Question

A two-slit arrangement with 60.3 μm separation between the slits are illuminated with 537.0-nm wavelength light. If a viewing screen is located 2.14 m from the slits, find the distance on the screen from the first dark fringe on one side of the central maximum to the second dark fringe on the other side.

Answer 1

Answer:

38.12 mm

Explanation:

In the interference pattern created by the diffraction of light through a double slit, the position of the maxima (bright fringes) are given by

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

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen from the slits

d is the separation between the slits

On the other hand, the position of the minima (dark fringes) is given by

$y_m = \frac{(m+\frac{1}{2})\lambda D}{d}$

where

(m+1) is the order of the minima

In this problem we have:

$d=60.3\mu m = 60.3\cdot 10^{-6} m$ is the separation between the slits

$\lambda=537.0 nm = 537.0\cdot 10^{-9} m$ is the wavelength of light

D = 2.14 m is the position of the screen

So, distance of the first dark fringe (m=0) from the central maximum is

$y_1 = \frac{(1/2)\lambda D}{d}=\frac{(537.0\cdot 10^{-9})(2.14)}{2(60.3\cdot 10^{-6})}=9.53\cdot 10^{-3} m$

On the other side, the distance of the second dark fringe (m=1) from the central maximum is

$y_2= \frac{(3/2)\lambda D}{d}=\frac{3(537.0\cdot 10^{-9})(2.14)}{2(60.3\cdot 10^{-6})}=28.59\cdot 10^{-3} m$

Therefore, the  distance between the two dark fringes is:

$d' = y_1 + y_2 = 9.53\cdot 10^{-3} + 28.59\cdot 10^{-3} = 38.12 \cdot 10^{-3} m = 38.12 mm$