Question

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wavelength of 576 nm and the interference pattern observed on a screen 3.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fourth order bright fringe on the screen? μm (b) What is the difference in path lengths from the two slits to the location of the fourth dark fringe on the screen, away from the center of the pattern? μm

Answer 1

Answer:

$1.152\ \mu m$

$1.44\ \mu m$

Explanation:

d = Gap between slits = 0.142 mm

$\lambda$ = Wavelength of light = 576 nm

L = Distance between light and screen = 3.5 m

m = Order = 2

Difference in path length is given by

$\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m$

The difference in path lengths is $1.152\ \mu m$

For dark fringe the difference in path length is given by

$\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m$

The difference in path length is $1.44\ \mu m$