NCERT CLASS 12 PHYSICS CHAPTER 10 WAVE OPTICS
NCERT CLASS 12 PHYSICS
CHAPTER 10 – WAVE OPTICS
1. IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
Q1. What is wave optics?
Answer: Wave optics is the branch of optics that explains the behaviour of light by considering its wave nature. It explains phenomena such as interference, diffraction and polarisation.
Q2. What are the two main theories that explain the nature of light?
Answer:
Wave theory: According to this theory, light behaves as a wave.
Particle theory: According to this theory, light consists of particles called photons.
Wave optics explains several phenomena that cannot be explained by geometrical ray optics alone.
Q3. What is a wavefront?
Answer: A wavefront is an imaginary surface joining all the points of a wave that are in the same phase of vibration at a particular instant.
Q4. What are the different types of wavefronts?
Answer: There are three main types of wavefronts:
Spherical wavefront: It is produced by a point source of light. The wavefront consists of concentric spherical surfaces.
Cylindrical wavefront: It is produced by a linear source of light.
Plane wavefront: It is produced when the source is very far away. A small portion of a spherical wavefront at a large distance can be considered plane.
Q5. What is a ray of light?
Answer: A ray of light is a line that represents the direction of propagation of light energy. A ray is always perpendicular to the wavefront.
Q6. What is Huygens’ principle?
Answer: Huygens’ principle states that:
Every point on a given wavefront acts as a source of secondary wavelets.
These secondary wavelets spread out in the forward direction with the speed of light in that medium.
The forward envelope or common tangent to these secondary wavelets gives the new position of the wavefront after a certain time.
Q7. What are secondary wavelets?
Answer: The small wave disturbances produced by every point on a primary wavefront are called secondary wavelets. They spread in the medium and help in constructing the new wavefront.
Q8. What is the difference between a wavefront and a ray?
Answer:
A wavefront is a surface joining points having the same phase.
A ray represents the direction of propagation of light.
Rays are perpendicular to wavefronts.
Q9. What is the principle of superposition of waves?
Answer: The principle of superposition states that when two or more waves travel through the same medium simultaneously, the resultant displacement at any point is equal to the vector sum of the individual displacements produced by each wave.
Q10. What is interference of light?
Answer: Interference of light is the phenomenon in which two or more coherent light waves superpose to produce a redistribution of intensity, forming alternate bright and dark regions.
Q11. What are coherent sources of light?
Answer: Two sources are said to be coherent if they emit light waves of the same frequency and maintain a constant phase difference between them.
Q12. Why are two independent sources of light generally not coherent?
Answer: Two independent sources emit light through separate atomic processes. Their phase difference changes randomly with time. Therefore, they do not produce a stable interference pattern.
Q13. What is constructive interference?
Answer: Constructive interference occurs when two waves meet in the same phase. Their amplitudes add up and produce maximum intensity.
The condition for constructive interference is:
Path difference = nλ
Where n = 0, 1, 2, 3, ...
Q14. What is destructive interference?
Answer: Destructive interference occurs when two waves meet in opposite phases. Their amplitudes cancel each other and produce minimum intensity.
The condition for destructive interference is:
Path difference = (2n + 1)λ/2
Where n = 0, 1, 2, 3, ...
Q15. What is Young’s double-slit experiment?
Answer: Young’s double-slit experiment demonstrates the interference of light. In this experiment, light from a single source passes through two narrow, closely spaced slits. The slits act as coherent sources and produce alternate bright and dark fringes on a screen.
Q16. What is fringe width?
Answer: Fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes in an interference pattern.
It is represented by β.
Formula:
β = λD/d
Where:
λ = Wavelength of light
D = Distance between the slits and the screen
d = Separation between the two slits
Q17. What happens to fringe width when the wavelength of light is increased?
Answer: Fringe width increases because fringe width is directly proportional to wavelength.
β ∝ λ
Q18. What happens to fringe width when the distance between the slits and the screen is increased?
Answer: Fringe width increases because it is directly proportional to the distance between the slits and the screen.
β ∝ D
Q19. What happens to fringe width when the separation between the two slits is increased?
Answer: Fringe width decreases because it is inversely proportional to the separation between the two slits.
β ∝ 1/d
Q20. What is diffraction of light?
