Q#28
(a) Apparatus used to produce interference fringes is shown in Fig. 6.1. The apparatus is
not drawn to scale.
Fig. 6.1 (not to scale)
Laser light is incident on two slits. The laser provides light of a single wavelength.
The light from the two slits produces a fringe pattern on the screen. A bright fringe is
produced at C and the next bright fringe is at B. A dark fringe is produced at P.
(i) Explain why one laser and two slits are used, instead of two lasers, to produce a
visible fringe pattern on the screen. [1]
(ii) State the phase difference between the waves that meet at
1. B [1]
2. P [1]
(iii) 1. State the principle of superposition. [2]
2. Use the principle of superposition to explain the dark fringe at P. [1]
(b) In Fig. 6.1 the distance from the two slits to the screen is 1.8 m. The distance CP is
2.3 mm and the distance between the slits is 0.25 mm.
Calculate the wavelength of the light provided by the laser. [3]
Solution:
(a)
(i) To produce coherent sources or constant phase difference.
(ii)
1. 360o / 2π allow n × (360o) or n × (2π)
{A bright fringe is seen due to constructive interference. So, the waves from the slits should be in phase. The phase difference is thus 360o / 2π or a multiple of it – that is, the 2 waves are completely in phase. A phase difference of 360o / 2π means no phase difference.
Note that the phase difference of 360° means that the waves are in phase. It produces the same effect as two waves having a phase difference of 0 (that is, they are also in phase).
2. 180o / π allow (n × 360o) – 180o or (n × 2π) – π
{A dark fringe is seen due to destructive interference. So, the waves from the slits should be out of phase. The phase difference is thus 180o / π or (n × 360o) – 180o or (n × 2π) – π – that is, the 2 waves are completely out of phase.}
(iii) 1. The principle of superposition states that when waves overlap / meet, the (resultant) displacement is the sum of displacements of each wave.
2. At P, there is a crest on a trough
(b)
{CP is the width between a dark fringe and a bright fringe. In the equation below, the fringe width is the distance between 2 successive bright fringe or 2 successive dark fringe. So, we need to consider twice the distance CP.}
Wavelength λ = ax / D
Wavelength λ = (2× 2.3×10-3) × (0.25×10-3) / 1.8
Wavelength λ = 639 nm
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