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Wave Optics NCERT Solutions and Important Questions
You will learn how light bends around corners, overlaps to create beautiful patterns, and behaves like a wave. This topic holds incredible weight in your CBSE 2026 board exam and paves the way for advanced optics in competitive exams like JEE and NEET. Let's make this seemingly complex chapter super simple together!
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Chapter NameWave Optics (Chapter 10)
SubjectPhysics
ClassClass 12
BoardCBSE (2026-27 Syllabus)
Important TopicsHuygens' Principle, YDSE, Diffraction
Difficulty LevelModerate to High
Exam Weightage~8-9 Marks (Combined with Ray Optics)
Learning Objectives
After completing this chapter, students will be able to:
- Explain Huygens' Principle and use it to prove the laws of reflection and refraction.
- Distinguish between coherent and incoherent addition of light waves.
- Understand the conditions required for constructive and destructive interference.
- Calculate fringe width and analyze patterns in Young's Double Slit Experiment (YDSE).
- Explain the phenomenon of diffraction through a single slit and find the width of the central maximum.
Key Concepts & Formulas
- Wavefront: The locus of all particles of a medium vibrating in the same phase at any given instant. Types include spherical, cylindrical, and plane wavefronts.
- Huygens' Principle: Each point on a primary wavefront acts as a source of secondary spherical disturbances called wavelets. The forward envelope of these wavelets gives the new wavefront position.
- Coherent Sources: Sources of light that emit waves with zero or a constant phase difference over time. Necessary for sustained interference.
- Interference: The redistribution of light energy due to the superposition of light waves coming from two coherent sources.
- Path Difference Conditions:
Bright Fringes (Maximal Intensity): $\Delta x = n\lambda$ (where $n=0, 1, 2 \dots$)
Dark Fringes (Minimal Intensity): $\Delta x = (2n-1)\frac{\lambda}{2}$ (where $n=1, 2, \dots$) - Fringe Width ($\beta$): The distance between two consecutive bright or dark fringes.
$$\beta = \frac{\lambda D}{d}$$
(Where $\lambda$ is wavelength, $D$ is distance to the screen, and $d$ is distance between slits). - Diffraction Width of Central Maximum:
$$W = \frac{2\lambda D}{a}$$
(Where $a$ is the width of the slit).
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Full NCERT Solutions
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.
Answer:
Step 1: Identify given values.
Wavelength in air, $\lambda_a = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
Refractive index of water, $n = 1.33 = \frac{4}{3}$
Speed of light in vacuum, $c = 3 \times 10^8 \text{ m/s}$
Step 2: Find the absolute frequency ($f$).
Frequency stays constant across media:
$$f = \frac{c}{\lambda_a} = \frac{3 \times 10^8}{589 \times 10^{-9}} = 5.09 \times 10^{14} \text{ Hz}$$
Step 3: Analyze Reflected Light (a).
Since the light stays in the same medium (air):
* Speed: $v = c = 3 \times 10^8 \text{ m/s}$
* Wavelength: $\lambda = \lambda_a = 589 \text{ nm}$
* Frequency: $f = 5.09 \times 10^{14} \text{ Hz}$
Step 4: Analyze Refracted Light (b).
When light shifts to a denser medium (water), its speed and wavelength change, but its frequency remains identical.
* Frequency: $f = 5.09 \times 10^{14} \text{ Hz}$
* Speed: $v = \frac{c}{n} = \frac{3 \times 10^8}{1.33} = 2.25 \times 10^8 \text{ m/s}$
* Wavelength: $\lambda_w = \frac{\lambda_a}{n} = \frac{589 \text{ nm}}{1.33} \approx 442.8 \text{ nm}$
Step 1: Identify given values.
Wavelength in air, $\lambda_a = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
Refractive index of water, $n = 1.33 = \frac{4}{3}$
Speed of light in vacuum, $c = 3 \times 10^8 \text{ m/s}$
Step 2: Find the absolute frequency ($f$).
Frequency stays constant across media:
$$f = \frac{c}{\lambda_a} = \frac{3 \times 10^8}{589 \times 10^{-9}} = 5.09 \times 10^{14} \text{ Hz}$$
Step 3: Analyze Reflected Light (a).
Since the light stays in the same medium (air):
* Speed: $v = c = 3 \times 10^8 \text{ m/s}$
* Wavelength: $\lambda = \lambda_a = 589 \text{ nm}$
* Frequency: $f = 5.09 \times 10^{14} \text{ Hz}$
Step 4: Analyze Refracted Light (b).
When light shifts to a denser medium (water), its speed and wavelength change, but its frequency remains identical.
* Frequency: $f = 5.09 \times 10^{14} \text{ Hz}$
* Speed: $v = \frac{c}{n} = \frac{3 \times 10^8}{1.33} = 2.25 \times 10^8 \text{ m/s}$
* Wavelength: $\lambda_w = \frac{\lambda_a}{n} = \frac{589 \text{ nm}}{1.33} \approx 442.8 \text{ nm}$
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:
Step 1: Case (a) - Point Source.
Shape: Spherical Wavefront. Light spreads out equally in all directions from a single localized point source.
Step 2: Case (b) - Convex Lens.
Shape: Plane Wavefront. Diverging rays from the focus turn into a perfectly parallel beam after passing through the convex lens.
Step 3: Case (c) - Distant Star.