Answer: Diffraction is the phenomenon in which light bends or spreads around the edges of an obstacle or through a narrow opening into the region of geometrical shadow.
Diffraction becomes significant when the size of the obstacle or opening is comparable to the wavelength of light.
Q21. What is the difference between interference and diffraction?
Answer:
Interference:
It generally results from the superposition of waves coming from two or more coherent sources.
It produces alternate bright and dark fringes.
In the ideal case of two equal-amplitude waves, bright fringes have equal maximum intensity.
Diffraction:
It results from the superposition of wavelets originating from different parts of the same wavefront.
It produces a broad central maximum and weaker secondary maxima.
The intensity of secondary maxima decreases away from the centre.
Q22. What is the central maximum in single-slit diffraction?
Answer: The broad and brightest central bright region formed in a single-slit diffraction pattern is called the central maximum. It is wider than the secondary maxima.
Q23. What is the condition for minima in single-slit diffraction?
Answer: For a slit of width a, the condition for diffraction minima is:
a sin θ = nλ
Where:
a = Width of the slit
θ = Angle of diffraction
λ = Wavelength of light
n = 1, 2, 3, ...
Q24. What is polarisation of light?
Answer: Polarisation is the phenomenon in which the vibrations of the electric field of light are restricted to a particular direction perpendicular to the direction of propagation.
Q25. What is unpolarised light?
Answer: Light in which the electric field vibrations occur randomly in different directions perpendicular to the direction of propagation is called unpolarised light.
Q26. What is plane-polarised light?
Answer: Light in which the electric field vibrations are restricted to one particular direction perpendicular to the direction of propagation is called plane-polarised light.
Q27. What is a Polaroid?
Answer: A Polaroid is an optical material that allows light vibrations along a particular direction to pass through while absorbing or reducing the perpendicular component.
Q28. What is an analyser?
Answer: An analyser is a Polaroid used to examine whether light is polarised and to determine the direction of its plane of polarisation.
Q29. State Malus’ law.
Answer: Malus’ law states that when plane-polarised light of intensity I₀ passes through an analyser, the transmitted intensity is given by:
I = I₀ cos² θ
Where θ is the angle between the transmission axes of the polariser and analyser.
Q30. What happens when unpolarised light passes through an ideal Polaroid?
Answer: The intensity of unpolarised light becomes half after passing through an ideal Polaroid.
I = I₀/2
Q31. How does polarisation prove that light is a transverse wave?
Answer: Polarisation is possible only for transverse waves because their vibrations are perpendicular to the direction of propagation. Since light can be polarised, it demonstrates the transverse nature of light waves.
Q32. What is the relation between the speed, frequency and wavelength of light?
Answer:
The relation is:
v = νλ
Where:
v = Speed of light in the medium
ν = Frequency of light
λ = Wavelength of light
Q33. What happens to the frequency of light when it enters another medium?
Answer: The frequency of light remains unchanged when it enters another medium. However, its speed and wavelength change.
Q34. Why does the wavelength of light decrease when it enters an optically denser medium?
Answer: The speed of light decreases in an optically denser medium, while its frequency remains constant.
Since:
v = νλ
A decrease in speed at constant frequency results in a decrease in wavelength.
Q35. What is the relation between refractive index and speed of light?
Answer:
The absolute refractive index of a medium is:
n = c/v
Where:
n = Refractive index
c = Speed of light in vacuum
v = Speed of light in the medium
2. NCERT EXERCISE QUESTIONS AND ANSWERS
Question 10.1
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
Solution:
Given:
Wavelength of incident light:
λ = 589 nm
λ = 589 × 10⁻⁹ m
Speed of light in air:
c = 3 × 10⁸ m/s
Refractive index of water:
n = 1.33
First, calculate the frequency of incident light.
Formula:
ν = c/λ
Substituting the values:
ν = (3 × 10⁸)/(589 × 10⁻⁹)
ν = 5.09 × 10¹⁴ Hz
(a) Reflected light
The reflected light travels in air. Therefore, its wavelength, frequency and speed remain the same as those of the incident light.
Wavelength = 589 nm
Frequency = 5.09 × 10¹⁴ Hz
Speed = 3 × 10⁸ m/s
(b) Refracted light
When light enters water, its frequency remains unchanged.
