# Problems on Wave Theory of Light

### Calculation of Angle of Refraction, Angle of Deviation, Wavelength and Frequency in Medium

#### Example – 1:

• A ray of light is incident on a glass slab making an angle of 25° with the surface. Calculate the angle of refraction in glass and velocity of light in glass, if the refractive index of glass and velocity of light are 1.5 and 3 x 108 m/s respectively.
• Solution:
• Given: Glancing angle = ig = 25o, Refractive index = μ = 1.5, Velocity of light in air = ca = 3 x 108  m/s
• To Find: Angle of refraction = r = ?, Velocity of light in glass = cg = ?

Angle of incidence = 90o – glancing angle = 90o – 25o = 65o.

μ = sin i / sinr

∴  Sin r = sin i / μ = sin 65o / 1.5 = 0.9063/1.5 = 0.6042

∴  angle of refraction = r = sin-1 (0.6042)

∴  Angle of refraction =  37o11’

Now, μ = ca/cg

∴  c g = ca/m = 3 x 108 /1.5 = 2 x 108 m/s

Ans: Angle of refraction = 37O10’ and velocity of light in glass = 2 x 108 m/s

#### Example – 2:

• A ray of light is incident on a glass slab making an angle of 60o with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air 1.5 and 3 x 108 m/s respectively.
• Solution:
• Given: Glancing angle = ig = 25o, Refractive index = μ = 1.5, Velocity of light in air = ca = 3 x 108  m/s
• To Find: Angle of refraction = r = ?, Velocity of light in glass = cg = ?

Angle of incidence = 90o – glancing angle = 90o – 60o = 30o.

μ = sin i / sinr

∴  Sin r = sin i / μ = sin 30o / 1.5 = 0.5/1.5 = 0.3333

∴ Angle of refraction = r = sin-1 (0.3333)

∴  Angle of refraction = 19o28’

Now, μ = ca/cg

∴  c g = ca/m = 3 x 108 /1.5 = 2 x 108 m/s

Ans: Angle of refraction = 19O28’ and velocity of light in glass = 2 x 108 m/s

#### Example – 3:

• A plane wavefront is made incident at an angle of 30° on the surface of the glass. Calculate angle of refraction if R.I. of glass is 1.5. Also find angle of deviation.
• Solution:
• Given: Angle of incidence = i =  30o, Refractive index = μ = 1.5,
• To Find: Angle of refraction = r = ? Angle of deviation = δ =?

μ = sin i / sinr

∴  Sin r = sin i / μ = sin 30o / 1.5 = 0.5/1.5 = 0.3333

∴  Angle of refraction = r = sin-1 (0.3333)

∴  Angle of refraction  = 19o28’

δ  = i – r = 30o  – 19o28’ = 10o32’

Ans: Angle of refraction = 19O28’ and angle of deviation = 10o32’

#### Example – 4:

• The wave number of beam of light in air is 2.5 x 106 per metre. What is the wavelength in glass if R.I. of the glass is 1.5.
• Solution:
• Given: Wave number in air  = 2.5 x 106 per metre, Refractive index = μg = 1.5,
• To Find: Wavelength in glass = λg = ?

Wave length in air = λ= 1/Wave number in air = 1/ 2.5 x 106  = 4 x 10-7 m

μg = λa / λg

∴  λg = λ/ μg = 4 x 10-7 / 1.5 = 2.667  x 10-7  m

∴  λg = 2.667  x 10-7  x 1010  = 2667 Å

Ans: Wavelength of light in glass is  2667 Å

#### Example – 5:

• What is the wave number of a beam of light in air if its frequency is 14 x 1014Hz?  c = 3 x 108 m/s.
• Solution:
• Given: Frequency in air = νa = 14 x 1014 Hz,
• Velocity of light in air = ca = 3 x 108  m/s
• To Find: Wave number in air = ?

We have = ca = νa λa

1/λ= νa /ca  = 14 x 1014 / 3 x 10 = 4.67 x 106  m-1

∴  Wave number = 1/λ=  4.67 x 106  m-1

Ans: Wave number in air is 4.67 x 106  m-1

#### Example – 6:

• The wavelength of monochromatic light is 5000 A.U. What will be its wave number in a  medium of R.I. 1.5?
• Solution:
• Given: Wavelength in air = λa = 5000 Å = 5000 x 10-10 m = 5 x 10-7 m, Refractive index of medium = 1.5
• To Find: Wave number in medium = ?

