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

Answer:

492.183 nm

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

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

Answer 2

Answer:

The wavelength of the light is 446.8 nm.

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

$y=\dfrac{m\lambda D}{d}$

$\lambda=\dfrac{yd}{mD}$

Put the value into the formula

$\lambda=\dfrac{1.35\times10^{-3}\times0.705\times10^{-3}}{1\times2.13}$

$\lambda=4.468\times10^{-7}\ m$

$\lambda=446.8\ nm$

Hence, The wavelength of the light is 446.8 nm.