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Question 920: [Waves > Double Slit]

A double-slit interference experiment is set up as shown.

Fringes are formed on screen. Distance between successive bright fringes is found to be 4 mm.

Two changes are then made to the experimental arrangement. The double slit is replaced by another double slit which has half the spacing. The screen is moved so that its distance from the double slit is twice as great.

What is now the distance between successive bright fringes?

A 1 mm                       B 4 mm                       C 8 mm                       D 16 mm

Reference: Past Exam Paper – June 2006 Paper 1 Q28

Solution 920:

Answer: D.

For double-slit experiment: λ = ax / D

Where λ = wavelength, a = slit separation, x = fringe separation, D = distance of the screen from the slits.

The wavelength is the same is both experiments since the same red light source is used.

1^st experiment:

x = 4mm; a = slit separation; D = distance of the screen from the slits

λ = 4a / D

2^nd experiment:

Slit separation = a / 2; Distance of the screen from the slits = 2D

λ = (a/2)x / 2D = ax / 4D

Since the wavelength is the same in both cases, the 2 equations obtained can be equated.

ax / 4D = 4a / D

New fringe separation, x = 4 × 4 = 16mm

Question 921: [Waves > Intensity]

A health inspector is measuring the intensity of a sound. Near a loudspeaker his meter records an intensity I. This corresponds to an amplitude A of the sound wave. At another position the meter gives an intensity reading of 2I.

What is the corresponding sound wave amplitude?

A A / √2                      B √2 A                        C 2 A                          D 4 A

Reference: Past Exam Paper – June 2005 Paper 1 Q25

Solution 921:

Answer: B.

The intensity I is proportional to (amplitude, A)².

In other words, the amplitude A is proportional the √I.

When the intensity changes from I to 2I, the amplitude is now proportional to √(2I) = √2 (√I).

Recall from above that A is proportional to √I and vice versa – that is √I is also proportional to A.  So, the corresponding sound wave amplitude is √2 A.

Question 922: [Electromagnetism]

(a) A uniform magnetic field has constant flux density B. A straight wire of fixed length carries a current I at an angle θ to the magnetic field, as shown in Fig.1.

(i) The current I in the wire is changed, keeping the angle θ constant. 

On Fig.2, sketch a graph to show the variation with current I of the force F on the wire.

(ii) The angle θ between the wire and the magnetic field is now varied. The current I is kept constant.

On Fig.3, sketch a graph to show the variation with angle θ of the force F on the wire.

(b) A uniform magnetic field is directed at right-angles to the rectangular surface PQRS of a slice of a conducting material, as shown in Fig.4.

Electrons, moving towards the side SR, enter the slice of conducting material. The electrons enter the slice at right-angles to side SR.

(i) Explain why, initially, the electrons do not travel in straight lines across the slice from side SR to side PQ.

(ii) Explain to which side, PS or QR, the electrons tend to move.

Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q6

Solution 922:

(a)

(i) The graph is a straight line with positive gradient, passing through the origin

(ii)

{F = BIL sinθ }

For the graph, the maximum force is shown at θ = 90^o and zero force is shown at θ = 0^o. The graph is a reasonable curve with the force being about half the maximum at 30^o. 

(b)

(i) There is a force on the electrons due to the magnetic field. The force on the electrons is normal to the magnetic field and to the direction of the electrons. 

(ii)

A quote / mention of (Fleming’s) left hand rule.

The electron moves towards QR.

Question 923: [Waves > Intensity]

A source of sound of constant power P is situated in an open space. Intensity I of sound at distance r from this source is given by

I = P / 4πr².

How does the amplitude a of the vibrating air molecules vary with the distance r from the source?

A a ∝ 1 / r                   B a ∝ 1 / r²                  C a ∝ r                         D a ∝ r²

Reference: Past Exam Paper – June 2011 Paper 11 Q24 & Paper 13 Q22

Solution 923:

Answer: A.

From the equation given, the intensity I is inversely proportional to r².

In terms of amplitude a, the intensity I is proportional to a².

Thus, a² is inversely proportional to r². Amplitude a is inversely proportional to r.