# Chapter 10 – Wave Optics Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 10 - Wave Optics Questions and Answers.

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Class 12 Physics NCERT book solutions for Chapter 10 - Wave Optics Questions and Answers.

(a) Reflected light?

λ=589nm=589×109mλ=589nm=589×109m

Speed of light in air is

c=3×108mc=3×108m

Refractive index of water is

μ=1.33μ=1.33

In this case, the ray will reflect in the same medium as that of the incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the equal to that of the incident ray.

Frequency of light can be given by the relation,

ν=cλν=cλ

Where,

ν=ν=

Frequency of light

c=c=

Speed of light

λ=λ=

Wavelength of light

⇒ν=3×108589×10−9⇒ν=3×108589×10−9

⇒ν=5.09×1014Hz⇒ν=5.09×1014Hz

Hence, the speed, frequency, and wavelength of the reflected light are

3×108m/s3×108m/s

5.09×1014Hz5.09×1014Hz

and

589nm589nm

respectively.

Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident light or reflected light in air.

Frequency of the refracted light ray,

ν=5.09×1014Hzν=5.09×1014Hz

Speed of light in water is related to the refractive index of water as given in the formula below:

v=cμv=cμ

⇒v=3×1081.33⇒v=3×1081.33

⇒v=2.26×108m/s⇒v=2.26×108m/s

The formula below gives the relation of wavelength of light in water and the speed and frequency of light,

λ=vνλ=vν

⇒λ=2.26×1085.09×1014⇒λ=2.26×1085.09×1014

⇒λ=444.007×10−9m⇒λ=444.007×10−9m

⇒λ=444.01nm⇒λ=444.01nm

(a) Light diverging from a point source.

μ=1.5μ=1.5

Speed of light,

c=3.0×108m/sc=3.0×108m/s

Speed of light in glass is given by the formula,

v=cμv=cμ

⇒v=3×1081.5=2×108m/s⇒v=3×1081.5=2×108m/s

Hence, the speed of light in glass is

2×108m/s2×108m/s

The refractive index of a violet component of white light is more than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass as speed and refractive index are inversely related to each other.

Hence, violet light travels slower as compared to red light in a glass prism.

d=0.28mm=0.28×10−3md=0.28mm=0.28×10−3m

Distance between the slits and the screen,

D=1.4mD=1.4m

Distance between the central fringe and the fourth

(n=4)(n=4)

fringe,

u=1.2cm=1.2×10−2mu=1.2cm=1.2×10−2m

In case of a constructive interference, the relation for the distance between the two fringes can be given as:

u=nλDdu=nλDd

where,

n=n=

Order of fringes

=4=4

λ=λ=

Wavelength of light used

λ=udnDλ=udnD

⇒λ=1.2×10−2×0.28×10−34×1.4⇒λ=1.2×10−2×0.28×10−34×1.4

⇒λ=6×10−7⇒λ=6×10−7

⇒λ=600nm⇒λ=600nm

Hence, the wavelength of the light is

600nm600nm

I1I1

and

I2I2

. Their resultant intensities can be evaluated as:

I′=I1+I2+2I1I2−−−−√cosϕI′=I1+I2+2I1I2cosϕ

Where,

ϕ=ϕ=

The phase difference between two waves for monochromatic light waves,

Since

I1=I2I1=I2

So

I′=2I1+2I1cosϕI′=2I1+2I1cosϕ

The formula for phase difference can be given as:

Phasedifference=2πλ×PathdifferencePhasedifference=2πλ×Pathdifference

Since, path difference is

λλ

Phase difference is

ϕ=2πϕ=2π

I′=2I1+2I1=4I1I′=2I1+2I1=4I1

Given,

I1=K′4I1=K′4

…… (1)

When path difference

=λ3=λ3

phase difference,

ϕ=2π3ϕ=2π3

Hence, resultant intensity,

IR=I1+I1+2I1I1−−−−√cos2π3IR=I1+I1+2I1I1cos2π3

⇒IR=2I1+2I2(−12)=I1⇒IR=2I1+2I2(−12)=I1

Using equation (1), we can state that

IR=I1=K4IR=I1=K4

Hence, for monochromatic light waves, the intensity of light at a point where the path difference is λ3λ3 is K4K4 units.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650nm650nm

Wavelength of the first light beam,

λ1=650nmλ1=650nm

Wavelength of second light beam,

λ2=520nmλ2=520nm

Distance of the slits from the screen

=D=D

Distance between the two slits

=d=d

Distance of the

nthnth

bright fringe on the screen from the central maximum is given by the formula below,

x=nλ1(Dd)x=nλ1(Dd)

For the third bright fringe,

n=3n=3

x=3×650×Dd=1950(Dd)nmx=3×650×Dd=1950(Dd)nm

, which is nothing but the distance of the third bright fringe on the screen from the central maximum.

