Question

A beam of light of wavelength 400 nm from a
distant source falls on a single slit 2.00 mm wide
and the resulting diffraction pattern is observed on
screen 2 m away. The distance between the first
dark fringes on either side of the central fringe is...

ps : required ASAP!​

Answer 1

Solution :

• Wavelength = 400 nm
• Width of the slit = 2 mm
• Distance between the screen and the slit = 2 m

Let the distance of the first dark fringe from the light fringe be y

Then , the distance between the first dark fringe on the either sides of the central fringe = 2y

We know that ,

➽ d sin θ = n λ

As θ is very small

∴ sin θ ≈ θ

➽ θ = tan θ = y / D

hence ,

➽ d y /D = n λ

On rearranging ,

➽ y = nλD /d

➽ y = 1 x 400 x 10-⁹  x 2 / 2 x 10-³

➽ y = 400 x 10 -⁶

➽ y = 400 μm

Now ,

The distance between the first dark fringe on the either sides of the central fringe

= 2y

= 2 x 400 μm

= 800 μm

= 0.8 mm

The distance between the first dark fringe on the either sides of the central fringe is 0.8 mm

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