1 answer

A single-slit diffraction pattern is formed when light of λ = 759.0 nm is passed through...

Question:

A single-slit diffraction pattern is formed when light of λ = 759.0 nm is passed through a narrow slit. The pattern is viewed on a screen placed one meter from the slit. What is the width of the slit (mm) if the width of the central maximum is 2.49 cm?

Please show all work!


Answers

The width b of central maxima is given by

b = 2Dk/a

Where D is the distrance of screen from slit, k is the wavelength of light and a is width of the slit.

Rearranging the terms, we can write

a= 2Dk/b ...eqn 1

Given, k= 759.0 nm = 759*10-9 m

D = 1 m

b = 2.49 cm = 0.0249 m

Putting all these values in eqn 1, we get

a= 2*1*759*10-9/0.0249

= 1518*10-9/0.0249

= 60963.8554*10-9 m

= 0.060*10-3 m

a = 0.060 mm

Which is the required slit width.

.

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