In this article, we shall study brief history of study of light, sources of light, andterminology of optics.

History of Study of Light:

• Light is a form of that energy which simulates our vision. Around 400 B.C., it was proposed that particles were emitted by eye when the object is seen.
• In 1660 The great British physician  Sir Issac Newton proposed Corpuscular theory to explain the propagation of light. He assumed the particle nature of light.
• In 1680, A Dutch (Holand) scientist and contemporary of Newton, Christan Huygens (1629 – 1695) proposed Wave theory of light.
• The rectilinear propagation of light is due to the fact that the wavelength of light may be much smaller than the dimensions of openings and obstacles casting sharp shadows. Due to Newton’s clout on scientists in that era, the Huygens’s wave theory of light remained in a dump for almost century. Newton’s theory was challenged on the basis of Huygens’s wave theory of light by Thomas Young (1773 – 1829) in 1801 using his Double slit experiment.
• These experiments clearly established that light coming from two coherent sources interfere and produce maxima and minima depending upon path difference between the two waves. This phenomenon is known as the interference of light.
• Augustin Jean Fresnel (1788 – 1827)  performed a series of experiments to study the diffraction of light and disapproved Newton’s theory and supported Huygens’s wave theory of light.
• The exact nature of light waves was not known, because only mechanical waves were known at that time. For the propagation of mechanical waves, the medium is required. But light waves were found to travel in a vacuum. Hence in 1860 Maxwell proposed that the light are electromagnetic waves and do not require any medium for propagation.
• In 1888 Hertz and in 1900 Hallwachs and Lenard observed that when light falls on the metal surface, electrons are ejected and that the kinetic energy of emitting electrons does not depend on the intensity of light used.  This observation is known as the photoelectric effect. This effect was explained by Albert Einstein in 1905 on the basis of the particle model of light.
• Now it is accepted by the scientific community of the world that light has a dual nature particle as well as wave. A particular nature either wave or particle depends on circumstances.

Phenomenon Associated With Light:

Rectilinear Propagation of Light:

Light travels in a straight lines unless it is reflected by a polished surface or the medium of propagation is changed.

Reflection of Light:

When light rays are incident on a polished  surface, they are sent back in the same medium such that the angle of incidence is equal to angle of reflection and the incident ray, the reflected ray and the normal at the point of incidence lie in same plane.

Refraction of Light:

When light ray traveling in one optically active medium enters another optically active medium then the light ray deviates from its path. This phenomenon is known as refraction of light.

Diffraction of Light:

When a light wave is obstructed by an obstacle, then the rays round the corner. This phenomenon is called diffraction of light.

Interference of Light:

Light coming from two coherent sources interfere and produce maxima and minima depending upon path difference between the two waves. Thus alternate bright and dark regions can be obtained on the screen kept in the path of the two waves. This phenomenon is known as interference of light.

Polarization of Light:

The electric field in a light wave vibrate in the direction perpendicular to the direction of propagation of the wave. But there are infinite directions which are perpendicular to the direction of propagation of the wave. Thus the wave can vibrate in any plane perpendicular to the direction of propagation of the wave. Such wave is called non polarised wave.

By certain arrangements the non-polarized light wave is made to vibrate in one and only one plane, then the light is called polarised light.

Scattering of Light:

When a parallel beam of light passes through a gas, a part of it appears in directions other than the incident direction. This phenomenon is known as scattering of light.

Spectrum of Light

Visible Spectra:

The part of the spectrum of light which is visible to the human eye is called visible spectra. The frequency of visible light varies from  3.8 x 10¹⁴ Hz to 7.8 x 10¹⁴ Hz. The corresponding wavelengths are 380 nm and 780 nm.

The colour sensation of the human eye is related to the wavelength of light. The light close to 780 nm appears red and light close to 380 nm appears violet. The human eye is more sensitive to yellow and green light.