μm = λa / λm

1/λ= μma  = 1.5 / 5 x 10-7  = 3 x 106  m-1

∴  Wave number in medium = 1/λ=  3 x 106 m-1

Ans: Wave number in medium is 3 x 106 m-1

#### Example – 7:

• The wavelength of a beam of light in glass is 4400 Å. What is its wavelength in air, if refractive index of glass is 1.5
• Solution:
• Given: Wavelength in glass = λg = 4400 Å, Refractive index of medium = 1.5
• To Find: Wavelength in air = λa = ?

μg = λa / λg

∴ λ= μλg  = 1.5 x 4400  = 6600 Å

Ans: Wavelength in air is 6600 Å

#### Example – 8:

• The speed of light in air is 3 x 108 m/s and that in diamond is 1.4 x 108 m/s. Find the R.I. of diamond.
• Solution:
• Given: Speed of light in air  = ca = 3 x 108 m/s, Speed of light in diamond  = cd = 1.4 x 108 m/s,
• , Refractive index of medium = 1.5
• To Find: Refractive index of diamond = μd = ?

μd = ca / cd  = 3 x 10/ 1.4 x 108 = 2.142

Ans: Refractive index of diamond is 2.142

#### Example – 9:

• The velocity of light in diamond is 1.25  x 10m/s. Find the refractive index of diamond w.r.t. water. R.I. of water w.r.t. air is 1.33. Speed of light in air is 3 x 108 m/s
• Solution:
• Given: Speed of light in diamond  = cd = 1.25 x 108 m/s, Speed of light in air  = ca = 3 x 108 m/s, R.I. of water = μw = 1.33
• To Find: 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. water is given by

wμd = cw / cd   =  2.25 x 108 /1.25 x 108 =   1.8

Ans: Refractive index of diamond w.r.t. water is 1.8

#### Example – 10:

• The refractive indices of glycerine and diamond with respect to air are 1.4 and 2.4 respectively. Calculate the speed of light in glycerine and in diamond. From these results calculate the refractive index of diamond w.r.t. glycerine. c= 3 x 108 m/s,
• Solution:
• Given: Refractive index for glycerine = μg = 1.4, Refractive index for diamond = μd = 2.4, Speed of light in air  = ca = 3 x 108 m/s, R.I. of water = μw = 1.33
• To Find: Speed of light in glycerine = cg = ? The speedof light in diamond = cd = ?, Refractive index of diamond w.r.t. glycerine = gμd  = ?

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

μg = ca / c

cg = ca / μ= 3 x 10/1.4 = 2.143 x 108  m/s

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

μd = ca / c

cd = ca / μ= 3 x 10/2.4 = 1.25 x 108  m/s

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

gμd = cg / cd   =  2.143 x 108 /1.25 x 108 =   1.71

Ans: Speed of light in glycerine is 2.143 x 10m/s and that in diamond is 1.25 x 10m/s.

Refractive index of diamond w.r.t. glycerine is 1.71

#### Example – 11:

• The wavelength of light in water is 4000 Å and in the glass is 2500 Å Find the refractive index of glass w.r.t. water.
• Solution:
• Given: SWavelength in water  = λw = 4000 Å, Wavelength in glass = λ= 2500 Å
• To Find: Refractive index of glass w.r.t. water = wμg  = ?

μw = λw / λg  = 4000 / 2500 = 1.6

Ans: Refractive index of glass w.r.t. water is 1.6

#### Example – 12:

• The refractive indices of two media are 1.5 and 1.7. Calculate the velocity of light in these two media.
• Solution:
• Given: Refractive index of first medium  = μ1 = 1.5, Refractive index of second medium  = μ2 = 1.7, Speed of light in air  = ca = 3 x 108 m/s,
• To Find: Velocity of light in the two media = c1 = ? and c2 = ?

