(a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.

OR

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10^-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

(a) In Young’s double slit experiment we consider two coherent sources S₁ and S₂ which are separated by a distance d and the distance of the screen from the sources is D.

The path difference between the waves reaching the point P is,

x = S₂P – S₁P

In ΔS₂PF

In ΔS₁PE

On expanding the above term,

For the bright fringes the path difference is an integral multiple of wavelength, i.e path difference is nλ.

We get,

The fringe width ()is defined as the distance between the consecutive bright and dark fringe,

The separation between the centres of two consecutive dark interference fringe width of a bright fringe.

Therefore,

Β₁=β₂

All the bright and ark fringes are equal width.

(b) Let the amplitude of wave of the slit 1 be A₁ and of slit 2 be A₂

Also, the maximum amplitude will come when constructive interference happens,

A_max = A₁ + A₂

And, the minimum amplitude will come when destructive interference happens,

A_min = A₁ - A₂

Also, we know that Intensity of radiation is proportional to the square of its amplitude, so,

We can argue that the intensity of the radiation is directly proportional to the width of the slit, so,

The ratio of the width of the slits is 16:1.

OR

When a single slit is illuminated by a monochromatic source of light, a pattern of alternate pattern of dark and bright regions is obtained, whose intensity gets weaker as going away from the center. The intensity at any point on the screen is due the path difference between the waves emanating from different parts of the wave-front at the slit.

Consider a point P on screen at which wavelets moving in a direction making an angle θ with CM are bought to focus by the lens.

For the right angle Δ LNQ

NQ=LN sin θ

NQ=a sin θ

Suppose BN =λ and θ= θ₁

Then, the above equation becomes

λ=a sin θ₁

In general, for n th minimum at point p.

Sin θ_n=nλ/a

For small angle sin θn=θn=tan θn=y_n/D

On comparing the above we get,

Width of the secondary maximum,

Width of the secondary minimum,

We get,

β'=β

Therefore, the fringe width is the same for the secondary maximum and minimum.

b) Given: -

Wavelength of first sodium light, λ₁ = 590 nm

Wavelength of second sodium light, λ₂ = 596 nm

Width of the single slit, d = 2 × 10^-6 m

Distance of the screen, D=1.5 m

Formula: -

The nth maxima occurs at,

Where

λ is the wavelength

D is the distance between the screen and slit

d is the thickness of the slit

Calculations: -

So, the separation between the two cases is,

Conclusion: -

The Separation between the maximas is 6.75 mm.