Colour Wavelengths:  Red (620 – 780 nm) Orange (590 – 620 nm) Yellow (570 – 590 nm) Green (500 – 570 nm) Blue (450 – 500 nm) Violet (380 – 450 nm).

Non Visible Spectra:

Light waves having a wavelength above 780 nm or having a frequency less than 3.8 x 1014 Hz are said to be lying in the infrared region.

Light waves having a wavelength above 380 nm or having frequency more than 7.8 x 1014 Hz are said to be lying in the ultraviolet region.

Characteristics of Light:

• When a monochromatic light travels from one medium to another, its wavelength changes but its frequency remains constant.
• In the electromagnetic waves the angle between electric field vector and magnetic field vector is 90°.
• In the propagation of electromagnetic waves, the angle between the direction of propagation and the plane of polarisation is zero.
• In the propagation of electromagnetic waves, the angle between the plane of vibration and the plane of polarisation is 90°.
• The oscillating electric and magnetic vector of an electromagnetic wave are oriented along mutually perpendicular direction and are in phase.
• Electromagnetic waves transport energy, momentum, information.
• Electromagnetic waves do not carry any charge.Energy of visible light is low (few eV)

Sources of Light:

Sun is major source of light. Some artificial sources of light are as follows

Thermal Sources :

These sources light of all wavelengths in the visible region i.e. having wavelengths in the range from 390 nm to 760 nm. Example: incandescent Bulb, Gas Discharge Tubes. A gas discharge tube emits light of a few wavelengths band. Example: Neon tube gives a characteristic red light.

Luminescent Sources:

These sources emit light partly in the visible region and partly in the ultraviolet region. Example: Fluorescent tube.

Phenomenon of Luminescene:

The phenomenon of emission of light after absorbing some electromagnetic radiations is called luminescene.

Types of Luminescene: 

• Electroluminescence: It is a phenomenon of emission of light by electrical means.
• Chemiluminescence: It is a phenomenon of emission of light by chemical reactions.
• Bioluminescence: It is a phenomenon of emission of light by biochemical reactions.

Photometry:

The branch of physics which deals with the measurement of light energy or with the comparisons of illuminating power of the sources or with the comparisons of illumination of the surfaces is called photometry.

 Efficiency of Light Source:

When a source of light emits energy, the whole of the radiant energy does not lie in the visible region. A small amount of energy lies in the nonvisible region i.e. infrared and ultraviolet region.

The efficiency of a light source is defined as the ratio of output power in the visible region to the input electrical power.

Luminous flux is measured in lumen (lm) and input electric power is measured in watts (W), hence unit of efficiency of a light source is lumen per watt (lm W-1)

Terminology:

• Optical Medium: The medium capable of transmitting light is called an optical medium.
• Isotropic medium: The homogeneous medium, which has the same properties in all the directions is called an isotropic medium.
• Monochromatic light: A light having one single wavelength is called as monochromatic light (Mono means one and chroma means colour).
• Incident ray: The light ray, which is falling on reflecting or refracting surface is called the incident ray.
• Point of incidence: The point at which the incident ray cuts the reflecting or refracting surface is called the point of incidence.
• Normal: A perpendicular drawn to the surface at the point of incidence is called the normal.
• Angle of incidence: The angle made by the incident ray with the normal at the point of incidence is called the angle of incidence.

Science > Physics > Wave Theory of Light >Photometry and Sources of Light

The post Photometry and Sources of Light appeared first on The Fact Factor.

In this article, we shall study to solve the problems on the calculation of the refractive index, angle of refraction, the wavelength of light and wavenumber of waves using snell’s law and the definition of the refractive index of the medium.

Example – 16: 

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_a = 4500 Å. = 4.5 x 10^-7  m, Refractive index of glass = μ = 1.55, c_a = 3 x 10⁸ m/s.

To Find: Frequency = ν =?, Wavelength in glass = λ_g  =?

