Question

A youngs interference experiment is performed with blu-green argon laser light .The slits is 0.5mm, and the screen is located 3,30m from the slits .the first bright fringe is located 3.40mm from the center of the interference pattern. what is the wavelength of the argon laser light?

Answer 1

Answer:

515 nm

Explanation:

The equation that gives the position of the maxima (bright fringes) in the interference pattern produced on a distant screen by the diffraction of light through two slits is

$y=\frac{m\lambda D}{d}$

where

y is the distance of the maximum (bright fringe) from the central maximum

m is the order of the maximum

$\lambda$ is the wavelength of the light used

D is the distance of the screen

d is the separation between the slits

In this problem we have:

$d=0.5 mm = 0.5\cdot 10^{-3} m$ is the separation between the slits

D = 3.30 m is the distance of the screen

m = 1 (we are analyzing the 1st bright fringe)

$y=3.40 mm = 3.40\cdot 10^{-3} m$ is the distance of the 1st bright fringe from the central maximum

Solving for $\lambda$, we find the wavelength of the light used:

$\lambda=\frac{yd}{mD}=\frac{(3.40\cdot 10^{-3})(0.5\cdot 10^{-3})}{(1)(3.30)}=5.15\cdot 10^{-7} m = 515 nm$