Therefore:
Frequency = 5.09 × 10¹⁴ Hz
Speed of light in water:
v = c/n
v = (3 × 10⁸)/1.33
v = 2.26 × 10⁸ m/s
Wavelength in water:
λ' = λ/n
λ' = 589/1.33
λ' = 442.86 nm
λ' ≈ 443 nm
Final Answer:
(a) Reflected light:
Wavelength = 589 nm
Frequency = 5.09 × 10¹⁴ Hz
Speed = 3 × 10⁸ m/s
(b) Refracted light:
Wavelength ≈ 443 nm
Frequency = 5.09 × 10¹⁴ Hz
Speed = 2.26 × 10⁸ m/s
Question 10.2
What is the shape of the wavefront in each of the following cases?
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) Light diverging from a point source
The wavefront is spherical because light spreads in all directions from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus
The wavefront is plane because the convex lens converts the diverging light into a parallel beam when the point source is placed at its principal focus.
(c) Light from a distant star intercepted by the Earth
The wavefront is approximately plane because the star is extremely far away. A small portion of its spherical wavefront reaching the Earth appears almost flat.
Final Answer:
(a) Spherical wavefront.
(b) Plane wavefront.
(c) Plane wavefront.
Question 10.3
(a) The refractive index of glass is 1.5. What is the speed of light in glass?
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Solution:
(a) Speed of light in glass
Given:
Refractive index of glass:
n = 1.5
Speed of light in vacuum:
c = 3 × 10⁸ m/s
Formula:
n = c/v
Therefore:
v = c/n
Substituting the values:
v = (3 × 10⁸)/1.5
v = 2 × 10⁸ m/s
Answer:
Speed of light in glass = 2 × 10⁸ m/s.
(b) Does the speed of light depend on colour?
Yes, the speed of light in glass depends on its colour because the refractive index of glass is different for different wavelengths.
Violet light has a higher refractive index than red light.
Since:
v = c/n
The speed of violet light is less than the speed of red light.
Answer: Violet light travels slower than red light in a glass prism.
Question 10.4
In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Determine the wavelength of light used.
Solution:
Given:
Distance between the two slits:
d = 0.28 mm
d = 0.28 × 10⁻³ m
Distance between slits and screen:
D = 1.4 m
Distance between central bright fringe and fourth bright fringe:
x₄ = 1.2 cm
x₄ = 1.2 × 10⁻² m
Order of bright fringe:
n = 4
Formula for the position of the nth bright fringe:
xₙ = nλD/d
Therefore:
λ = xₙd/nD
Substituting the values:
λ = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4)
λ = (3.36 × 10⁻⁶)/5.6
λ = 0.6 × 10⁻⁶ m
λ = 6 × 10⁻⁷ m
Converting into nanometres:
λ = 600 nm
Final Answer:
Wavelength of light = 6 × 10⁻⁷ m or 600 nm.
Question 10.5
In a Young's double-slit experiment, the intensity at a point where the path difference is λ is K units. What is the intensity at a point where the path difference is λ/3?
Solution:
Given:
Intensity at path difference λ:
I₁ = K
Path difference at the second point:
Δx = λ/3
The phase difference is given by:
φ = 2πΔx/λ
Substituting:
φ = 2π(λ/3)/λ
φ = 2π/3
The resultant intensity for two coherent waves of equal intensity is:
I = 4I₀ cos²(φ/2)
For the first point, path difference = λ.
Therefore:
φ₁ = 2π
K = 4I₀ cos²(π)
K = 4I₀
For the second point:
I₂ = 4I₀ cos²(π/3)
I₂ = 4I₀ × (1/2)²
I₂ = I₀
Since K = 4I₀:
I₀ = K/4
Therefore:
Final Answer:
Intensity at path difference λ/3 = K/4 units.
Question 10.6
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young's double-slit experiment.
(a) Find the distance of the third bright fringe for the wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Solution:
Given:
First wavelength:
λ₁ = 650 nm
Second wavelength:
λ₂ = 520 nm
Distance between slits = d
Distance between slits and screen = D
(a) Distance of the third bright fringe for wavelength 650 nm
Formula for the position of the nth bright fringe:
xₙ = nλD/d
For the third bright fringe:
n = 3
Therefore:
x₃ = 3λ₁D/d
Substituting:
x₃ = 3 × 650 × 10⁻⁹ × D/d
x₃ = 1950 × 10⁻⁹ × D/d
x₃ = 1.95 × 10⁻⁶ × D/d m
Answer:
Distance of the third bright fringe:
x₃ = 1.95 × 10⁻⁶ D/d m
(b) Least distance where bright fringes of both wavelengths coincide
For bright fringes to coincide, their positions must be equal.