Solution:

c_a = ν_a λ_a

∴  ν_a  = c_a/λ_a = 3 x 10⁸/4.5 x 10^-7 = 6.667 x 10¹⁵ Hz

μ_g = λ_a / λ_g 

∴  λ_g  = λ_a /μ_g  = 4500/1.55 = 2903 Å

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

Example 17:

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.

Given: Frequency of light in air = ν_a = 5.4 x 10¹⁴ Hz, Angle of incidence = i = 24°, Angle of refraction = r = 18°, Velocity of light in air = c_a = 3 x 10⁸ m/s.

To Find: Refractive index = μ =? Wavelength of red light in air and medium, λ_m = ?, λ_a = ?

Solution:

Refractive index of medium

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

In air c_a = ν_a λ_a

∴ λ_a =  c_a / ν_a   = 3 x 10⁸  / 5.4 x 10¹⁴  = 5.555 x 10^-7 m

∴ λ_a =  5555 x 10^-10 m = 5555 Å

Now, μ_m = λ_a / λ_m 

λ_m = λ_a / μ_m = 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 18:

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.

Given: Wavelength of light in air = λ_a = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m, Refractive index of medium = μ = 1.6, Velocity of light in air = c_a = 3 x 10⁸ m/s.

To Find: velocity in medium = c_m=? Frquency in medium = ν_m = ?, Wavelength in medium, λ_m = ?,

Solution:

We have, μ = c_a / c_m

c_m = c_a / μ_m = 3 x 10⁸ /1.6 = 1.875 x 10⁸ m/s

In air c_a = ν_a λ_a

∴  ν_a  =  c_a / λ_a   = 3 x 10⁸  / 6 x 10^-7  = 5  x 10¹⁴ Hz

If medium changes, frequency remains the same, ν_m = ν_a   = 5  x 10¹⁴ Hz

Now, μ = λ_a / λ_m

λ_m = λ_a / μ_m = 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 19:

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.

Given: Wavelength of light in air = λ_a = 5000 Å = 5000 x 10^-10 m = 5 x 10^-7 m, Refractive index of water = μ = 4/3, Velocity of light in air = c_a = 3 x 10⁸ m/s. Frequency in air = ν_a = 6 x 10¹⁴ Hz

To Find: Frquency in water = ν_w =? Wavelength in water, λ_w = ?,

Solution:

If medium changes, frequency remains the same, ν_w =  ν_a   = 6  x 10¹⁴ Hz

Now, μ = λ_a / λ_m

λ_m = λ_a / μ_m = 5000/(4/3) = 3750 Å

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

Example 20:

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.

Given: Wavelength of light in glass = λ_g = 4000 Å = 4000 x 10^-10 m = 4 x 10^-7 m, Refractive index of glass w.r.t. water = _wμ_g = 9/8, Velocity of light in glass = c_g = 2 x 10⁸ m/s.

To Find: velocity in water = c_w =? Wavelength in water, λ_w = ?,

Solution:

we have, _wμ_g = v_w / v_g

v_w = v_g x  _wμ_g = 2  x 10⁸ x (9/8) = 2.25 x 10⁸  m/s

Now, _wμ_g = λ_w / λ_g

λ_w = λ_a x _wμ_g = 4000x(9/8) = 4500 Å

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

Example 21:

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. Also find wave number in glass. c = 3 x 10⁸ m/s.

Given: Frequency of light in air = ν_g = 4 x 10¹⁴ Hz, Refractive index of glass = μ = 1.5, Velocity of light in air = c_a = 3 x 10⁸ m/s.

To Find: Change in wavelength of light = | λ_a – λ_g | =?

Solution:

We have c_a = ν_a λ_a

For air, λ_a = c/ν

λ_a = c_a/ν_a= 3x 10⁸  / 4 x 10¹⁴  =  7.5 x 10^-7  m

λ_a = 7500 x 10^-10  m = 7500 Å

Now, μ = λ_a / λ_g

λ_g = λ_a / μ= 7500/ 1.5= 5000 Å

Change in wavelength = λ_a – λ_g = 7500 – 5000 = 2500 Å

Wave number in glass = 1/λ_g = 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 22:

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.

