Question

Question: Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm . Calculate the wavelength of the light.

Short answer

Light of 506 nm is used.

Step-by-step solution

Given Data

• Width of slit, $a=0.750mm=0.750\times {10}^{-3}\hspace{0.33em}m$
• Distance between screen and slit, R = 2.00 m
• Distance of first minima from central maxima, $y=1.35mm=1.35\times {10}^{-3}m$

Concept

Diffraction is a process by which a beam of light spreads when passing through a narrow passage or across the edge of an obstacle accompanied by interference between the waveforms.

Wavelength of light source

By using the equation:

$$\begin{gathered} \sin \theta =\frac{m\lambda }{a} \\ \lambda =\frac{a\sin \theta }{m} \end{gathered}$$

Here, a is the slit width, $\lambda$is the wavelength of light used and is the diffraction angle.

For first minima, m = 1

$\lambda =a\sin \theta$

To obtain the value of θ

$$\begin{gathered} tan\theta =\frac{y}{R} \\ \theta =ta{n}^{-1}\frac{y}{R} \\ \theta =ta{n}^{-1}\left(\frac{1.35\times {10}^{-3}\hspace{0.33em}m}{2.0m}\right) \\ \theta =0.0387° \end{gathered}$$

Therefore, the wavelength of source is given by

$$\begin{gathered} \lambda =a\sin \theta \\ =0.750\times \sin \left(0.0387°\right) \\ =5.06\times {10}^{-7}m \\ =506nm \end{gathered}$$

Therefore, the wavelength of the light source is 506 nm.