Question

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm .

1. Calculate the wavelength of the light (answer in nm).

Answer 1

The wavelength of the light is 446.8 nm.

Step-by-step explanation:

Given that,

Width = 0.705 mm

Distance = 2.13 m

Diffraction pattern = 1.35 mm

Number of order = 1

We need to calculate the wavelength of light

Using formula of wavelength

`y=(m\lambda D)/(d)`

`\lambda=(yd)/(mD)`

Put the value into the formula

`\lambda=(1.35*10^(-3)*0.705*10^(-3))/(1*2.13)`

`\lambda=4.468*10^(-7)\ m`

`\lambda=446.8\ nm`

Hence, The wavelength of the light is 446.8 nm.

Answer 2

492.183 nm

Step-by-step explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

`tan\theta=(x)/(l)\\\Rightarrow \theta=tan^(-1)(1.35* 10^(-3))/(2.13)\\\Rightarrow \theta=0.04^(\circ)`

`dsin\theta=m\lambda\\\Rightarrow \lambda=(dsin\theta)/(m)\\\Rightarrow \lambda=(0.705* 10^(-3)* sin0.04)/(1)\\\Rightarrow \lambda=4.92183* 10^(-7)=492.183\ nm`

The wavelength of the light is 492.185 nm