Question

You illuminate a slit with a width of 0.0537 mm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.77 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

Answer 1

Answer:

Width of central maximum will be $7.423\times 10^{-12}m$

Explanation:

We have given width of the slit $D =0.0537mm=5.37\times 10^{-5}m$

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

Distance of screen from slit D = 2.77 m

Angle to first minimum $sin\Theta =\frac{\lambda }{d}=\frac{719\times 10^{-19}}{0.0537\times 10^{-3}}=1.34\times 10^{-12}$

So width of the central maximum

w= $2y=2\times Dsin\Theta =2\times 2.77\times 1.34\times 10^{-12}=7.423\times 10^{-12}m$

So width of central maximum will be $7.423\times 10^{-12}m$