Given: Difference in velocities = | c_g – c_w | = 2.5 x 10⁷ m/s, R.I. of water = μ_w = 4/3, R.I. of glass = μ_g = 1.5

To Find: Velocity of light in air = c_a =?

Solution:

μ_g (1.5) > μ_w (4/3)

Hence, c_w  >  c_g

c_w  –  c_g = 2.5 x 10⁷

We have, for water μ_w = c_a/c_w  and for glass μ_g = c_a/c_g

Hence for water c_w = c_a/μ_w  and for glass c_g = c_a/μ_g

c_w  –  c_g = 2.5 x 10⁷

∴  c_a/μ_w  –  c_a/μ_g = 2.5 x 10⁷

∴  c_a(1/μ_w  –  1/μ_g) = 2.5 x 10⁷

∴  c_a(3/4 –  2/3) = 2.5 x 10⁷

∴  c_a(1/12) = 2.5 x 10⁷

∴  c_a = 12 x 2.5 x 10⁷ = 3 x 10⁸  m/s

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

Example – 23:

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.

Given: Refractive index of red colour = μ_r = 1.325, Refractive index of violet colour = μ_v = 1.3334, Velocity of light in air = c_a = 3 x 10⁸ 

To Find: Difference in velocities = | c_r – c_v | =?

Solution:

μ_v (1.334) > μ_r (1.325)

Hence, c_r > c_v

We have for red light μ_r = c_a/c_r and for violet light μ_v = c_a/c_v

Hence for red light c_r = c_a /μ_r and for violet light c_v = c_a/μ_v

c_r – c_v  = c_a /μ_r  –  c_a/μ_v

c_r – c_v  = c_a (1/μ_r  – 1/μ_v)

c_r – c_v  =3 x 10⁸(1/1.325  – 1/1.334)

c_r – c_v  =3 x 10⁸(0.7547  – 0.7496)

c_r – c_v  = 3 x 10⁸(0.0051) = 1.53 x 10⁸ m/s

Ans: Difference in velocities of red and violet colour in water is 1.53 x 10⁸ m/s

Example – 24:

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.

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 =?

Solution:

We have to find ratio CD/AB’

μ = sin i / sinr

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

∴ r = sin^-1 (0.5773) = 35^o16’

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 35^o16’ / cos 60^o

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 – 25:

A parallel beam of monochromatic light is incident on a glass 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 45^o / 1.5 = 0.7070/1.5 = 0.4713

∴ r = sin^-1 (0.4713) = 28^o7’

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 28^o7’ / cos 45^o

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 – 26:

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.

Given: λ₁ = 4560 Å, λ₂ = 3648 Å.

To Find: c_a/c_m =?

Solution:

We have for air c_a = ν_aλ_a   ………….. (1)

for medium  c_m = ν_mλ_m  ………….. (2)

Dividing equation (1) by (2)

c_a / cm = λ_a / λ_m = 4560/3648 = 1.25

If medium changes, the frequency remains the same. Hence ν_a = ν_m

Hence for red light c_r = c_a /μ_r and for violet light c_v = c_a/μ_v

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

Example – 27:

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 10⁸ m/s.

Given: Wavelength of light = λ_a = 6400 Å = 6400 x 10^-10 m, μ = 1.6, c = 3 x 10⁸ m/s.

To Find: Distance of source from surface =?

Solution:

μ_g = c_a / c_g

∴ c_g = c_a / μ_g  = 3 x 10⁸ / 1.6 = 1.875 x 10⁸ m/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 10⁸ x 2.667 x 10^-10

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

We have for air c_a = ν_aλ_a 

ν_a = c_a / λ_a = 3 x 10⁸ / 6400 x 10^-10 =4.69 x 10¹⁴

μ_g = λ_a / λ_g

λ_g = λ_a / μ_g  = 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

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Science > Physics > Wave Theory of Light > Numerical Problems on wave Theory of Light

The post Numerical Problems on Wave Theory of Light – 02 appeared first on The Fact Factor.

