Question

a glass plate of thickness 1.5um and refractive index 1.5 is introduced between one of the slits and screen in a youngs double slit experiment. if the wavelength of the monochromatic source used is 0.75um, the phase difference between the interfering waves at the centre of the screen is equal to

a. 6pi
b. 3pi
c. pi
d. 2pi

Answer 1

Answer:

your answer is A

Explanation:

may it helps you

Answer 2

The option is (d): The phase difference between the interfering waves at the center of the screen is equal to 2 pi.

Given:

• The plate of the thickness(t) 1.5 μm= $1.5*10^{-6}$
• Glass plate refractive index(μ) 1.5
• monochromatic source used i.e wavelength of light is 0.75um=$0.75*10^{-6}$

To find:

The phase difference between the interfering waves at the center of the screen.

Solution:

We know that phase difference is given by the formula,

δ = (2π/λ)(Δx)                                      ...(I)

Where,

δ- Phase Difference

Δx- Optical Path Difference

Optical path difference is given by the formula,

Δx=(μ-1)t

Δx=$(1.5-1)*(1.5*10^{-6})$

Δx= $0.5*(1.5*10^{-6})$

Δx= $0.75*10^{-6})$

From (I),

δ = (2π/λ)

   =$\frac{2\pi }{0.75*10^{-6}} 0.75*10^{-6})$    

   =$2\pi$

Hence, The option is (d): The phase difference between the interfering waves at the center of the screen is equal to 2π.

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