Physics – Wave Theory of Light Textual Solved and Unsolved Problems

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Example – 01:

  • The refractive indices of water and diamond with respect to air are 4/3 and 2.42 respectively. Calculate the speed of light in water and in diamond. From these results calculate the refractive index of diamond w.r.t. water. c= 3 x 108 m/s,
  • Solution:
  • Given: Refractive index for water = μw = 4/3 = 1.33 , Refractive index for diamond = μd = 2.4, Speed of light in air  = ca = 3 x 108 m/s,
  • To Find: Speed of light in water = cw = ? The speedof light in diamond = cd = ?, Refractive index of diamond w.r.t. water = wμd  = ?

Refractive index of water w.r.t. air is given by

μw = ca / c

cw = ca / μ= 3 x 10/1.33 = 2.25 x 108  m/s

Refractive index of diamond w.r.t. air is given by



μd = ca / c

cd = ca / μ= 3 x 10/2.42 = 1.24 x 108  m/s

Refractive index of diamond w.r.t. water is given by

wμd = cw / cd   =  2.25 x 108 /1.24 x 108 =   1.815



Ans: Speed of light in water is 2.25 x 10m/s and that in diamond is 1.25 x 10m/s.

Refractive index of diamond w.r.t. water is 1.815

Example – 02: 

  • The wavelength of blue light in air is 4500 Å. What is its frequency? If the refractive index of glass for blue light is 1.55, what will be the wavelength of blue light in glass?
  • Given: Wavelength in air = λ = 4500 Å. =  4.5 x 10-7  m, Refractive index of glass = μ = 1.55, ca = 3 x 108 m/s.
  • To Find: Frequency = ν = ?, Wavelength in glass = λ =?
  • Solution:

ca = νa λa

∴  νa  = caa = 3 x 108/4.5 x 10-7 = 6.667 x 1015 Hz

μg = λa / λ



∴  λ = λag  = 4500/1.55 = 2903 Å

Ans: The frequency and wavelength of blue light are 6.667 x 1015 Hz and 2903 Å respectively.

Example – 03:

  • Determine the change in wavelength of a ray of light during its passage from air to glass if the refractive index of glass is 1.5 and the frequency of the ray is 4 x 1014 Hz. Find the wave number of light in glass, c = 3 x 10m/s.
  • Solution:
  • Given: Frequency of light in air = ν=  4 x 1014 Hz,  Refractive index of glass = μ = 1.5, Velocity of light in air = ca = 3 x 108 m/s.
  • To Find: Change in wavelength of light = | λ – λ| = ?, wave number in glass = ?

We have ca = νa λa

For air,  λa = caa

λa = caa= 3x 108  / 4 x 1014  =  7.5 x 10-7  m



λ= 7500 x 10-10  m = 7500 Å

Now, μ = λa / λg

λg = λa / μ = 7500 / 1.5= 5000 Å = 5 x 10-7 m

Change in wavelength =  λ – λ = 7500 – 5000 = 2500 Å

Wave number in glass = 1/λg = 1 /5 x 10-7  = 2 x 106 m-1.



Ans:  Change in the wavelength of light is 2500 Å, Wavenumber in glass = 2 x 106 m-1.

Example – 04:

  • A parallel beam of monochromatic light is incident on glass slab at an angle of incidence of 60°. Find the ratio of the width of the beam in the glass to that in the air if the refractive index of glass is 1.5.
  • Solution:
  • Given: Angle of incidence = i = 60°, Refractive index of glass = μ = 1.5.
  • To Find: The ratio of the width of the beam in the glass to that in the air =?

Refractive index

We have to find ratio CD/AB’

μ = sin i / sinr



∴  Sin r = sin i / μ = sin 60o / 1.5 = 0.8660/1.5 = 0.5773

∴  Angle of refraction = r = sin-1 (0.5773) = 35o16’

In Δ AB’C, cos i = AB’ / AC  ………….. (1)

In Δ ADC, cos r = CD / AC  ………….. (2)

Dividing equation (2) by (1)

cos r / cos i = CD/AB’



CD/AB’ = cos 35o16’ / cos 60o

CD/AB’ = 0.8165 / 0.5 = 1.633

Ans: The ratio of the width of the beam in the glass to that in the air is 1.633

Example – 05:

  • Find the refractive index of glass if the angle of incidence at which the polarization of light reflected from the surface of the glass is 58°.
  • Given: the polarization angle = ip = 58°
  • To Find: Refractive index of the medium = μ = ?
  • Solution:

μ = tan ip = tan 58° = 1.6003



Ans: Refractive index of the medium is 1.6003

Example – 06:

  • In a glass plate of refractive index is 1.732 is to be used as a polarizer, what would be the polarizing angle and angle of refraction.
  • Given: Refractive index of the medium = μ = 1.732
  • To find: polarising angle = ip = ?, angle of refraction = r =?
  • Solution:

μ = tan ip = 1.732

∴  ip = tan-1(1.732) = 60°

By snaell’s law

μ = sini/sinr

sin r = sini/μ = sin 60°/1.732 = 0.8660/1.732 = 0.5



∴  r = sin-1(0.5) = 30°

Ans: Polarizing angle is 60° and angle of refraction is 30°

Unsolved  Problems

Example – 01:

  • The refractive indices of water for red and violet colours are 1.325 and 1.334 respectively. Find the difference between velocities of these two colours in the water.
  • Solution:
  • Given: Refractive index of red colour = μr = 1.325, Refractive index of violet colour = μv = 1.3334,  Velocity of light in air = ca = 3 x 108 
  • To Find: Difference in velocities = | c – c| = ?

 μ(1.334) > μ(1.325)

Hence, c >  cv

We have for red light  μr = ca/cr  and for violet light  μv = ca/cv

Hence for red light  cr = cr  and for violet light  cv = cav

c – c = cr  –  cav

c – c = c(1/μr  – 1/μv)

c – c =3 x 108(1/1.325  – 1/1.334)

c – c =3 x 108(0.7547  – 0.7496)

c – c = 3 x 108(0.0051) = 1.53 x 10m/s

Ans: Difference in velocities of red and violet colour in water is 1.53 x 10m/s

Example – 02:

  • Red light of wavelength 6400 Å. in air has a wavelength of 4000 Å . in glass. If the wavelength of the violet light in air is 4400 Å . What is its wavelength in glass?
  • Solution:
  • Given: Wavelength in air for red light = λar = 6400 Å, Wavelength in glass for red light = λgr = 4000 Å, Wavelength in air for violet light = λav = 4400 Å,
  • To Find: Wavelength in glass for violet light = λgv = ?

Consider red light

μr = λar / λgr   = 6400 /4000 = 1.6

Assuming the refractive index for both colours is same i.e. μr = μv

Consider violet light

μv = λav / λgv   

λgv= λav /  μv

=   4400 / 1.6 = 2750 Å

Ans: Wavelength of violet colour in the glass is 2750 Å  

Example – 03:

  • The width of a plane incident wavefront is found to be doubled in the denser medium. If it makes an angle of 20° with the surface, calculate the refractive index for the denser medium. (Ans: 1.288)

Example – 04:

  • The difference in velocities of a light ray in glass and in water is 0.25 x 108  m/s. R.I. of water and glass are 4/3 and 1.5 respectively. Find c.
  • Solution:
  • Given: Difference in velocities = | c – c|  = 0.25 x 108  m/s = 2.5 x 107  m/s, R.I. of water = μw = 4/3, R.I. of glass = μ= 1.5 = 3/2
  • To Find: Velocity of light in air =  ca = ?

 μ(1.5) > μ(4/3)

Hence, c >  cg

c –  cg = 2.5 x 107

We have, for water μw = ca/cw  and for glass μg = ca/cg

Hence for water cw = caw  and for glass cg = cag

c –  cg = 2.5 x 107

∴  caw  –  cag = 2.5 x 107

∴  ca(1/μw  –  1/μg) = 2.5 x 107

∴  ca(3/4  –  2/3) = 2.5 x 107

∴  ca(1/12) = 2.5 x 107

∴  c= 12 x 2.5 x 10= 3 x 108  m/s

Ans:  Speed of light = c = 3 x 108  m/s

Example – 05:

  • A ray of light travelling through air, falls on the surface of a glass slab at an angle i. it is found that the angle between the reflected and refracted ray is 90°. If the speed of light in glass is 2 x 108 m/s, find the angle of incidence. c = 3 x 10m/s.
  • Given:  angle between the reflected and refracted ray is 90°, speed of light in glass is = cg = 2 x 108, speed of light in air = ca = 3 x 10m/s
  • To find: angle of incidence = i = ?
  • Solution:

The refractive index of glass is given by

μg = ca/cg = 3 x 10/ 2 x 10= 1.5

Now the angle between the reflected and refracted ray is 90°

Hence the angle of incidence is equal to the angle of polarization.

μ = tan ip = 1.5

∴  i= tan-1(1.5) = 56°19′

Ans: Angle of incidence is 56°19′

Example – 06:

  • If the critical angle of the medium is sin-1 (3/5), find the polarizing angle.
Science > Physics > Wave Theory of LightYou are Here
Physics Chemistry  Biology  Mathematics

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