In this article, we shall study to solve the problems on the calculation of the refractive index, angle of refraction, the wavelength of light and wavenumber of waves using snell’s law and the definition of the refractive index of the medium.

Example 01:

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 the glass, if the refractive index of glass and velocity of light are 1.5 and 3 x 10⁸ m/s respectively.

Given: Glancing angle = ig = 25^o, Refractive index = μ = 1.5, Velocity of light in air = c_a = 3 x 10⁸ m/s

To Find: Angle of refraction = r =? Velocity of light in glass = c_g = ?

Solution:

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

μ = sin i / sinr

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

∴ r = sin^-1 (0.6042)

∴ Angle of refraction = 37^o10’

Now, μ = c_a/c_g

∴ c_g = c_a/ μ = 3 x 10⁸ /1.5 = 2 x 10⁸ m/s

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

Example 02:

A ray of light is incident on a glass slab making an angle of 60^o 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.

Given: Glancing angle = i_g = 60^o, Refractive index = μ = 1.5, Velocity of light in air = c_a = 3 x 10⁸ m/s

To Find: Angle of refraction = r =? Velocity of light in glass = c_g = ?

Solution:

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

μ = sin i / sinr

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

∴ r = sin^-1 (0.3333)

∴ Angle of refraction = 19^o28’

Now, μ = c_a/c_g

∴ c_g = c_a/ μ = 3 x 10⁸ /1.5 = 2 x 10⁸ m/s

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

Example 03:

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.

Given: Angle of incidence = i = 30^o, Refractive index = μ = 1.5,

To Find: Angle of refraction = r =? Angle of deviation = δ =?

Solution:

μ = sin i / sinr

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

∴ r = sin^-1 (0.3333)

∴ Angle of refraction = 19^o28’

δ = i – r = 30^o – 19^o28’ = 10^o32’

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

Example 04:

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?

Given: Wave number in air = 2.5 x 10⁶ per metre, Refractive index = μ_g = 1.5,

To Find: Wavelength in glass = λ_g =?

Solution:

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

μ_g = λ_a / λ_g

∴ λ_g = λ_a / μ_g = 4 x 10^-7 / 1.5 = 2.667  x 10^-7  m

∴ λ_g = 2.667  x 10^-7  x 10¹⁰  = 2667 Å

Ans: Wavelength of light in glass is 2667 Å

Example 05:

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.

Given: Frequency in air = ν_a = 14 x 10¹⁴ Hz, Velocity of light in air = c_a = 3 x 10⁸ m/s

To Find: Wave number in air =?

Solution:

We have = c_a = ν_a λ_a

1/λ_a = ν_a /c_a = 14 x 10¹⁴ / 3 x 10⁸ = 4.67 x 10⁶ m^-1 

∴ Wave number = 1/λ_a = 4.67 x 10⁶ m^-1 

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

Example 06:

The wavelength of monochromatic light is 5000 A.U. What will be its wave number in a medium of R.I. 1.5?

Given: Wavelength in air = λ_a = 5000 Å = 5000 x 10^-10 m = 5 x 10^-7 m, Refractive index of medium = 1.5

To Find: Wave number in medium =?

Solution:

μ_m = λ_a / λ_m

1/λ_m = μ_m /λ_a = 1.5/ 5 x 10^-7 = 3 x 10⁶  m^-1 

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

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

Example 07:

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?

Given: Wavelength in glass = λ_g = 4400 Å, Refractive index of medium = 1.5

To Find: Wavelength in air = λ_a =?

Solution:

μ_g = λ_a / λ_g

∴ λ_a = μ_g λ_g  = 1.5 x 4400 = 6600 Å

Ans: Wavelength in air is 6600 Å 

Example 08:

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.

Given: Speed of light in air = c_a = 3 x 10⁸ m/s, Speed of light in diamond = c_d = 1.4 x 10⁸ m/s, Refractive index of medium = 1.5

To Find: Refractive index of diamond = μ_d =?

