Question

A slit of width 0.1 mm is illuminated with monochromatic light of wavelength 682.22 nm, and a diffraction pattern is formed on a screen 1.05 m away from the slit. What is the width of the central maximum in millimeters?

Answer 1

`\begin{equation*} 14.33mm \end{equation*}`

Step-by-step explanation

Parameters given:

Width of the slit, a = 0.1 mm = 1 * 10^(-4) m

Wavelength of the light, λ = 682.22 nm = 682 * 10^(-9) m

Distance of the screen from the slit, D = 1.05 m

To find the width of the central maximum of the diffraction pattern, apply the formula:

`\alpha=(2\lambda D)/(a)`

Therefore, for the given diffraction pattern, the width of the central maximum is:

`\begin{gathered} \alpha=(2*682.22*10^(-9)*1.05)/(1*10^(-4)) \\ \\ \alpha=0.01433m=14.33mm \end{gathered}`

That is the answer in millimeters.