Interference fringes may be observed using a light-emitting laser to illuminate a double slit.

Question 17

(a) Interference fringes may be observed using a light-emitting laser to illuminate a double slit.

The double slit acts as two sources of light.

Explain

(i) the part played by diffraction in the production of the fringes, [2]

(ii) the reason why a double slit is used rather than two separate sources of light. [1]

(b) A laser emitting light of a single wavelength is used to illuminate slits S1 and S2, as shown in Fig. 6.1.

Fig. 6.1 (not to scale)

An interference pattern is observed on the screen AB. The separation of the slits is 0.48 mm. The slits are 2.4 m from AB. The distance on the screen across 16 fringes is 36 mm, as illustrated in Fig. 6.2.

Fig. 6.2

Calculate the wavelength of the light emitted by the laser. [3]

(c) Two dippers D1 and D2 are used to produce identical waves on the surface of water, as illustrated in Fig. 6.3.

Fig. 6.3 (not to scale)

Point P is 7.2 cm from D1 and 11.2 cm from D2.

The wavelength of the waves is 1.6 cm. The phase difference between the waves produced at D1 and D2 is zero.

(i) State and explain what is observed at P. [2]

(ii) State and explain the effect on the answer to (c)(i) if the apparatus is changed so that, separately,

1. the phase difference between the waves at D1 and at D2 is 180°,

2. the intensity of the wave from D1 is less than the intensity of that from D2.

 [2]

[Total: 10]

Reference: Past Exam Paper – June 2017 Paper 22 Q6

Solution:

(a)

(i)

The waves at (each) slit/aperture spread (into the geometric shadow) and the wave(s) overlap.

(ii) The waves/light from the double slit are coherent/have a constant phase difference

(b)

x = λD / a

{λ = xa / D

Separation between 16 fringes = 36×10^-3 m

Fringe separation x = (36×10^-3 / 16) m}

λ = (36 × 10^-3 × 0.48 × 10^-3) / (16 × 2.4)

λ = 4.5 × 10^-7 m                                 

(c)

(i)

There is no movement of the water as the path difference is 2.5λ

{Wavelength λ = 1.6 cm

Path difference = path D₂P – path D₁P = 11.2 – 7.2 = 4.0 cm

In terms of λ, path difference = 4.0 / 1.6 = 2.5λ

A path difference of 2.5λ means that destructive interference occurs.

So, at P, no movement of the water would be observed (no disturbance / no ripples).}

(ii)

1. The surface/water/P vibrates/ripples as (waves from the two dippers) arrive in phase   

{When there were initially in phase (as in part(i)), destructive interference occurred. So, if they are now out of phase, constructive interference would occur.}

2. The surface/water/P vibrates/ripples as amplitudes/displacements are no longer equal/do not cancel

{We return to the same phase difference (no phase difference) as in part (i) where destructive interference occurred.

If the intensity of one of them is changed, its amplitude would also change.

Let say amplitude from D₁ = 0.5 A and amplitude from D₂ = 1 A

Upon superposition of the waves, resultant = 1A – 0.5A = 0.5 A

So, even if destructive interference occurs, the resultant still has an amplitude. So, the particles at P would vibrate as the displacements no longer cancel out.}