Therefore:
n₁λ₁D/d = n₂λ₂D/d
Cancelling D/d:
n₁λ₁ = n₂λ₂
Substituting the wavelengths:
650n₁ = 520n₂
Dividing by 130:
5n₁ = 4n₂
The smallest non-zero integer values satisfying this condition are:
n₁ = 4
n₂ = 5
Therefore, the fourth bright fringe of wavelength 650 nm coincides with the fifth bright fringe of wavelength 520 nm.
The least distance is:
x = 4λ₁D/d
x = 4 × 650 × 10⁻⁹ × D/d
x = 2600 × 10⁻⁹ × D/d
x = 2.6 × 10⁻⁶ D/d m
Final Answer:
(a) Distance of third bright fringe:
x₃ = 1.95 × 10⁻⁶ D/d m
(b) Least distance where bright fringes coincide:
x = 2.6 × 10⁻⁶ D/d m
3. IMPORTANT QUESTIONS AND ANSWERS
Q1. Explain Huygens’ principle.
Answer:
Huygens’ principle explains how a wavefront moves from one position to another.
According to this principle:
Every point on a wavefront acts as a source of secondary wavelets.
These wavelets spread in the medium with the speed of the wave.
The forward envelope of the secondary wavelets forms the new wavefront.
This principle is used to explain the laws of reflection and refraction of light.
Q2. State and explain the laws of reflection of light using Huygens’ principle.
Answer:
The two laws of reflection are:
The angle of incidence is equal to the angle of reflection.
The incident ray, reflected ray and normal at the point of incidence lie in the same plane.
Using Huygens’ principle:
Consider a plane wavefront AB incident on a reflecting surface.
Let point A touch the reflecting surface first. During the time taken by point B to reach point C, a secondary wavelet from A travels a distance equal to the distance travelled by the incident wave.
If the speed of light is v and the time taken is t:
BC = vt
AE = vt
The two right-angled triangles formed in the construction have equal hypotenuse and one corresponding side equal. Therefore, they are congruent.
Hence:
∠i = ∠r
Thus, the angle of incidence is equal to the angle of reflection.
The incident wavefront, reflected wavefront and the normal lie in the same plane. Therefore, both laws of reflection are satisfied.
Q3. State and explain Snell's law of refraction using Huygens’ principle.
Answer:
Snell's law states that for a given pair of media:
n₁ sin i = n₂ sin r
Where:
n₁ = Refractive index of the first medium
n₂ = Refractive index of the second medium
i = Angle of incidence
r = Angle of refraction
Explanation using Huygens’ principle:
Consider a plane wavefront AB incident on the boundary separating two media.
Let:
v₁ = Speed of light in the first medium
v₂ = Speed of light in the second medium
t = Time taken by the wavefront to travel
During time t:
BC = v₁t
AE = v₂t
From the geometry of the wavefront construction:
sin i = BC/AC
sin r = AE/AC
Dividing the two equations:
sin i/sin r = BC/AE
Substituting:
sin i/sin r = v₁t/v₂t
sin i/sin r = v₁/v₂
Since:
n₁/n₂ = v₂/v₁
Therefore:
n₁ sin i = n₂ sin r
Hence, Snell's law is proved.
Q4. Explain Young's double-slit experiment.
Answer:
Young's double-slit experiment demonstrates the wave nature of light through interference.
Experimental arrangement:
A monochromatic source of light illuminates two narrow, closely spaced slits.
The two slits act as coherent sources.
Light waves from both slits travel towards a screen.
The waves superpose and form alternate bright and dark fringes.
Observation:
Bright fringes are formed due to constructive interference.
Dark fringes are formed due to destructive interference.
The central fringe is bright when the path difference at the centre is zero.
This experiment proves that light exhibits wave behaviour.
Q5. Derive the expression for fringe width in Young's double-slit experiment.
Answer:
Consider two coherent slits S₁ and S₂ separated by a distance d.