Solution:

μ_d = c_a / c_d  = 3 x 10⁸ / 1.4 x 10⁸ = 2.142

Ans: Refractive index of diamond is 2.142

Example 09:

The velocity of light in diamond is 1.25 x 10⁸ m/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

Given: Speed of light in diamond = c_d = 1.25 x 10⁸ m/s, Speed of light in air = c_a = 3 x 10⁸ m/s, R.I. of water = μ_w = 1.33

To Find: Refractive index of diamond w.r.t. water = _wμ_d =?

Solution:

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

μ_w = c_a / c_w 

c_w = c_a / μ_w = 3 x 10⁸ /1.33 = 2.25 x 10⁸  m/s

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

_wμ_d = c_w / c_d   =  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,

Given: Refractive index for glycerine = μ_g = 1.4, Refractive index for diamond = μ_d = 2.4, Speed of light in air = c_a = 3 x 10⁸ m/s, R.I. of water = μ_w = 1.33

To Find: Speed of light in glycerine = c_g =? The speedof light in diamond = c_d =?, Refractive index of diamond w.r.t. glycerine = _gμ_d  = ?

Solution:

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

μ_g = c_a / c_g 

c_g = c_a / μ_g = 3 x 10⁸ /1.4 = 2.143 x 10⁸  m/s

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

μ_d = c_a / c_d 

c_d = c_a / μ_d = 3 x 10⁸ /2.4 = 1.25 x 10⁸  m/s

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

_gμ_d = c_g / c_d   =  2.143 x 10⁸ /1.25 x 10⁸ = 1.71

Ans: Speed of light in glycerine is 2.143 x 10⁸ m/s and that in diamond is 1.25 x 10⁸ m/s, Refractive index of diamond w.r.t. glycerine is 1.71

Example – 11:

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,

Given: Refractive index for water = μ_w = 4/3 = 1.33, Refractive index for diamond = μ_d = 2.4, Speed of light in air = c_a = 3 x 10⁸ m/s,

To Find: Speed of light in water = c_w =? The speed of light in diamond = c_d =?, Refractive index of diamond w.r.t. water = _wμ_d  = ?

Solution:

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

μ_w = c_a / c_w 

c_w = c_a / μ_w = 3 x 10⁸ /1.33 = 2.25 x 10⁸  m/s

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

μ_d = c_a / c_d 

c_d = c_a / μ_d = 3 x 10⁸ /2.42 = 1.24 x 10⁸  m/s

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

_wμ_d = c_w / c_d   =  2.25 x 10⁸ /1.24 x 10⁸ = 1.815

Ans: Speed of light in water is 2.25 x 10⁸ m/s and that in diamond is 1.25 x 10⁸ m/s and Refractive index of diamond w.r.t. water is 1.815

Example 12:

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.

Given: SWavelength in water = λ_w = 4000 Å, Wavelength in glass = λ_g = 2500 Å

To Find: Refractive index of glass w.r.t. water = _wμ_g = ?

Solution:

μ_w = λ_w / λ_g  = 4000 / 2500= 1.6

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

Example 13:

The refractive indices of two media are 1.5 and 1.7. Calculate the velocity of light in these two media.

Given: Refractive index of first medium = μ₁ = 1.5, Refractive index of second medium = μ₂ = 1.7, Speed of light in air = c_a = 3 x 10⁸ m/s,

To Find: Velocity of light in the two media = c₁ =? c₂ =?

Solution:

Consider first medium, μ₁ = c_a / c₁ 

c₁ = c_a / μ₁ = 3 x 10⁸ /1.5 = 2 x 10⁸ m/s

Consider second medium, μ₂ = c_a / c₂ 

c₂ = c_a / μ₂ = 3 x 10⁸ /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 14:

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?

Given: Wavelength in air for red light = λ_ar = 6400 Å, Wavelength in glass for red light = λ_gr = 4000 Å, Wavelength in air for violet light = λ_av = 4400 Å,

To Find: Wavelength in glass for violet light = λ_gv =?

