Question

Determine the wavelength of a monochromatic beam of light that impinges on a double slit with a slit separation of 2.80 mm. Bright fringes are produced on a screen 1.20 m from the double slit. The brights are separated by .024m.

Answer 1

Answer:

$\lambda=5.60\times 10^{-5}\ m$

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

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

For small angle , $sin\theta=tan\theta$

So,  

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

For two fringes:

The formula is:

$\Delta y=L\times \frac {\lambda}{d}\times \Delta m$

For first and second bright fringe,  

$\Delta m=1$

Given that:

$\Delta y=0.024\ m$

d = 2.80 mm

L = 1.20 m  

Also,  

1 mm = 10⁻³ m

d = 2.80×10⁻³ m

Applying in the formula,  

$0.024=1.20\times \frac {\lambda}{2.80\times 10^{-3}}\times 1$

$\lambda=5.60\times 10^{-5}\ m$