Question

In Young's double slit experiment, using monochromatic light of wavelength λ, the

intensity of light at a point on the screen where path difference is λ, is K units. Find out

the intensity of light at a point where path difference is λ/3.​

Answer 1

Answer:

For monochromatic light, I

1

+I

2

is the intensity.

I

′

=I

1

+I

2

+2

I

1

I

2

cosθ

Phase difference is given by:

ϕ=2π×

λ

path difference

ϕ=2π×

λ

λ

=2π

So, the new intensity can be obtained as: ⇒I

′

=4I

1

given I

′

=K,I

1

=K/4,path difference=λ/3

phase difference is 2π×1/3

Hence, I=I

1

+I

2

+2

I

1

I

2

cos2π/3

⇒I=I

1

=K/4

Answer 2

Given info : In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units.

To find : Find put the intensity of light at a point where path difference is λ/3.

solution : phase difference is given by,

Φ = 2π × path difference/λ

case 1 : given path difference = λ

so phase difference, Φ = 2π × λ/λ = 2π

let I₁ = I₂ = I₀

so, I = I₁ + I₂ + 2√(I₁I₂) cosΦ

⇒ k = I₀ + I₀ + 2√(I₀I₀)cos2π

= 4I₀

so, I₀ = K/4 ------- (1)

case 2 : path difference = λ/3

so, phase difference, Φ' = 2π × (λ/3)/λ = 2π/3

now intensity, I' = I₀ + I₀ + 2√(I₀I₀) cos(2π/3)

= I₀

so, I' = k/4 [ from eq (1). ]

Therefore intensity of light at a point where path difference is λ/3, is k/4.