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

625.1 nm

Step-by-step explanation:

Wavelength of light,
`= \lambda`

Width of fringe,
`\beta = 0.555 mm = 0.555 * 10^(-3)m`

Distance of screen,
`D= 1.90 m`

separation of slits,
`d = 2.14 mm = 2.14 *10^(-3) m`

The formula for fringe width is :

`\beta = (\lambda D)/(d)`

so rearranging the equation to get wavelength

`\lambda = (\beta d)/(D)\\\lambda=( 0.555 * 10^(-3)m(2.14 *10^(-3) m))/( 1.90 m) \\\lambda=0.625105*10^(-6)m\\ \lambda=625.1 nm`

Therefore, the wavelength of the light is 625.1 nm

Answer 2

625 nm

Step-by-step explanation:

For constructive interference, the expression is:

`d* sin\theta=m* \lambda`

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

d is the distance between the slits.

The formula can be written as:

`sin\theta=\frac {\lambda}{d}* m` ....1

The location of the bright fringe is determined by :

`y=L* 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* sin\theta`

Using 1, we get:

`y=L* \frac {\lambda}{d}* m`

For two fringes:

The formula is:

`\Delta y=L* \frac {\lambda}{d}* \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* 10^(-3)\ m`

d = 2.14×10⁻³ m

Applying in the formula,

`0.555* 10^(-3)=1.90* \frac {\lambda}{2.14* 10^(-3)}* 1`

`\lambda=625* 10^(-9)\ m`

Also,

1 m = 10⁹ nm

So wavelength is 625 nm