Question

A single slit 1.3 mm wide is illuminated by 420-nm light. What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.0 m away? Express your answer using two significant figures.

Answer 1

26 cm

Step-by-step explanation:

Width of slit = 1.3 mm = d

Wavelength = λ = 420 nm

Distance to screen = 4 m

`x=(\lambda D)/(d)\\\Rightarrow x=(420* 10^(-9)* 4)/(1.3* 10^(-3))`

Width of central maximum is 2 times the distance from the center to first maxima

`2x=2* (420* 10^9* 4)/(1.3* 10^(-3))=0.00258 m = 2.58* 10^(-3)\ m =26\ cm`

Width of the central maximum is 26 cm