Let:
D = Distance between the slits and screen
λ = Wavelength of light
x = Distance of a point from the central fringe
d = Separation between the slits
For a point P on the screen, the path difference is approximately:
Δx = d sin θ
For a small angle θ:
sin θ ≈ tan θ
tan θ = x/D
Therefore:
Δx = dx/D
For a bright fringe, the condition is:
Δx = nλ
Therefore:
dx/D = nλ
xₙ = nλD/d
The distance between two successive bright fringes is:
β = xₙ₊₁ − xₙ
β = (n + 1)λD/d − nλD/d
Therefore:
β = λD/d
This is the expression for fringe width in Young's double-slit experiment.
Q6. Explain the conditions for constructive and destructive interference.
Answer:
Constructive interference:
It occurs when two waves meet in the same phase.
Condition:
Path difference = nλ
Phase difference = 2nπ
Result: Maximum intensity or bright fringe.
Destructive interference:
It occurs when two waves meet in opposite phases.
Condition:
Path difference = (2n + 1)λ/2
Phase difference = (2n + 1)π
Result: Minimum intensity or dark fringe.
Q7. What is diffraction of light? Explain single-slit diffraction.
Answer:
Diffraction is the bending or spreading of light waves around the edges of an obstacle or through a narrow opening.
When monochromatic light passes through a narrow slit, the light spreads and produces a diffraction pattern on a screen.
The pattern consists of:
A broad and bright central maximum.
Dark regions on both sides of the central maximum.
Weaker secondary maxima on either side.
For a slit of width a, the condition for minima is:
a sin θ = nλ
Where n = 1, 2, 3, ...
Q8. Write the expression for the width of the central maximum in single-slit diffraction.
Answer:
For the first minimum:
a sin θ = λ
For small angles:
sin θ ≈ θ
Therefore:
θ = λ/a
The angular width of the central maximum is:
2θ = 2λ/a
If the screen is at a distance D, the linear width is:
Width of central maximum = 2λD/a
Q9. Explain the difference between interference and diffraction.
Answer:
| Interference | Diffraction |
|---|---|
| It is produced by the superposition of waves from coherent sources. | It is produced by the superposition of wavelets from different parts of a wavefront. |
| It is commonly demonstrated using two slits. | It can be demonstrated using a single narrow slit. |
| For equal-amplitude waves, bright fringes have equal maximum intensity. | The central maximum is the brightest and widest. |
| Bright and dark fringes are formed. | A broad central maximum and weaker secondary maxima are formed. |
Q10. What is polarisation? Explain how it proves the transverse nature of light.
Answer:
Polarisation is the phenomenon in which the vibrations of the electric field of light are restricted to a particular direction perpendicular to the direction of propagation.
Light waves are transverse electromagnetic waves. Their electric and magnetic fields vibrate perpendicular to the direction of propagation.
Only transverse waves can be polarised. Therefore, the phenomenon of polarisation proves the transverse nature of light.
Q11. State Malus’ law and write its mathematical expression.
Answer:
Malus’ law states that when plane-polarised light passes through an analyser, the intensity of transmitted light is directly proportional to the square of the cosine of the angle between the transmission axes of the polariser and analyser.
Mathematical expression:
I = I₀ cos² θ
Where:
I = Transmitted intensity
I₀ = Initial plane-polarised intensity
θ = Angle between the transmission axes
Q12. What happens when the axes of a polariser and analyser are parallel?
Answer:
When the transmission axes are parallel:
θ = 0°
Using Malus’ law:
I = I₀ cos² 0°
I = I₀
Therefore, the transmitted intensity is maximum.
Q13. What happens when the axes of a polariser and analyser are perpendicular?
Answer:
When the transmission axes are perpendicular:
θ = 90°
Using Malus’ law:
I = I₀ cos² 90°
I = 0
Therefore, no light is transmitted through an ideal analyser.
4. IMPORTANT NUMERICAL QUESTIONS
Q1. The wavelength of light in air is 600 nm. Calculate its wavelength in a medium of refractive index 1.5.
Solution:
Given:
λ = 600 nm
n = 1.5
Formula:
λ' = λ/n
λ' = 600/1.5
λ' = 400 nm
Answer: Wavelength in the medium = 400 nm.
Q2. The speed of light in a medium is 2 × 10⁸ m/s. Calculate its refractive index.
