# Physics 9702 Doubts | Help Page 167

__Question 835: [Waves > Interference > Double Slits]__
Fringes of separation y are observed
on screen 1.00 m from a Young’s slit arrangement that is illuminated by yellow
light of wavelength 600 nm.

At which distance from the slits
would fringes of same separation y be observed when using blue light of
wavelength 400 nm?

A 0.33 m B 0.67 m C
0.75 m D 1.50 m

**Reference:**

*Past Exam Paper – November 2004 Paper 1 Q28*

__Solution 835:__**Answer: D.**

For double slits: Separation of
slits, a = Dλ / w

where D is the distance of the slits
from the screen, λ is the
wavelength and w is the fringe spacing

Fringe separation = y = Dλ / a =
1.00 × 600 / a

The same Young’s slit arrangement is
being used. So, the slit separation a is the same.

For the blue light,

Wavelength λ = 400nm, Fringe separation = y, Slit separation = a and D = ???

Distance D = ay / λ = a (1.00 × 600 / a) /
400 = 3/2 = 1.50m

__Question 836: [Waves > Superposition]__
The three waves shown in each
diagram have same amplitude and frequency but differ in phase.

They are added together to give a
resultant wave.

In which case is the resultant wave
zero?

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q26*

__Solution 836:__**Answer: A.**

For the resultant wave to be zero,
the sum of displacements of the waves at any position should be zero.

For 2 waves (having non-zero
displacements), the resultant wave can only be zero at a point if one of the
wave has appositive displacement while the other has a negative displacement.

For the 3 waves shown, let’s look at
the beginning (the left-most part) so that we do not get confused. The question
has been simplified a lot since in all of the 4 choices, at least one wave is a zero
displacement at the left-most position. Thus, at this position, we only need to
find where the resultant of the 2 other waves is zero.

In C and D, both waves have negative
displacements. This cannot give a resultant zero wave. [C
and D are incorrect]

As for B, 2 of the 3 waves already
have zero displacement while the other wave has a negative displacement.
Obviously, a single wave with non-zero displacement cannot give a resultant zero
wave on its own. [B is incorrect]

For choice A, one wave has zero
displacement, another wave has positive displacement and the last one has
negative displacement. Here, a resultant zero wave can be obtained.

__Question 837: [Waves > Diffraction]__
Plane wavefronts in ripple tank pass
through a gap as shown.

Which property of the wave will be
different at Q compared with P?

A velocity

B frequency

C amplitude

D wavelength

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q28*

__Solution 837:__**Answer: C.**

Speed of wave: v = fλ

The frequency of the wave depends only
on the source generating it (e.g. a vibrator) and thus is constant.

The wavelength of the wave can be
identified by the separation of the wavefronts (lines in the diagram). In a
ripple tank, the wavelength of the wave would change if the wave enters from a
region of deep water to a region of shallow water, or vice versa. This is not
the case here – the water wave is only passing through a gap. In the diagram,
it can be clearly seen that the separation of the wavefronts is constant, so
the wavelength is constant.

Speed v = fλ

Since both the frequency f and the
wavelength λ are constant, the speed of the velocity (velocity) is also constant.

Only the amplitude of the wave
changes as the wave loses some energy as it passes through the gap.

__Question 838: [Measurement > Graphs]__
A light meter measures intensity I
of the light falling on it. Theory suggests that this varies as the inverse
square of the distance d.

**Reference:**

*Past Exam Paper – June 2006 Paper 1 Q4*

__Solution 838:__**Answer: D.**

Theory suggests that the intensity I
varies as the inverse square of the distance d.

I α
1 / d

^{2}
I = k (1 / d

^{2}) = k / d^{2}where k is a constant
Consider the graph of I (on y-axis)
against 1/d

^{2}(on x-axis).
Equation of a line: y = mx + x

Compare I = k (1 / d

^{2}) = k / d^{2}with the equation y = mx + c:
y = I

x = 1 / d

^{2}
m = k

c = 0

So, if a graph of I against 1/d

^{2}is plotted, it should be a straight line with a positive gradient (the constant k is positive) and starting at the origin (y-intercept, c = 0 – that is, it intercepts the y-axis at y=0). [D is correct]
Options A and B both have the graph
cutting an axis, which would not be the case for an inverse square law graph.

A graph of I against d

^{2}would be similar to a graph of y = 1 / x where y = I and x = d^{2}. Such a graph is a curve with a negative gradient and not touching the axes.
Type: graph y=1/x

at Google search to see to shape of
the graph.

__Question 839: [Current of Electricity > Resistance]__
Tensile strain may be measured by
change in electrical resistance of a strain gauge. A strain gauge consists of
folded fine metal wire mounted on a flexible insulating backing sheet. The strain
gauge is firmly attached to the specimen, so that the strain in the metal wire
is always identical to that in the specimen.

When the strain in the specimen is
increased, what happens to resistance of the wire?

A It decreases, because the length
decreases and the cross-sectional area increases.

B It decreases, because the length
increases and the cross-sectional area decreases.

C It increases, because the length
decreases and the cross-sectional area increases.

D It increases, because the length
increases and the cross-sectional area decreases.

**Reference:**

*Past Exam Paper – November 2005 Paper 1 Q33*

__Solution 839:__**Answer: D.**

Resistance R of wire = ρL / A

Strain = e/ L where e is the extension and L is the original length.

Total of wire = L + e

So, resistance R of wire = ρ (L+e) / A

An increase in the strain (extension) means increasing the total length of
the wire, Moreover, since the volume of the wire is constant, increasing the
length of the wire causes its cross-sectional area to decrease, so that its
volume always remains constant.

The length has been increase while the cross-sectional area has
decreased. Both of these changes causes the resistance to increase.

__Question 840: [Waves > Interference > Diffraction grating]__
Monochromatic light of wavelength
5.30 × 10

^{–7}m is incident normally on diffraction grating. The first order maximum is observed at angle of 15.4° to the direction of the incident light.
What is the angle between first and
second order diffraction maxima?

A 7.6° B 15.4°
C 16.7° D 32.0°

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q29*

__Solution 840:__**Answer: C.**

Diffraction grating: d sinθ

_{n}= nλ where n is an integer
When n = 1, θ

_{1}= 15.4°
Slit separation, d = λ / sin(15.4°) = 1.996×10

^{-6 }m
When n = 2,

θ

_{2}= sin^{-1 }(2λ / d) = 32.08°
Angle between 1

^{st}and 2^{nd}orders = 32.08 – 15.4 = 16.7°
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