Waves - Reflection and refraction



Home learning: Waves - Reflection and refraction

1. Fill out the table below to show the types of waves

| |Transverse or longitudinal? |Mechanical or electromagnetic? |

|A plucked guitar string |transverse |mechanical |

|Radio waves |transverse |EM |

|Sound from a radio |longitudinal |mechanical |

|Light from a torch |transverse |EM |

2. Water waves pass by at a rate of one every 7 seconds. Their wavelength is 5 metres. Calculate their speed. (hint: calculate their frequency first, then use v=fλ)

frequency = number of waves

time in seconds

f = 1 ÷ 7 = 0.14Hz

v = f λ = 0.14 * 5 = 0.71 = 0.7 ms-1 (1sf)

3. A wave generator makes 16 waves every 4.0 seconds. These 16 waves cover a distance of 16cm.

a) Calculate the frequency of the waves.

frequency = number of waves

time in seconds

f = 16 ÷ 4.0 = 4.0Hz

b) Calculate the period of the waves.

T = 1

f

T = 1 ÷ 4.0 = 0.25 s (2sf)

c) Calculate the wavelength of the waves.

16 waves cover 16cm, so 1 wave (which is one wavelength) covers 1.0cm

d) Calculate the speed of the waves.

v = f λ = 4.0 * 0.01 = 0.040 ms-1 (2sf)

4. Complete the diagram below to show how the waves are reflected from the boundary.

a) Label your diagram to show the direction of travel of the reflected waves.

b) Label the angle of incidence and angle of reflection.

5. The diagram shows waves arriving at a boundary between deep and shallow water. In the deep region a total of 10 waves are counted passing by in 4.0s. Their wavelength is 3.0cm in the deep water and 2.0cm in the shallow water.

a) Calculate the frequency of the waves in the deep region.

frequency = number of waves

time in seconds

f = 10 ÷ 4.0 = 2.5Hz (2sf)

b) What is the frequency of the waves in the shallow region?

2.5Hz (the frequency of the waves does not change from deep to shallow water, only the wavelength and speed change).

c) Calculate the wave speed in the shallow region.

v = f λ = 2.5 * 0.020 = 0.050 ms-1 (2sf)

6. The diagram shows a ripple tank set up to demonstrate refraction. As the waves travel from the deep region into the shallow region they refract and change direction.

[pic]

a) Draw the pattern showing the wave fronts as they pass into the shallow region.

b) Each incident wave takes 0.20s to travel the 15cm shown. Calculate the speed of the waves.

v = d

t

v = 0.15 ÷ 0.20 = 0.75 ms-1 (2sf)

c) Use the diagram to work out the wavelength of the incident waves.

3 waves cover 15cm (remember to count complete wavelengths, not number of crests). So 1 wave covers 5cm. The wavelength is 5.0cm (2sf)

d) Calculate the frequency of the incident waves.

f = v

λ

f = 0.75 ÷ 0.050 = 15Hz (2sf)

e) What is the frequency of the refracted waves?

15Hz (the frequency of the waves does not change from deep to shallow water, only the wavelength and speed change).

f) The wavelength of the refracted waves is measured to be 4.0cm. Use v=fλ to calculate the speed of the waves in the shallow region.

v = f λ = 15 * 0.040 = 0.60 ms-1 (2sf)

g) Use your values to verify that v1/v2 = λ1/λ2

v1/v2 = 0.75 ÷ 0.60 = 1.25

λ1/λ2 = 0.05 ÷ 0.04 = 1.25

Both are equal to 1.25, so v1/v2 = λ1/λ2 is proven to be true.

7. The diagram below shows a wave generator producing waves with a frequency of 20 Hz. The wave fronts are incident on a plane boundary which separates a shallow region from a deep region. at an angle of 30˚. The speed of the waves in the shallow region is 40 cm s-1 and in the deep region is 55 cm s-1.

[pic]

a) Calculate the angle of refraction in the deep region.

(Use vdeep/vshallow = sin Өdeep/sinӨshallow)

vdeep = 55 cm s-1

vshallow = 40 cm s-1

Өdeep = ?

Өshallow = 30˚

55 = sin Өdeep

40 sin30˚

sin30˚ x 55 = sin Өdeep

40

sin-1( sin30˚ x 55 ) = Өdeep

40

Өdeep = 43˚ (2sf)

-----------------------

DEEP

SHALLOW

15cm

30˚

SHALLOW

DEEP

SHALLOW

DEEP

direction of propagation

angle of incidence

angle of reflection

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