• 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 / d²

I = k (1 / d²) = k / d²    where k is a constant

Consider the graph of I (on y-axis) against 1/d² (on x-axis).

Equation of a line: y = mx + x

Compare I = k (1 / d²) = k / d² with the equation y = mx + c:

y = I

x = 1 / d²

m = k

c = 0

So, if a graph of I against 1/d² 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 d² would be similar to a graph of y = 1 / x where y = I and x = d². 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, θ₁ = 15.4°

Slit separation, d = λ / sin(15.4°) = 1.996×10^-6 m

When n = 2,

θ₂ = sin^-1 (2λ / d) = 32.08°

Angle between 1^st and 2^nd orders = 32.08 – 15.4 = 16.7°