Karnataka Common Entrance Test (CET) 2006 Questions on Interference and Diffraction of Waves

The following three questions appeared in Karnataka CET 2006 question paper:

(1) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. This is due to

(1) diffraction of signal received from the antenna

(2) interference of the direct signal received by the antenna with the weak signal reflected by the passing aircraft

(3) change of magnetic flux occurring due to the passing aircraft

(4) vibrations created by the passing aircraft.

The microwaves reaching the antenna directly and after reflection at the aircraft interfere. The interference pattern produced is unstable since the aircraft is moving. This results in a shaking picture [Option (2)].

(2) A beam of light of wave length 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

(1) 1.2 cm

(2) 1.2 mm

(3) 2.4 cm

(4) 2.4 mm

This is a popular question repeatedly asked in various entrance tests with changes in numerical values.

The diffraction bands are distributed symmetrically on either side of the central maximum and the angular separation of the centre of the first dark fringe from the centre of the central maximum is λ/a where λ is the wave length of light and a is width of the slit.

Therefore, the angular separation between the first dark fringes on either side of the central bright fringe is 2λ/a.

The distance (linear separation) between the first dark fringes on either side of the central bright fringe is D×2λ/a where D is the distance between the slit and the screen and we have

D×2λ/a = 2×(2×600×10^–9/10^–3) = 2.4×10^–3 m = 2.4 mm.

(3) If white light is used in the Newton’s ring experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light through the same point. This is due to

(1) 90º change of phase in one of the reflected waves

(2) 180º change of phase in one of the reflected waves

(3) 145º change of phase in one of the reflected waves

(4) 45º change of phase in one of the reflected waves

The mechanism of formation of Newton’s rings is the same as that of the production of colours in thin films and is shown in the figure. Usually the light beam falls normally on the air film included between the upper lens and the lower glass plate, but we have shown the rays slanting to make things clear.

Newton’s rings in the reflected system are formed by the interference of waves 1 and 2 (fig.) where as Newton’s rings in the transmitted system are formed by the interference of waves 3 and 4.

If the phase change of π introduced due to reflection at a denser medium is ignored, the difference between the path lengths of the rays 1 and 2 is the same as the difference between the path lengths of the rays 3 and 4.

In the case of waves 1 and 2 there is an additional path difference of λ/2 because of the phase difference π introduced due to reflection at B. In the case of waves 3 and 4 there is an additional path difference of λ because of the phase difference 2π introduced due to reflections at B and C. If the condition for brightness is satisfied for one colour in the case of the interfering waves in the reflected system, the condition for darkness will be satisfied for the same colour in the case of the interfering waves in the transmitted system. So the colours present in the reflected system will be absent in the transmitted system and vice versa. The basic reason for this is the 180º phase change [Option (2)].

You will find a useful post in this section at physicsplus.

Optics-Questions on Interference

The following MCQ appeared in Kerala Engineering Entrance 2005 test paper:

In the Young’s double slit experiment, the intensity of the central maximum is observed to be I₀. If one of the slits is covered, the intensity at the central maximum will become

(a) I₀/2 (b) I₀/√2 (c) I₀/4 (d) I₀ (e) I₀²

If the resultant amplitude (due to the two interfering waves) at the central maximum is ‘a’. we can write

I₀ α a², since the intensity is proportional to the square of the amplitude.

When one of the slits is covered, the amplitude is reduced to a/2. If ‘I’ is the intensity at the position of the central maximum now, we can write

I α (a/2)².

From the above, we obtain I = I₀/4 [Option (c)].

Now, consider the following question:

In a double slit interference pattern, the intensity at the centre of a bright fringe is I. The intensity at a point one quarter of the distance to the next bright fringe is

(a) I/2 (b) I/4 (c) I/8 (d) I (a) zero

At the centre of a bright fringe the waves arrive in phase. You may imagine that the photons starting from the two slits are in the same state of vibration when they reach the position of the centre of a bright fringe and that is why their amplitudes get added to produce maximum intensity. At the centre of the next bright fringe, the photons will have an extra phase difference of 2π, but this too is ‘in phase’ condition (for the same state of vibration).

At a point one quarter of the distance to the next bright fringe, the phase difference between the interfering photons will be 2π/4 = π/2.

If ‘a’ is the amplitude of each interfering wave, the resultant amplitude at the centre of a bright fringe is 2a and the intensity I is given by

I α 4a²

At a point one quarter of the distance to the next bright fringe, the amplitudes are added with a phase difference of π/2 and the resultant amplitude is √(a² + a²) = √2 a. The intensity (I') in this case is given by

I' α 2a²

From the above expressions, we obtain I' = I/2 [Option (a)].

[Note that the intensity (I) produced by two interfering waves of the same amplitude ‘a’ is given by I α 4a²cos²(δ/2) where ‘δ’ is the phase difference].

You will find more multiple choice questions with solution in this section at physicsplus: Multiple Choice Questions on Wave Optics and at physicsplus: Questions on Polarisation