Question

A laser beam is incident on two slits with a separation of 0.200 mm, and a screen is placed 5.00 m from the slits. If the bright interference fringes on the screen are separated by 1.58 cm, what is the wavelength of the laser light?

Answer 1

Answer:

632 nm

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:

$\Delta y=1.58\ cm$

d = 0.200 mm

L = 5.00 m  

Also,  

1 cm = 10⁻² m

1 mm = 10⁻³ m

So,  

$\Delta y=1.58\times 10^{-2}\ m$

d = 0.2×10⁻³ m

Applying in the formula,  

$1.58\times 10^{-2}=5.00\times \frac {\lambda}{0.2\times 10^{-3}}\times 1$

$\lambda=632\times 10^{-9}\ m$

Also,  

1 m = 10⁹ nm

So wavelength is 632 nm