Chapter 10 Wave Optics Solutions
Question - 11 : - The 6563 Å 
line emitted by hydrogen ina star is found to be red shifted by 15 Å. Estimate the speed with which thestar is receding from the Earth.
Answer - 11 : -
Wavelength of line emitted by hydrogen,
λ = 6563 Å
= 6563 ×10−10 m.
Star’sred-shift, 
Speed of light, 
Let thevelocity of the star receding away from the Earth be v.
The redshift is related with velocity as:
Therefore, the speed with whichthe star is receding away from the Earth is 6.87 × 105 m/s.
Question - 12 : - Explain how Corpuscular theorypredicts the speed of light in a medium, say, water, to be greater than thespeed of light in vacuum. Is the prediction confirmed by experimentaldetermination of the speed of light in water? If not, which alternative pictureof light is consistent with experiment?
Answer - 12 : -
Newton’s corpuscular theory oflight states that when light corpuscles strike the interface of two media froma rarer (air) to a denser (water) medium, the particles experience forces ofattraction normal to the surface. Hence, the normal component of velocityincreases while the component along the surface remains unchanged.
Hence,we can write the expression:
… (i)Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We havethe relation for relative refractive index of water with respect to air as:
Hence, equation (i) reduces to
But,
> 1
Hence, itcan be inferred from equation (ii) that v > c. This is not possible since thisprediction is opposite to the experimental results of c > v.
The wavepicture of light is consistent with the experimental results.
Question - 13 : - You have learnt in the text howHuygens’ principle leads to the laws of reflection and refraction. Use the sameprinciple to deduce directly that a point object placed in front of a planemirror produces a virtual image whose distance from the mirror is equal to theobject distance from the mirror.
Answer - 13 : -
Let an object at O be placed infront of a plane mirror MO’ at a distance r (asshown in the given figure).
A circle is drawn from the centre(O) such that it just touches the plane mirror at point O’. According toHuygens’ Principle, XY is the wavefront of incident light.
Ifthe mirror is absent, then a similar wavefront X’Y’ (as XY) would form behindO’ at distance r (as shown inthe given figure).
can be considered as a virtualreflected ray for the plane mirror. Hence, a point object placed in front ofthe plane mirror produces a virtual image whose distance from the mirror isequal to the object distance (r).
Question - 14 : - Let us list some of the factors,which could possibly influence the speed of wave propagation:
(i) Nature of the source.
(ii) Direction of propagation.
(iii) Motion of the source and/orobserver.
(iv) Wave length.
(v) Intensity of the wave.
On whichof these factors, if any, does
(a) The speed of light in vacuum,
(b) The speed of light in amedium (say, glass or water), depend?
Answer - 14 : -
(a) Thespeedof light in a vacuum i.e., 3 × 108 m/s(approximately) is a universal constant. It is not affected by the motion ofthe source, the observer, or both. Hence, the given factor does not affect thespeed of light in a vacuum.
(b) Out of the listed factors, thespeed of light in a medium depends on the wavelength of light in that medium.
Question - 15 : - For sound waves, the Dopplerformula for frequency shift differs slightly between the two situations: (i)source at rest; observer moving, and (ii) source moving; observer at rest. Theexact Doppler formulas for the case of light waves in vacuum are, however,strictly identical for these situations. Explain why this should be so. Wouldyou expect the formulas to be strictly identical for the two situations in caseof light travelling in a medium?
Answer - 15 : -
No
Sound waves can propagate onlythrough a medium. The two given situations are not scientifically identicalbecause the motion of an observer relative to a medium is different in the twosituations. Hence, the Doppler formulas for the two situations cannot be thesame.
In caseof light waves, sound can travel in a vacuum. In a vacuum, the above two casesare identical because the speed of light is independent of the motion of theobserver and the motion of the source. When light travels in a medium, theabove two cases are not identical because the speed of light depends on thewavelength of the medium.
Question - 16 : - In double-slit experiment usinglight of wavelength 600 nm, the angular width of a fringe formed on a distantscreen is 0.1º. What is the spacing between the two slits?
