Question

A beam of yellow light is made to pass through two slits that are 3.0 x 10−3 meters apart. On a screen 2.0 meters away from the slits, an interference pattern appears with bands of light separated by 3.9 x 10−4 meters. What is the wavelength of the light?

Answer 1

Answer:

585 nm

Explanation:

The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:

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

where

m is the order of the maximum

$\lambda$ is the wavelength

D is the distance of the screen from the slits

d is the separation between the slits

The distance between two consecutive bright fringes therefore is given by:

$\Delta y = \frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}$

In this problem we have:

$\Delta y = 3.9\cdot 10^{-4} m$ (distance between two bright fringes)

D = 2.0 m (distance of the screen)

d = 3.0 x 10−3 m (separation between the slits)

Solving for $\lambda$, we find the wavelength:

$\lambda=\frac{\Delta y d}{D}=\frac{(3.9\cdot 10^{-4})(3.0\cdot 10^{-3})}{2.0}=5.85\cdot 10^{-7} m = 585 nm$