nthnth

bright fringe due to wavelength

λ2λ2

and

(n−1)th(n−1)th

bright fringe due to wavelength

λ1λ1

coincide on the screen. Equate the conditions for bright fringes as follows:

nλ2=(n−1)λ1nλ2=(n−1)λ1

⇒520n=650n−650⇒520n=650n−650

⇒650=130n⇒650=130n

⇒n=5⇒n=5

Hence, the least distance from the central maximum can be attained by the relation:

x=nλ2Ddx=nλ2Dd

⇒x=5×520Dd=2600Ddnm⇒x=5×520Dd=2600Ddnm

Note: The value of

dd

and

DD

are not given in the question, hence the exact answer cannot be found.

Distance of the screen from the slits is given as,

D=1mD=1m

Wavelength of light used,

λ1=600nmλ1=600nm

Angular width of the fringe in air,

θ1=0.2∘θ1=0.2∘

Angular width of the fringe in water

=θ2=θ2

Refractive index of water is

4343

Refractive index is associated with angular width as:

μ=θ1θ2μ=θ1θ2

⇒θ2=34θ1⇒θ2=34θ1

⇒θ2=34×0.2=0.15∘⇒θ2=34×0.2=0.15∘

Therefore, the angular width of the fringe in this case

θθ

in water will reduce to

0.15∘0.15∘

Refractive index of glass is given

=1.5=1.5

Brewster angle

=θ=θ

Brewster angle is associated to refractive index as mentioned in the formula below:

tanθ=μtanθ=μ

⇒θ=tan−1(1.5)=56.31∘⇒θ=tan−1(1.5)=56.31∘

Hence, the Brewster angle for transition from air to glass is

56.31∘56.31∘

λ=5000A∘=5000×10−10mλ=5000A∘=5000×10−10m

Speed of light,

c=3×108m/sc=3×108m/s

Frequency of incident light is given by the formula,

ν=cλν=cλ

⇒ν=3×1085000×1010=6×1014Hz⇒ν=3×1085000×1010=6×1014Hz

The wavelength and frequency of incident light is equal to that of reflected ray.

Therefore, the wavelength of reflected light is

5000A∘5000A∘

and its frequency is

6×1014Hz6×1014Hz

When reflected ray is normal to incident ray, the addition of the angle of incidence, given as,

∠i∠i

and angle of reflection, given as,

∠r∠r

is

90∘90∘

According to the law of reflection, the angle of incidence is always the same as the angle of reflection. Hence, the sum can be written as:

∠i+∠r=90∘∠i+∠r=90∘

⇒∠i+∠i=90∘⇒∠i+∠i=90∘

⇒2∠i=90∘⇒2∠i=90∘

⇒∠i=90∘2=45∘⇒∠i=90∘2=45∘

Hence, the angle of incidence for the given condition in the question is

45∘45∘

(ZF)(ZF)

can be defined as the distance for which the ray optics is a good approximation. It can be expressed by the relation,

ZF=a2λZF=a2λ

Where,

Aperture width is

aa

Wavelength of light is

λλ

Now,

a=4mm=4×10−3ma=4mm=4×10−3m

λ=400nm=400×10−9mλ=400nm=400×10−9m

On substitution,

ZF=(4×10−3)2400×10−9=40mZF=(4×10−3)2400×10−9=40m

Thus, the distance for which the ray optics is a good approximation is

40m40m

Additional Exercises

HaHa

line emitted by hydrogen is given as,

λ=6563A∘λ=6563A∘

⇒λ=6563×−10−10m⇒λ=6563×−10−10m

Red shift of the star can be written as,

(λ′−λ)=15A∘=15×10−10m(λ′−λ)=15A∘=15×10−10m

Speed of light,

c=3×108m/sc=3×108m/s

Let us suppose that the velocity of the star receding away from the Earth be

vv

The red shift is associated with velocity as given:

λ′−λ=vcλλ′−λ=vcλ

⇒v=cλ×(λ′−λ)⇒v=cλ×(λ′−λ)

⇒v=3×108×15×10−106563×10−10=6.87×105m/s⇒v=3×108×15×10−106563×10−10=6.87×105m/s

Therefore, the speed with which the star is receding away from the Earth is

6.87×105m/s6.87×105m/s

Corpuscular theory of light given by Newton suggests that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, forces of attraction normal to the surface are experienced by the particles. Thus, the normal component of velocity increases while the component along the surface does not change.

