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.

Two Questions (MCQ) on Magnetic Effect of Electric Current

You will find many questions (with solution) involving magnetic fields on this blog, which can be accessed by trying a search for ‘magnetic field’ using the search box at the top of this page (or, you may click on the label ‘magnetic field’ below this post.

Today we will discuss two multiple choice questions on magnetic field produced by current carrying conductors:

(1) The adjoining figure shows a plane wire loop made of two semicircular portions of radii R₁ and R₂ and two straight portions. The wire loop contains a battery which drives a current I through the loop. What is the magnitude of the magnetic flux density at the common centre ‘O’ of the semicircular portions of this wire loop?

(a) (μ₀I/2π)(1/R₁ + 1/R₂)

(b) (μ₀I/4)(1/R₁ + 1/R₂)

(c) (μ₀I/4)(1/R₂ – 1/R₁)

(d) (μ₀I/4π)(1/R₂ – 1/R₁)

(e) Zero

The magnetic field produced at O by the straight portions of the loop is zero. The semicircular portion of smaller radius R₂ produces a magnetic field μ₀I/4R₂ which is directed normally into the plane of the figure (away from the reader).

[Note that a single turn plane circular coil of radius R produces a magnetic field μ₀I/2R at its centre].

The semicircular portion of larger radius R₁ produces a smaller magnetic field μ₀I/4R₁ which is directed normally outwards (towards the reader).

The resultant magnetic field at the centre O has magnitude (μ₀I/4R₂ – μ₀I/4R₁) = (μ₀I/4)(1/R₂ – 1/R₁), as given in option (c).

[The resultant field is directed normally into the plane of the figure (away from the reader). The above question can be modified to check your understanding of this fact also].

(2) A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram.

The resultant magnetic field at the centre of the coil is

(a) zero

(b) μ₀I/2R

(c) (¾)(μ₀I/2R)

(d) (¼ )(μ₀I/2R)

(e) (½)(μ₀I/2R)

This question appeared in KEAM 2009 (Engineering) question paper.

The magnetic field produced at the centre by the straight current leads is zero. So it is sufficient to consider the fields produced by the circular portions.

The currents through the branches are inversely proportional to the lengths of the branches while the magnetic fields for a given current are directly proportional to the lengths of the branches. The magnitudes of the fields due to the two branches are therefore equal. Since the fields due to the branches are perpendicular to the plane of the loop and oppositely directed, the net field at the centre is zero. This argument is enough for answering the above question.

If you want to make your argument more rigorous, you will proceed as follows:

Since the longer branch is made of three quarters of the circle and the shorter branch is made of one quarter of the circle, the resistance of the longer branch is 3 times that of the shorter branch. The currents through the shorter and longer branches are therefore given respectively by

I₁ = 3I/4 and

I₂ = I/4

The magnetic field produced at the centre by the shorter branch is directed normally into the plane of the coil (away from the reader) and has magnitude B₁ given by

B₁ = (¼)(μ₀I₁/2R) = 3μ₀I/32R, on substituting for I₁.

The magnetic field produced at the centre by the longer branch is directed normally outwards (towards the reader) and has magnitude B₂ given by

B₂ = (¾)(μ₀I₂/2R) = 3μ₀I/32R, on substituting for I₂.

The fields B₁ and B₂ get canceled and the resultant field at the centre is zero [Option (a)].

[Note that in all situations of branching of current at a circular coil, if the current leads are straight and point to the centre of the coil, the magnetic field at the centre will be zero (the angle shown need not necessarily be 90º)].

Now, find an interesting question in this section here.

“Be always at war with your vices, at peace with your neighbors, and let each New Year find you a better man.”

– Benjamin Franklin

Happy New Year…