) \\ &=500 \times 10^{-9} \mathrm{~m} \end{aligned}$$

Convert slit width from mille meter to meter

$$\begin{aligned} a &=(0.50 \mathrm{~mm})\left(\frac{1.0 \mathrm{~m}}{1,000 \mathrm{~mm}}\right) \\ &=0.0005 \mathrm{~m} \end{aligned}$$

Substitute $0.0005 \mathrm{~m}$ for $a, 500 \times 10^{-9} \mathrm{~m}$ for $\lambda$ and $2.0 \mathrm{~m}$ for $L$ in $w=\frac{2 \lambda L}{a}$.

$$\begin{aligned} w &=\frac{2\left(500 \times 10^{-9} \mathrm{~m}\right)(2.0 \mathrm{~m})}{0.0005 \mathrm{~m}} \\ &=4.0 \times 10^{-3} \mathrm{~m} \\ &=\left(4.0 \times 10^{-3} \mathrm{~m}\right)\left(\frac{1 \mathrm{~mm}}{10^{-3} \mathrm{~m}}\right) \\ &=4.0 \mathrm{~mm} \end{aligned}$$

Hence, the width of the central maximum on screen is ${4.0 \mathrm{~mm}}$.