Constructive Interference - General Physics - Solved Exam, Exams of Physics

This is the Solved Exam of General Physics which includes Density and Flotation, Volume of Block, Irregular Shaped Object, Unit of Density, Density of Ethanol, Electronic Balance, Weighing Scales etc. Key important points are: Constructive Interference, Destructive Interference, Coherent Sources, Wave Nature of Light, Wavelength of Radio Waves, Laws of Refraction of Light, Optical Structure, Focal Length

Typology: Exams

2012/2013

Uploaded on 02/19/2013

pankhadi
pankhadi 🇮🇳

4.7

(3)

48 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
2002 Question 7
(i) Constructive interference and destructive interference take place when waves from
two coherent sources meet.
Explain the underlined terms in the above statement.
Constructive Interference occurs when waves from two coherent sources meet to produce a
wave of greater amplitude.
Coherent Waves: Two waves are said to be coherent if they have the same frequency and are
in phase.
(ii) What is the condition necessary for destructive interference to take place when waves
from two coherent sources meet?
They must be out of phase by half a wavelength (this means that the crest of one wave will
be over the trough of the other.
(iii)Describe an experiment that demonstrates the wave nature of light.
Shine a laser through a diffraction grating; an interference pattern will be produced on a
screen, caused by interference of the light waves
(iv) Radio waves of frequency 30 kHz are received at a location 1500 km from a transmitter.
The radio reception temporarily “fades” due to destructive interference between the waves
travelling parallel to the ground and the waves reflected from a layer (ionosphere) of the
earth’s atmosphere, as indicated in the diagram.
Calculate the wavelength of the radio waves.
c = f λ λ = c/f = (3.0 × 108)/ (30000) = 104 m = 10 km
(v) What is the minimum distance that the reflected waves should travel for destructive
interference to occur at the receiver?
For destructive interference to occur the reflected wave must arrive out of phase, i.e. it must
have travelled half a wavelength more than the regular wave.
The regular wave will have travelled 1500 km and half a wavelength is 5 km therefore the
reflected wave must travel1500 km + 5 km = 1505 km.
(vi) The layer at which the waves are reflected is at a height h above the ground.
Calculate the minimum height of this layer for destructive interference to occur at the
receiver.
Use Pythagoras: h = 61000 m
2002 Question 12 (b)
(i) State the laws of refraction of light.
The incident ray, the normal and the refracted ray all lie on the same plane.
The ratio of the sin of the sin of the angle of incidence to the sin of the angle of refraction is a
constant.
(ii) Draw a labelled diagram showing the optical structure of the eye.
See diagram.
(iii)How does the eye bring objects at different distances into focus?
It can change the shape of the lens which in turn changes the focal length of the
lens.
(iv) The power of a normal eye is +60 m-1. A short-sighted person’s eye has a
power of +65 m-1. Calculate the power of the contact lens required to correct
the person’s short-sightedness.
pf2

Partial preview of the text

Download Constructive Interference - General Physics - Solved Exam and more Exams Physics in PDF only on Docsity!

2002 Question 7 (i) Constructive interference and destructive interference take place when waves from two coherent sources meet. Explain the underlined terms in the above statement. Constructive Interference occurs when waves from two coherent sources meet to produce a wave of greater amplitude. Coherent Waves: Two waves are said to be coherent if they have the same frequency and are in phase. (ii) What is the condition necessary for destructive interference to take place when waves from two coherent sources meet? They must be out of phase by half a wavelength (this means that the crest of one wave will be over the trough of the other. (iii)Describe an experiment that demonstrates the wave nature of light. Shine a laser through a diffraction grating; an interference pattern will be produced on a screen, caused by interference of the light waves (iv) Radio waves of frequency 30 kHz are received at a location 1500 km from a transmitter. The radio reception temporarily “fades” due to destructive interference between the waves travelling parallel to the ground and the waves reflected from a layer (ionosphere) of the earth’s atmosphere, as indicated in the diagram. Calculate the wavelength of the radio waves. c = f λ  λ= c/f = (3.0 × 10^8 )/ (30000) = 10^4 m = 10 km (v) What is the minimum distance that the reflected waves should travel for destructive interference to occur at the receiver? For destructive interference to occur the reflected wave must arrive out of phase, i.e. it must have travelled half a wavelength more than the regular wave. The regular wave will have travelled 1500 km and half a wavelength is 5 km therefore the reflected wave must travel1500 km + 5 km = 1505 km. (vi) The layer at which the waves are reflected is at a height h above the ground. Calculate the minimum height of this layer for destructive interference to occur at the receiver. Use Pythagoras: h = 61000 m

2002 Question 12 (b) (i) State the laws of refraction of light. The incident ray, the normal and the refracted ray all lie on the same plane. The ratio of the sin of the sin of the angle of incidence to the sin of the angle of refraction is a constant. (ii) Draw a labelled diagram showing the optical structure of the eye. See diagram. (iii)How does the eye bring objects at different distances into focus? It can change the shape of the lens which in turn changes the focal length of the lens. (iv) The power of a normal eye is +60 m-1. A short-sighted person’s eye has a power of +65 m-1. Calculate the power of the contact lens required to correct the person’s short-sightedness.

PTotal = P 1 + P 2 60 = P 1 + 65 Power = - 5 m (v) Calculate the focal length of the contact lens required to correct the person’s short- sightedness. P = 1/f  f =- 0.2 m