Answer - 16 : -
Wavelength of light used, λ = 6000 nm = 600 × 10−9 m
Angularwidth of fringe,
Angular width of a fringe isrelated to slit spacing (d) as:
Therefore, the spacing between the slits is
Answer the following questions:
(a) In a single slit diffractionexperiment, the width of the slit is made double the original width. How doesthis affect the size and intensity of the central diffraction band?
(b) In what way is diffraction fromeach slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle isplaced in the path of light from a distant source, a bright spot is seen at thecentre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7m partition wall in a room 10 m high. If both light and sound waves can bendaround obstacles, how is it that the students are unable to see each other eventhough they can converse easily.
(e) Ray optics is based on theassumption that light travels in a straight line. Diffraction effects (observedwhen light propagates through small apertures/slits or around small obstacles)disprove this assumption. Yet the ray optics assumption is so commonly used inunderstanding location and several other properties of images in optical instruments.What is the justification?
Answer - 17 : -
(a) In asingle slit diffraction experiment, if the width of the slit is made double theoriginal width, then the size of the central diffraction band reduces to halfand the intensity of the central diffraction band increases up to four times.
(b) The interference pattern in adouble-slit experiment is modulated by diffraction from each slit. The patternis the result of the interference of the diffracted wave from each slit.
(c) When a tiny circular obstacle isplaced in the path of light from a distant source, a bright spot is seen at thecentre of the shadow of the obstacle. This is because light waves arediffracted from the edge of the circular obstacle, which interferesconstructively at the centre of the shadow. This constructive interferenceproduces a bright spot.
(d) Bending of waves by obstacles bya large angle is possible when the size of the obstacle is comparable to thewavelength of the waves.
On theone hand, the wavelength of the light waves is too small in comparison to thesize of the obstacle. Thus, the diffraction angle will be very small. Hence,the students are unable to see each other. On the other hand, the size of thewall is comparable to the wavelength of the sound waves. Thus, the bending ofthe waves takes place at a large angle. Hence, the students are able to heareach other.
(e) The justification is that inordinary optical instruments, the size of the aperture involved is much largerthan the wavelength of the light used.
Question - 18 : - Two towers on top of two hillsare 40 km apart. The line joining them passes 50 m above a hill halfway betweenthe towers. What is the longest wavelength of radio waves, which can be sentbetween the towers without appreciable diffraction effects?
Answer - 18 : -
Distance between the towers, d = 40 km
Height ofthe line joining the hills, d =50 m.
Thus, theradial spread of the radio waves should not exceed 50 km.
Since thehill is located halfway between the towers, Fresnel’s distance can be obtainedas:
ZP =20 km = 2 × 104 m
Aperturecan be taken as:
a = d = 50 m
Fresnel’sdistance is given by the relation,
Where,
λ =Wavelength of radio waves
Therefore, the wavelength of theradio waves is 12.5 cm.
Question - 19 : - A parallel beam of light ofwavelength 500 nm falls on a narrow slit and the resulting diffraction patternis observed on a screen 1 m away. It is observed that the first minimum is at adistance of 2.5 mm from the centre of the screen. Find the width of the slit.
Answer - 19 : -
Wavelength of light beam, λ = 500 nm = 500 × 10−9 m
Distanceof the screen from the slit, D =1 m
For firstminima, n = 1
Distancebetween the slits = d
Distanceof the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10−3 m
It isrelated to the order of minima as:
Therefore, the width of the slitsis 0.2 mm.
Question - 20 : - Answer the following questions:
(a) When a low flying aircraft passesoverhead, we sometimes notice
a slightshaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text,the principle of linear superposition of wave displacement is basic tounderstanding intensity distributions in diffraction and interference patterns.What is the justification of this principle?
Answer - 20 : -
(a) Weakradar signals sent by a low flying aircraft can interfere with the TV signalsreceived by the antenna. As a result, the TV signals may get distorted. Hence,when a low flying aircraft passes overhead, we sometimes notice a slightshaking of the picture on our TV screen.
(b) The principle of linearsuperposition of wave displacement is essential to our understanding ofintensity distributions and interference patterns. This is becausesuperposition follows from the linear character of a differential equation thatgoverns wave motion. If y1 and y2 are the solutions of the second order waveequation, then any linear combination of y1 and y2 will also be the solution of the wave equation.