The yellow component of light from a helium discharge tube A

The yellow component of light from a helium discharge tube (A = 587.5 nm) is allowed to fall on a plate containing parallel slits that are 0.200 mm apart. A screen is positioned so that the second bright fringe in the interference pattern is a distance equal to 10 slit spacings rrom the central maximum. What is the distance between the plate and the screen?

Solution

Wavelength of light,lamda = 587.5 nm = 587.5 x10 ^-9 m

Separation fo the slits d = 0.2 x10 -3 m

Distance of the second bright fringe from central maximum y = 10 d

Let the distance between the plate and the screen be D.

Condition for second bright fringe is d sin(theta) = 2(lamda)

For small angles , sin(theta) ~ tan(theta) = y / D

So, condition becomes , d (y/D ) = 2(lamda)

                                    d(10d/D ) = 2(lamda)

From this D = 10 d ² / [2(lamda)]

                  =10(0.2x10 ^-3) ² /[2(587.5 x10 ^-9 )]

                  = 0.3404 m