# IBDP Physics 9.4 – Resolution: IB Style Question Bank HL Paper 2

Question

a.

Police use radar to detect speeding cars. A police officer stands at the side of the road and points a radar device at an approaching car. The device emits microwaves which reflect off the car and return to the device. A change in frequency between the emitted and received microwaves is measured at the radar device.

The frequency change Δf is given by

$\Delta f = \frac{{2fv}}{c}$

where f is the transmitter frequency, v is the speed of the car and c is the wave speed.

The following data are available.

 Transmitter frequency f = 40 GHz Δf = 9.5 kHz Maximum speed allowed = 28 m s–1

(i) Explain the reason for the frequency change.

(ii) Suggest why there is a factor of 2 in the frequency-change equation.

(iii) Determine whether the speed of the car is below the maximum speed allowed.[6]

b.

Airports use radar to track the position of aircraft. The waves are reflected from the aircraft and detected by a large circular receiver. The receiver must be able to resolve the radar images of two aircraft flying close to each other.

The following data are available.

 Diameter of circular radar receiver = 9.3 m Wavelength of radar = 2.5 cm Distance of two aircraft from the airport = 31 km

Calculate the minimum distance between the two aircraft when their images can just be resolved.[2]

## Markscheme

a.

i
mention of Doppler effect
OR
«relative» motion between source and observer produces frequency/wavelength change
Accept answers which refer to a double frequency shift.
Award [0] if there is any suggestion that the wave speed is changed in the process.

the reflected waves come from an approaching “source”
OR
the incident waves strike an approaching “observer”

increased frequency received «by the device or by the car»

ii
the car is a moving “observer” and then a moving “source”, so the Doppler effect occurs twice
OR
the reflected radar appears to come from a “virtual image” of the device travelling at 2 v towards the device

iii
ALTERNATIVE 1
For both alternatives, allow ecf to conclusion if v OR Δf are incorrectly calculated.

v = «$$\frac{{\left( {3 \times {{10}^8}} \right) \times \left( {9.5 \times {{10}^3}} \right)}}{{\left( {40 \times {{10}^9}} \right) \times 2}} =$$» 36 «ms–1»

«36> 28» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.

ALTERNATIVE 2
reverse argument using speed limit.

$$\Delta f =$$ «$$\frac{{2 \times 40 \times {{10}^9} \times 28}}{{3 \times {{10}^8}}} =$$» 7500 «Hz»

« 9500> 7500» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.

b.

$$x = \frac{{31 \times {{10}^3} \times 1.22 \times 2.5 \times {{10}^{ – 2}}}}{{9.3}}$$
Award [2] for a bald correct answer.

Award [1 max] for POT error.

100 «m»
Award [1 max] for 83m (omits 1.22).

Question

This question is about the motion of a ship and observing objects from it.

The sailors on the ship wear polarized sunglasses when observing the sea from the ship. Unpolarized light from the Sun is incident on the sea.

e.

A security camera on the ship captures an image of two green lamps on the shore. The lamps emit light of wavelength 520 nm.

The camera has a circular aperture of diameter 6.2 mm. The lamps are separated by 1.5 m. Determine the maximum distance between the camera and the lamps at which the images of the lamps can be distinguished.[3]

f.i.

Describe the polarization of the sunlight that is reflected from the sea.[2]

f.ii.

Outline how polarized sunglasses help to reduce glare from the sea.[3]

## Markscheme

e.

use of $$\theta = \frac{{1.22\lambda }}{d}$$;

$$\left( {\frac{{1.22 \times 520 \times {{10}^{ – 9}}}}{{6.2 \times {{10}^{ – 3}}}}} \right) = 1.02 \times {10^{ – 4}}$$;

distance $$= \left( {\frac{{1.5}}{{1.02 \times {{10}^{ – 4}}}} = } \right)1.47 \times {10^4}{\text{ m}}$$;

Award [2 max] if distance is $$1.79 \times {10^4}{\text{ }}m$$ (ignoring factor of 1.22) in second marking point.

f.i.

partially/partly;

plane/horizontally polarized;

f.ii.

the light from the sea is (predominantly) horizontally polarized;

the sunglasses are arranged to admit a particular/vertical plane of polarization;

hence polarized sunglasses absorb much of the reflected light/glare;