Question

Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum dark fringe appears 3 mm from the central maximum. What is the width of the slit?

Answer 1

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light $\lambda =550nm=550\times 10^{-9}m$

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So $x=3\times 10^{-3}m$

We have to find the width of the slit

For the first order wavelength is equal to $\lambda =\frac{x}{D}\times a$, here a width of slit

So $a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm$

So width of the slit will be equal to 1.47 mm