Solution:

Consider red light

μ_r = λ_ar / λ_gr   = 6400 /4000= 1.6

Assuming the refractive index for both colour is same i.e. μ_r = μ_v

Consider violet light

μ_v = λ_av / λ_gv   

λ_gv= λ_av /  μ_v = 4400/1.6 = 2750 Å 

Ans: Wavelength of violet colour in glass is 2750 Å 

Example 15:

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.

Given: Wavelength of red light in air = λ_a = 7000 Å,Angle of incidence = i = 61°, Angle of refraction = r = 34°,

To Find: Wavelength of red light in medium = λ_m =?

Solution:

Refractive index of medium

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

Consider first medium, μ_m = λ_a / λ_m 

λ_m = λ_a / μ_m = 7000/1.564 = 4476 Å

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

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Science > Physics > Wave Theory of Light > Numerical Problems on wave Theory of Light

The post Numerical Problems on wave Theory of Light – 01 appeared first on The Fact Factor.

In the last article, we had a brief idea of the Huygens wave theory of light. In this article, we shall study its use to explain the phenomena of reflection and refraction of light.

Huygens Principle:

Huygens Proposed that

• Every point on a wavefront behaves as if it is a secondary source of light sending secondary waves in all possible directions.
• The new secondary wavelets are more effective in the forward direction only.
• The envelope (tangent) of all the secondary wavelets at a given instant in the forward direction gives the new wavefront at that instant.

Applications of Huygens Wave Theory of Light:

Huygens Construction of Spherical wavefront:

Consider a point source O of light giving rise to a spherical wavefront ABCDE at any instant. According to the Huygens principle, each point on a wavefront acts as a secondary source of light producing secondary wavelets in all directions.

Let ‘c’ be the velocity of light in air and ‘t’ be the time after which position of the wavefront is to be found. During time ‘t’, the light wave will travel a distance of ‘ct’. To find the position and shape of the wavefront after time ‘t’, a number of spheroids of radius ‘ct’ are drawn with their centres A, B, C, D, and E.

The tangential spherical envelope (surface) joining the points A’, B’, C’, D’ and E’ in the forward direction is the new wavefront at that instant.  The secondary wavelets are effective only in the direction of wave normal.  Therefore, a backward wavefront is absent.

Huygens Construction of Plane Wavefront:

Consider a plane wavefront ABCDE at any instant. According to Huygens’ principle, each point on a wavefront acts as a secondary source of light producing secondary wavelets in all directions.

Let ‘c’ be the velocity of light in air and ‘t’ be the time after which position of the wavefront is to be found. During time ‘t’ the light wave will travel a distance of ‘ct’. To find the position and shape of the wavefront after time ‘t’, a number of spheroids of radius ‘ct’ are drawn with their centres A, B, C, D, and E.

The tangential envelope (surface) joining the points A’, B’, C’, D’ and E’ in the forward direction is the new wavefront at that instant.  The secondary wavelets are effective only in the direction of wave normal.  Therefore, a backward wavefront is absent.

Reflection of Light on the basis of Huygens Theory:

Laws  of Reflection:

• The angle of incidence is equal to the angle of reflection
• The incident ray and the reflected ray lie on either side of the normal at the point of incidence
• The incident ray, reflected ray and the normal at the point of incidence lie in the same plane.

Explanation:

Consider a plane wavefront bounded by parallel rays PA and QB, traveling through air be obliquely incident on a plane reflecting surface XY. At an instant when the plane wavefront AB just touches the reflecting surface, point A’ becomes a secondary source to send out backward secondary wavelets.

As the incident plane, wavefront proceeds further let ‘t’ be the time required by end B of this wavefront to reach the reflecting surface at C. if ‘c’ is the velocity of light in the air then, BC = c t.

