Question

(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double-slit experiment. 

(b) In the diffraction due to a single-slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first-order minimum and the third-order maximum on one side of the screen. The distance between the slit and the screen is 1.5 m. 

Answer 1

(a) (i) Difference between interference and diffraction phenomena:

Interference	Diffraction
1. Interference of light is the phenomenon of redistribution of light energy in a medium due to superposition of light waves from two coherent sources.	1. Diffraction of light is the phenomenon of bending of light around corners of an obstacle in the path of light due to which light penetrates into geometrical shadow of obstacle.
2. All bright fringes have equal intensity.	2. Intensity of successive bright fringes decreases.
3. All dark fringes have zero intensity.	3. Intensity of dark fringes is not zero.
4. Fringes obtained through interference of monochromatic light have equal width.	4. Fringes obtained through diffraction of monochromatic light do not have equal width.

(ii) Intensity at a point of the interference pattern in Young’s double-slit experiment: Consider the following figure, where S is monochromatic source of light. Also, A and B are two narrow slits which generate two coherent sources of light P is any point on the screen.

Let the displacement of the waves be represented by

where a and b are amplitudes of two waves and ϕ is the phase difference. According to the principle of superposition, the resultant displacement is equal to the vector sum of the displacement of individual waves. Therefore,

y = y₁ + y₂ = a sin ω_t + b sin (ω_t + ϕ)

Using the trigonometric identity: sin(a + b) = sina cosb + cosa sinb, we get

y = a sin ωt + B sin ωt Cos ϕ + b sin ϕ  Cos ωt 

= Sin ωt (a + b Cos ϕ) + b sin ϕ Cos ωt

Substituting 

a + b Cos ϕ = R Cos θ

and b Sin ϕ R Sin θ

Therefore,

y = Sin ωt R Cos θ + Cos ωt R Sin θ

= R (Sin ωt  + Cos θ + Cos ωt sin θ)

= R Sin (ωt + θ)

Thus, the resultant wave has amplitude R and phase angle θ. Now, squaring Eqs. (1) and (2) and adding, we get

We know that the intensity is directly proportional to the square of amplitude of wave. Therefore, 

Also

Therefore, the intensity of light at a point of interference is

(b) Given:

• Aperture of slit is a = 3 mm. 

• Wavelength of incident light, λ = 620 nm 

• Distance between screen and slit, D = 1.5 m.

For diffraction minimum: a sinθ . n λ

For small θ = n λ .

For diffraction maxima:

 For first-order minima: n = 1. Therefore,

For third-order maxima: n = 3.

Separation:

Therefore,

Substituting values of λ, , D and a, we get

So, the separation between first-order minima and third order maxima is 0.775 mm.