Question

You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.11 m from the slit. What is the width, in centimeters, of the pattern?

Answer 1

Answer:

4.3 cm

Explanation:

We are given that

Width,d=70.3$\mu m=70.3\times10^{-6} m$

$1\mu m=10^{-6} m$

Wavelength,$\lambda=719 nm=719\times 10^{-9} m$

$1nm=10^{-9} m$

$r=2.11 m$

We have to find the width in cm of the pattern.

The angle for the first minimum m=1

$sin\theta_{min}\approx \theta=\frac{\lambda}{d}=\frac{719\times 10^{-9}}{70.3\times 10^{-6}}=0.0102 rad$

$y=r\theta=2.11\times 0.0102=0.0215 m$

The width of the pattern=$2y=2\times 0.0215=0.043 m=0.043\times 100=4.3 cm$