During this time, the secondary wavelet, originating from point A spreads over, a hemisphere. In time ‘t’ the radius of this spherical wavelet is AD = c t

Draw a tangent from point C to the secondary spherical wavelet to meet it at D. As C and D are in the same phase, the tangent CD represents the reflected plane wavefront after time ‘t’.

From the ray diagram:

∠ PA’N = ∠ BCN’ = i, angle of incidence

∠ B’CA’ = 90° – i ,

∠ NA’D =  r, angle of reflection

∠ DA’C = 90° – r

In Δ A’B’C and Δ CDA’

The angle between plane wavefront and wave normal is 90°

∠ A’B’C = ∠ A’DC  =  90°

B’C = A’D = ct

A’C is common to both the triangles

Hence, A’B’C and DCDA’ are congruent. (Hypotenuse side theorem)

∠ B’CA’ = ∠ DA’C (CACT)

∴ 90° – i = 90° – r

∴        i = r

Thus the angle of incidence is equal to angle of reflection,

From the diagram, it can be seen that the incident ray and the reflected ray lie on either side of the normal at the point of incidence. Similarly the incident ray, the reflected ray and the normal at the point of incidence lie in the same plane i.e. plane of the paper. Hence laws of reflection are proved. Thus the phenomenon of reflection is explained on the basis of the Huygens’ wave theory of light.

Reflection of Light on the Basis of Huygens Theory:

Laws of Refraction:

• The ratio of the sine of the angle of incidence to the sine of the angle of refraction is always constant and is equal to the refractive index of the medium.  This law is known as Snell’s law.
• The incident ray and the refracted ray lie on the opposite side of the normal at the point of incidence.
• The incident ray, refracted ray and the normal at the point of incidence lie in the same plane.

Explanation:

Consider a plane wavefront bounded by parallel rays PA and QB, traveling through a rarer medium of refractive index μ1 be obliquely incident on a plane refracting surface XY. At an instant when the plane wavefront AB just touches the refracting surface at point A’, point A’ becomes a secondary source to send out secondary wavelets in the second medium of refractive index μ2.

As the incident plane, wavefront proceeds further let ‘t’ be the time required by end B of this wavefront to reach the reflecting surface at B’. If c₁ is the velocity of light in the air then, B’C = c1 t.

During this time, the secondary wavelet, originating from point A’ spreads over, a hemisphere in the second medium. Let c2 be the speed of the light in the second medium. In time ‘t’ the radius of this spherical wavelet is A’D = c₂ t

Draw a tangent from point C to the secondary spherical wavelet to meet it at D. As C and D are in the same phase, the tangent CD represents the refracted plane wavefront after time ‘t’.

Construct NN’, normal to the refracting surface at A’.

From ray diagram:

∠AA’N  +  ∠ NA’B’ =  90°    ……. (1)

∠NA’B’  + ∠ B’A’C =  90°    ……. (2)

From (1) & (2)

∠ AA’N  + ∠ NA’B’ = ∠ NA’B’ + ∠ B’A’C

∴ ∠ AA’N = ∠ B’A’C  =  i

∠N’A’D  +  ∠ DA’C =  90°    ….. (3)

∠DA’C  +  ∠ A’CD =  90°     …. (4)

From (3) & (4)

∠ N’A’D + ∠ DA’C = ∠ DA’C + ∠ A’CD

∴ ∠ N’A’D = ∠ A’CD =  r

The angle between plane wavefront and wave normal is 90°

∠ A’B’C = ∠ A’CD = 90°

Thus Snell’s law is proved. From the diagram, it can be seen that the incident ray and the refracted ray lie on the opposite side of the normal at the point of incidence. Similarly the incident ray, the refracted ray and the normal at the point of incidence lie in the same plane i.e. plane of the paper. Hence laws of refraction are proved. Thus the phenomenon of refraction is explained on the basis of the Huygens’ wave theory of light.

For an optically denser medium n > 1, hence the velocity of light in an optically rarer medium is more than optically denser medium.

Next Topic: Numerical Problems on the wave Theory