Question

Monochromatic light falls on a screen 1.90 m from two slits separated by 2.14 mm. The first- and second-order bright fringes are separated by 0.555 mm. What is the wavelength of the light

Answer 1

Answer:

625 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:

$\Delta y=0.555\ mm$

d = 2.14 mm

L = 1.90 m

Also,

1 mm = 10⁻³ m

So,

$\Delta y=0.555\times 10^{-3}\ m$

d = 2.14×10⁻³ m

Applying in the formula,

$0.555\times 10^{-3}=1.90\times \frac {\lambda}{2.14\times 10^{-3}}\times 1$

$\lambda=625\times 10^{-9}\ m$

Also,

1 m = 10⁹ nm

So wavelength is 625 nm

Answer 2

Answer:

625.1 nm

Explanation:

Wavelength of light, $= \lambda$

Width of fringe,$\beta = 0.555 mm = 0.555 \times 10^{-3}m$

Distance of screen,$D= 1.90 m$

separation of slits, $d = 2.14 mm = 2.14 \times10^{-3} m$

The formula for fringe width is :

$\beta = \frac{\lambda D}{d}$

so rearranging the equation to get wavelength

$\lambda = \frac{\beta d}{D}\\\lambda=\frac{ 0.555 \times 10^{-3}m(2.14 \times10^{-3} m)}{ 1.90 m} \\\lambda=0.625105\times10^{-6}m\\ \lambda=625.1 nm$

Therefore, the wavelength of the light is 625.1 nm