Consider first medium, μ1 = ca / c

c1 = ca / μ= 3 x 10/1.5 = 2 x 108  m/s

Consider second medium, μ2 = ca / c

c2 = ca / μ= 3 x 10/1.7 = 1.76 x 108  m/s

Ans: Velocity of light in first medium is 2 x 108  m/s and in second medium is 1.76 x 108  m/s

#### Example – 13:

• 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 refracytive index for both colour 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 glass is 2750 Å

#### Example – 14:

• A beam of red light (7000 Å) is passing from air into a medium making an angle of incidence of 61° and angle of refraction of 34°. Find the wavelength of red light in the medium.
• Solution:
• Given: Wavelength of red light in air = λ= 7000 Å, Angle of incidence = i = 61°, Angle of refraction = r = 34°,
• To Find: Wavelength of red light in medium = λ= ?

Refractive index of medium

μ = sin i / sinr = sin 61° / sin34° = 0.8746 / 0.5592= 1.564

Consider first medium, μm = λa / λm

λm = λa / μ= 7000 /1.564 = 4476 Å

Ans:  wavelength of red light in the medium is 4476 Å.

#### Example – 15:

• A ray of light travels from air to liquid by making an angle of incidence 24° and angle of refraction of 18°. Find R.I. of liquid. Determine the wavelength of light in air and in liquid if the frequency of light is 5.4 x 1014 Hz, c = 3 x 108 m/s.
• Solution:
• Given: Frequency of light in air = ν= 5.4 x 1014 Hz,  Angle of incidence = i = 24°, Angle of refraction = r = 18°, Velocity of light in air = ca = 3 x 108 m/s.
• To Find: Refractive index = μ = ?, Wavelength of red light in air and medium, λ= ?, λ= ?

Refractive index of medium

μ = sin i / sinr = sin 24° / sin18° = 0.4067 / 0.3090= 1.316

In air  ca = ν λa

∴  λ =  ca / νa   = 3 x 108  / 5.4 x 1014   = 5.555 x 10-7 m

∴  λ=  5555 x 10-10 m = 5555 Å

Now, μm  = λa / λm

λm = λa / μ= 5555 /1.316 = 4221 Å

Ans:  Refractive index of medium = 1.316, wavelength in air is 5555 Å,

wavelength in medium is 4221 Å

#### Example – 16:

• Monochromatic light of wavelength 6000 Å enters glass of R.I. 1.6. Calculate its velocity, frequency and wavelength in glass. c = 3 x 108 m/s.
• Solution:
• Given: Wavelength of light in air = λ= 6000 Å = 6000 x 10-10 m = 6 x 10-7 m,  Refractive index of medium = μ = 1.6, Velocity of light in air = ca = 3 x 108 m/s.
• To Find: velocity in medium  = cm= ?, Frquency in medium = νm = ?, Wavelength in medium, λ= ?,

We have, μ = ca / cm

cm = ca / μ= 3 x 108 /1.6 = 1.875 x 108 m/s

In air  ca = ν λa

∴  νa  =  ca / λa   = 3 x 108  / 6 x 10-7   = 5  x 1014 Hz

If medium changes, frequency remains the same,  νm  =  ν  = 5  x 1014 Hz

Now, μ = λa / λm

λm = λa / μ= 6000 /1.6 = 3750 Å

Ans:  Velocity of light in medium = 1.875 x 108 m/s, frequency in medium is 5  x 1014 Hz

wavelength in medium is 3750 Å

#### Example – 17:

• Light of wavelength 5000 A.U. is incident on water surface of R.I. 4/3. . Find the frequency and wavelength of light in water if its frequency in air is 6 x 1014 Hz.
• Solution:
• Given: Wavelength of light in air = λ= 5000 Å = 5000 x 10-10 m = 5 x 10-7 m,  Refractive index of water = μ = 4/3, Velocity of light in air = ca = 3 x 108 m/s. Frequency in air = νa = 6 x 1014 Hz
• To Find: Frquency in water = νw = ?, Wavelength in water, λ= ?,