Clearly, the expression can be written as:

csini=vsinrcsini=vsinr

…… (1)

Where,

i=Angleofincidencei=Angleofincidence

r=Angleofreflectionr=Angleofreflection

c=Velocityoflightinairc=Velocityoflightinair

v=v=

Velocity of light in water

We have the expression for relative refractive index of water with respect to air as:

μ=vcμ=vc

Hence, equation (1) reduces to

⇒vc=sinisinr=μ⇒vc=sinisinr=μ

…… (2)

But

μ>1μ>1

Hence, it can be observed from equation (2) that

v>cv>c

This is not possible because this prediction is opposite to the experimental results of

c>vc>v

Clearly, the wave picture of light is consistent with the experimental results.

OO

be placed in front of a plane mirror

MO′MO′

at a distance

rr

A circle is drawn from the centre (

OO

) such that it just touches the plane mirror at point

O′O′

on the mirror.

According to Huygens’ Principle,

XYXY

is the wavefront of incident light.

If the mirror is not present, then a similar wavefront

X′Y′X′Y′

(as

XYXY

) would form behind the point

O′O′

at distance

rr

X′Y′X′Y′

can be considered as a virtual reflected ray for the plane mirror.

Therefore, a point object placed in front of the plane mirror generates a virtual image whose distance from the mirror is the same as the object distance (rr)

(i) Nature of the source.

(ii) Direction of propagation.

(iii) Motion of the source and/or observer.

(iv) Wavelength.

(v) Intensity of the wave.

On which of these factors, if any, does

(a) The speed of light in vacuum,

c=3×108m/sc=3×108m/s

(approximately) is a universal constant.

The motion of the source, the observer, or both, does not affect the speed of light. Hence, the above given factors do not affect the speed of light in a vacuum.

(a) source at rest; observer moving, and

(b) source moving; observer at rest.

The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Sound waves need a medium for propagation. The two given circumstances are not identical scientifically since the motion of an observer relative to a medium is not similar in the two situations. Hence, the doppler formulas for the two situations are different.

In case of light waves, they can travel in vacuum. In vacuum, the above two situations are identical because the speed of light is not dependent of the motion of the observer and the motion of the source when light travels in a medium. The above two cases are not identical because the speed of light depends on the wavelength of the medium.

λ=600nm=600×10−9mλ=600nm=600×10−9m

Angular width of fringe is,

θ=0.1∘=0.1×π180=3.141800radθ=0.1∘=0.1×π180=3.141800rad

Angular width of a fringe is related to slit spacing (

dd

) as:

θ=λdθ=λd

⇒d=λθ⇒d=λθ

⇒d=600×10−93.141800=3.44×10−4m⇒d=600×10−93.141800=3.44×10−4m

Hence, the spacing between the slits is

3.44×10−4m3.44×10−4m

(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

On the contrary, the wavelength of the light waves is too small in comparison to the size of the obstacle. Hence, the angle of diffraction will be less. Hence, the students are not able to see each other. On the contrary, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a great angle. Therefore, the students are able to hear each other.

Distance between the towers is given as

d=40kmd=40km

Height of the line joining the hills is

50m50m

Therefore, the radial spread of the radio waves should not exceed

50km50km

Since the hill is located halfway between the towers, Fresnel’s distance can be written as:

Zp=20km=20×104mZp=20km=20×104m

Aperture can be written as:

a=d=50ma=d=50m

Fresnel’s distance is given by the expression,

Zp=a2λZp=a2λ

Where,

λ=λ=

Wavelength of radio waves

λ=a2Zpλ=a2Zp

⇒λ=(50)22×104=1250×10−4=0.1250m⇒λ=(50)22×104=1250×10−4=0.1250m

⇒λ=12.5cm⇒λ=12.5cm

Therefore, the wavelength of the radio waves is

12.5cm12.5cm

Wavelength of light beam is

λ=500nm=500×10−9mλ=500nm=500×10−9m

Distance of the screen from the slit,

D=1mD=1m

For first minima,

n=1n=1

Distance between the slits

=d=d

Distance of the first minimum from the centre of the screen can be given as:

x=2.5mm=2.5×10−3mx=2.5mm=2.5×10−3m

It is related to the order of minima as given in the formula:

nλ=xdDnλ=xdD

⇒d=nλDx⇒d=nλDx

⇒d=1×500×10−9×12.5×10−3⇒d=1×500×10−9×12.5×10−3

⇒d=2×10−4m=0.2mm⇒d=2×10−4m=0.2mm

Clearly, the width of the slits is

0.2mm0.2mm

(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

y1y1

and

y2y2

are the solutions of the second order wave equation, then a linear combination of

y1y1

and

y2y2

and thus, will also be the solution of the wave equation.

dd

is divided into

nn

smaller slits.

Width of each slit can be expressed as

d′=dnd′=dn

Angle of diffraction is given by the formula,

θ=ddλd=λdθ=ddλd=λd

Now, each of these infinitesimally small slits transmit zero intensity in direction

θθ

. Therefore, the combination of these slits would give zero intensity.

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