A Light of wavelength 650 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is 47 mm from the central fringe on a screen 1.4 m away. What is the separation of the two slits?

Answer :

Answer:

5.81 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

[tex]d\times sin\theta=m\times \lambda[/tex]

d is the distance between the slits.

The formula can be written as:

[tex]sin\theta=\frac {\lambda}{d}\times m[/tex] ....1

The location of the bright fringe is determined by :

[tex]y=L\times tan\theta[/tex]

Where, L is the distance between the slit and the screen.

For small angle , [tex]sin\theta=tan\theta[/tex]

So,  

Formula becomes:

[tex]y=L\times sin\theta[/tex]

Using 1, we get:

[tex]y=L\times \frac {\lambda}{d}\times m[/tex]

Given that:

m = 3, y = 47 mm (Location of the fringe)

Since,  

1 mm = 0.001 m

y = 0.047 m

Given L =1.4 m

λ = 650 nm  

Since, 1 nm = 10⁻⁹ m

So,  

λ = 650 × 10⁻⁹ m

Applying the formula as:

[tex]0.047\ m=1.4\ m\times \frac {650\times 10^{-9}\ m}{d}\times 3[/tex]

⇒ d, distance between the slits = 5.81 × 10⁻⁵ m