Question

A Young’s double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light

λ=700 nm in vacuum. Find the fringe-width of the pattern formed on the screen.

Answer 1

Answer:

Fringe-width of the pattern formed on the screen is 0.90 mm.

Explanation:

Given:

Separation between two slits d=0.28 mm=0.28×10⁻³ m

Distance between screen and slit (D) = 48 cm = 0.48 m

Wavelength of the red light,λa=700 nm in vaccum =700 ×10⁻⁹m

Let the wavelength of red light in water = λω

We known that refractive index of water (μw =4/3),

μw = Speed of light in vacuum / Speed of light in the water

μw=va/vω=λa/λω

⇒4/3=λa/λω

⇒λω= 3λa/4=3×700/4

=525 nm

So, the fringe width of the pattern is given by

β=λωD/d

 =525×10⁻⁹×0.48/0.28×10⁻³  

=9×10⁻⁴

=0.90 mm

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.

Answer 2

Answer:

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