If medium changes, frequency remains the same,  νw  =  ν  = 6  x 1014 Hz

Now, μ = λa / λm

λm = λa / μ= 5000 /(4/3) = 3750 Å

Ans:  Frequency in medium is 6  x 1014 Hz, wavelength in water is 3750 Å

#### Example – 18:

• The R.I. of glass w.r.t. water is 9/8. If velocity and wavelength of light in glass are 2 x 108  m/s and 4000 Å respectively, find its velocity and wavelength in water.
• Solution:
• Given: Wavelength of light in glass = λ= 4000 Å = 4000 x 10-10 m = 4 x 10-7 m,  Refractive index of glass w.r.t. water = wμg = 9/8, Velocity of light in glass = cg = 2 x 108 m/s.
• To Find: velocity in water = cw = ?, Wavelength in water, λ= ?,

we have, wμ= vw / vg

vw = vg x  wμg = 2  x 108 x (9/8) = 2.25 x 108  m/s

Now, wμg = λw / λg

λw = λa x wμg = 4000 x(9/8) = 4500 Å

Ans:  Velocity of light in water  is 2.25 x 108  m/s, wavelength in water is 4500 Å

#### Example – 19:

• Find 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.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 = | λ – λ| = ?

We have ca = νa λa

For air,  λa = c/ν

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

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

Ans:  Change in wavelength of light is 2500 Å

#### Example – 20:

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

μ(1.5) > μ(4/3)

Hence, c >  cg

We have ca = νa λand for glass cg = νg λ

For air,  λa = c/ν

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

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

Ans:  Change in wavelength of light is 2500 Å

#### Example – 21:

• 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 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 – 22:

• 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 =?

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 – 23:

• A parallel beam of monochromatic light is incident on a glas slab at an angle of 45°. Find the ratio of width of beam in glass to that in air if R.I. for glass is 1.5.
• Solution:
• Given: Angle of incidence = i = 45°, 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 =?

We have to find ratio CD/AB’

μ = sin i / sinr

∴  Sin r = sin i / μ = sin 45o / 1.5 = 0.7070/1.5 = 0.4713

∴  Angle of refraction = r = sin-1 (0.4713) = 28o7’

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 28o7’ / cos 45o

CD/AB’ = 0.8820 / 0.7070 = 1.25

Ans: The ratio of the width of the beam in the glass to that in the air is 1.25

#### Example – 24:

• The wavelength of a certain light in air and in a medium is 4560 Å and 3648 Å respectively. Compare the speed of light in air with its speed in the medium.
• Solution:
• Given: λ1 = 4560 Å, λ2 = 3648 Å.
• To Find: ca/cm =?

We have for air  ca = νaλa   ………….. (1)

for medium  cm = νmλm  ………….. (2)

Dividing equation (1) by (2)

ca / cm = λa / λ= 4560/3648 = 1.25

If medium changes, the frequency remains the same. Hence  ν=  νm

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

Ans: The ratio of the speed of light in the air with its speed in the medium is 1.25

#### Example – 25:

• Light of wavelength 6400 Å is incident normally on a plane parallel glass slab of thickness 5 cm and μ  = 1.6. The beam takes the same time to travel from the source to the incident surface as it takes to travel through the slab. Find the distance of the source from the incident surface. What is the frequency and wavelength of light in glass?  c = 3 x 10m/s.
• Solution:
• Given: λa = 6400 Å = 6400 x 10-10 m, μ  = 1.6, c = 3 x 10m/s. .
• To Find: Distance of source from surface =?

μg = ca / cg

∴ c = ca / μg  = 3 x 10/ 1.6 = 1.875 x 10m/s

Let t be the time taken by light to travel through glass slab.

Distance travelled in glass = Speed in glass x time

time = distance /speed = (5 x 10-2)/(1.875 x 108) = 2.667 x 10-10 s

Distance of source from slab = speed x time = 3 x 10x 2.667 x 10-10

Distance of source from slab = speed x time = 8 x 10-2 m = 8 cm

We have for air  ca = νaλa

νa = ca / λ= 3 x 10/ 6400 x 10-10 =4.69 x 1014

μg = λa / λg

λg = λa / μg  = 6400/1.6 = 4000 Å

Ans: Distance of source from slab = 8 cm; frequency in glass = 4.69 1014 Hz, wavelength in glass =4000 Å