Solution:
Given:
c = 3 × 10⁸ m/s
v = 2 × 10⁸ m/s
Formula:
n = c/v
n = (3 × 10⁸)/(2 × 10⁸)
n = 1.5
Answer: Refractive index = 1.5.
Q3. In Young's double-slit experiment, the wavelength of light is 500 nm, the distance between the slits is 0.5 mm and the screen is 2 m away. Calculate the fringe width.
Solution:
Given:
λ = 500 × 10⁻⁹ m
d = 0.5 × 10⁻³ m
D = 2 m
Formula:
β = λD/d
β = (500 × 10⁻⁹ × 2)/(0.5 × 10⁻³)
β = 2 × 10⁻³ m
β = 2 mm
Answer: Fringe width = 2 mm.
Q4. In Young's double-slit experiment, the fringe width is 2 mm. If the distance between the slits and screen is 1 m and the wavelength is 500 nm, calculate the separation between the slits.
Solution:
Given:
β = 2 × 10⁻³ m
D = 1 m
λ = 500 × 10⁻⁹ m
Formula:
β = λD/d
Therefore:
d = λD/β
d = (500 × 10⁻⁹ × 1)/(2 × 10⁻³)
d = 2.5 × 10⁻⁴ m
d = 0.25 mm
Answer: Separation between the slits = 0.25 mm.
Q5. The angle between the transmission axes of a polariser and analyser is 60°. Calculate the transmitted intensity if the incident plane-polarised light has intensity 100 W/m².
Solution:
Given:
I₀ = 100 W/m²
θ = 60°
Formula:
I = I₀ cos² θ
I = 100 × cos² 60°
I = 100 × (1/2)²
I = 100 × 1/4
I = 25 W/m²
Answer: Transmitted intensity = 25 W/m².
5. IMPORTANT FORMULAS
Speed of light in a medium:
v = c/n
Refractive index:
n = c/v
Wave relation:
v = νλ
Wavelength in a medium:
λ' = λ/n
Frequency:
ν = c/λ
Snell's law:
n₁ sin i = n₂ sin r
Phase difference:
φ = 2πΔx/λ
Condition for constructive interference:
Δx = nλ
Condition for destructive interference:
Δx = (2n + 1)λ/2
Intensity due to two coherent waves of equal individual intensity I₀:
I = 4I₀ cos²(φ/2)
Position of nth bright fringe:
xₙ = nλD/d
Fringe width:
β = λD/d
Condition for single-slit diffraction minima:
a sin θ = nλ
Angular width of central maximum:
2θ ≈ 2λ/a
Linear width of central maximum:
W = 2λD/a
Malus’ law:
I = I₀ cos² θ
Intensity after an ideal Polaroid for unpolarised incident light:
I = I₀/2
6. IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Wavelength | Metre (m) |
| Frequency | Hertz (Hz) |
| Speed of light | Metre per second (m/s) |
| Distance | Metre (m) |
| Slit separation | Metre (m) |
| Fringe width | Metre (m) |
| Intensity | Watt per square metre (W/m²) |
| Phase difference | Radian (rad) |
| Angle | Radian (rad) |
| Refractive index | No unit |
7. QUICK REVISION POINTS
Wave optics explains the wave nature of light.
A wavefront is a surface joining points having the same phase.
Rays are perpendicular to wavefronts.
Huygens’ principle explains the propagation of wavefronts.
Every point on a wavefront acts as a source of secondary wavelets.
The forward envelope of secondary wavelets gives the new wavefront.
The angle of incidence is equal to the angle of reflection.
Snell's law is n₁ sin i = n₂ sin r.
The frequency of light remains unchanged when it enters another medium.
The speed and wavelength of light change when it enters another medium.
Interference is caused by the superposition of coherent waves.
Constructive interference produces bright fringes.
Destructive interference produces dark fringes.
Young's double-slit experiment demonstrates the wave nature of light.
Fringe width is β = λD/d.
Fringe width increases with wavelength and screen distance.
Fringe width decreases when slit separation increases.
Diffraction is the bending or spreading of light around obstacles or through narrow openings.
The central maximum in single-slit diffraction is the brightest and widest.
Polarisation demonstrates the transverse nature of light.
Malus’ law is I = I₀ cos² θ.
For crossed ideal Polaroids, transmitted intensity is zero.
Comments
Post a Comment