# Physics 9702 Doubts | Help Page 189

__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 / 4πr

^{2}.
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 / r

^{2}C a ∝ r D a ∝ r^{2}**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 r

^{2}.
In terms of amplitude a, the intensity
I is proportional to a

^{2}.
Thus, a

^{2}is inversely proportional to r^{2}. Amplitude a is inversely proportional to r.
Please consider answering ALL of the following questions before October:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.10(c)(ii),Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

42/O/N/09 Q.5(a),Q.7(b)(ii)

41/M/J/10 Q.6(a),Q.7(a)

51/M/J/10 Q.2(d)

42/O/N/10 Q.3(c)

For 42/O/N/09 Q.5(a), go to

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2009-paper-42-worked.html

q23 q32, 33, 31 june 2005

ReplyDeleteFor Q23, see solution 952 at

Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-196.html

Solution 922 bii) Sir according to Flemings left hand rule the thumb is pointing towards side PS so how come electrons will move towards QR? please explain..

ReplyDeleteIn Fleming's left hand rule, the middle finger gives the direction of current, and current is OPPOSITE to the direction of flow of electrons.

DeleteThanks,a lot I totally understand now..

Delete