Two light rays having the same wavelength λ in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index n1 while the second ray travels a path of length L2 through a medium of refractive index n2. The two waves are then combined to produce interference. The phase difference between the two waves is

Question

Two light rays having the same wavelength λ in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index n1 while the second ray travels a path of length L2 through a medium of refractive index n2. The two waves are then combined to produce interference. The phase difference between the two waves is (A) (2 π/λ) (L2 - L1) (B) (2 π/λ) (n1 L1 - n2 L2) (C) (2 π/λ) (n2 L1 - n1 L2) (D) (2 π/λ) big( (L1/n1) - (L2/n2) big)

Tardigrade Admin · Accepted Answer

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index

μ

in time t is given by

d = v t ... (i)

where v is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

Δ = c t

= c . \frac{d}{v} [f ro m Eq . (i)]

= d \frac{c}{v} = μ d ... (ii)

(S in ce, μ = \frac{c}{v})

This distance is the equivalent distance in vacuum and is called optical path.
Here, optical path for first ray =

n_{1} L_{1}

Optical path for second ray =

n_{2} L_{2}

Path difference =

n_{1} L_{1} - n_{2} L_{2}

Now, phase difference

= \frac{2 π}{λ} \times

path difference

= \frac{2 π}{λ} \times (n_{1} L_{1} - n_{2} L_{2})

$\frac{2 π}{λ} (L_{2} - L_{1})$

$\frac{2 π}{λ} (n_{1} L_{1} - n_{2} L_{2})$

$\frac{2 π}{λ} (n_{2} L_{1} - n_{1} L_{2})$

$\frac{2 π}{λ} (\frac{L _{1}}{n _{1}} - \frac{L _{2}}{n _{2}})$

λ2π​(L2​−L1​)

λ2π​(n1​L1​−n2​L2​)

λ2π​(n2​L1​−n1​L2​)

λ2π​(n1​L1​​−n2​L2​​)

$\frac{2 π}{λ} (L_{2} - L_{1})$

$\frac{2 π}{λ} (n_{1} L_{1} - n_{2} L_{2})$

$\frac{2 π}{λ} (n_{2} L_{1} - n_{1} L_{2})$

$\frac{2 π}{λ} (\frac{L _{1}}{n _{1}} - \frac{L _{2}}{n _{2}})$