Shape: Plane Wavefront. Over vast astronomical distances, a tiny section of an incredibly massive spherical wavefront behaves essentially as a flat plane.
Step 1: Case (a) - Point Source.
Shape: Spherical Wavefront. Light spreads out equally in all directions from a single localized point source.
Step 2: Case (b) - Convex Lens.
Shape: Plane Wavefront. Diverging rays from the focus turn into a perfectly parallel beam after passing through the convex lens.
Step 3: Case (c) - Distant Star.
Shape: Plane Wavefront. Over vast astronomical distances, a tiny section of an incredibly massive spherical wavefront behaves essentially as a flat plane.
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?
Answer:
Step 1: Calculate speed in glass (a).
Given $n = 1.5$ and $c = 3 \times 10^8 \text{ m/s}$.
$$v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}$$
Step 2: Determine color dependency (b).
No, the speed depends on color because the refractive index of a material varies with wavelength ($\lambda$).
Step 3: Compare Red and Violet.
According to Cauchy's formula, the index is higher for shorter wavelengths ($n_v > n_r$). Consequently, violet light travels slower than red light inside a glass prism.
Step 1: Calculate speed in glass (a).
Given $n = 1.5$ and $c = 3 \times 10^8 \text{ m/s}$.
$$v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}$$
Step 2: Determine color dependency (b).
No, the speed depends on color because the refractive index of a material varies with wavelength ($\lambda$).
Step 3: Compare Red and Violet.
According to Cauchy's formula, the index is higher for shorter wavelengths ($n_v > n_r$). Consequently, violet light travels slower than red light inside 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 measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Answer:
Step 1: Identify Given Values.
Distance between slits, $d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}$
Distance to screen, $D = 1.4 \text{ m}$
Order of fringe, $n = 4$
Position of 4th bright fringe, $x_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$
Step 2: Apply the Position Formula.
The position formula for a bright fringe is:
$$x_n = \frac{n\lambda D}{d}$$
Step 3: Substitute and solve for $\lambda$.
$$1.2 \times 10^{-2} = \frac{4 \times \lambda \times 1.4}{0.28 \times 10^{-3}}$$
$$1.2 \times 10^{-2} = \frac{5.6 \times \lambda}{0.28 \times 10^{-3}}$$
$$\lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{5.6}$$
$$\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}$$
The wavelength of the light used is 600 nm.
Step 1: Identify Given Values.
Distance between slits, $d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}$
Distance to screen, $D = 1.4 \text{ m}$
Order of fringe, $n = 4$
Position of 4th bright fringe, $x_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$
Step 2: Apply the Position Formula.
The position formula for a bright fringe is:
$$x_n = \frac{n\lambda D}{d}$$
Step 3: Substitute and solve for $\lambda$.
$$1.2 \times 10^{-2} = \frac{4 \times \lambda \times 1.4}{0.28 \times 10^{-3}}$$
$$1.2 \times 10^{-2} = \frac{5.6 \times \lambda}{0.28 \times 10^{-3}}$$
$$\lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{5.6}$$
$$\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}$$
The wavelength of the light used is 600 nm.
Question 10.5: In Young's double-slit experiment using mono-chromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$ is K units. What is the intensity of light at a point where path difference is $\lambda/3$?
Answer:
Step 1: State the General Formula.
The general formula for net intensity from two identical slits is:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
Where phase difference $\phi = \frac{2\pi}{\lambda} \times \text{Path Difference}$.
Step 2: Evaluate Case 1 (Path Difference = $\lambda$).
$$\phi_1 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$$
$$I_1 = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0(-1)^2 = 4I_0$$
We are given $I_1 = K$, so $K = 4I_0$.
Step 3: Evaluate Case 2 (Path Difference = $\lambda/3$).
$$\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \ (120^\circ)$$
$$I_2 = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$
Since $\cos(60^\circ) = \frac{1}{2}$:
$$I_2 = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \times \frac{1}{4} = I_0$$
Step 4: Express final intensity in terms of K.
Since $4I_0 = K$, we have $I_0 = \frac{K}{4}$. Thus, the new intensity is $\frac{K}{4}$.
Step 1: State the General Formula.
The general formula for net intensity from two identical slits is:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
Where phase difference $\phi = \frac{2\pi}{\lambda} \times \text{Path Difference}$.
Step 2: Evaluate Case 1 (Path Difference = $\lambda$).
$$\phi_1 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$$
$$I_1 = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0(-1)^2 = 4I_0$$
We are given $I_1 = K$, so $K = 4I_0$.
Step 3: Evaluate Case 2 (Path Difference = $\lambda/3$).
$$\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \ (120^\circ)$$
$$I_2 = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$
Since $\cos(60^\circ) = \frac{1}{2}$:
$$I_2 = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \times \frac{1}{4} = I_0$$
Step 4: Express final intensity in terms of K.
Since $4I_0 = K$, we have $I_0 = \frac{K}{4}$. Thus, the new intensity is $\frac{K}{4}$.
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 on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? (Take $d = 2 \text{ mm}$ and $D = 1.2 \text{ m}$)
Answer:
Step 1: Identify Given Values.
$\lambda_1 = 650 \times 10^{-9} \text{ m}$, $\lambda_2 = 520 \times 10^{-9} \text{ m}$
$d = 2 \times 10^{-3} \text{ m}$, $D = 1.2 \text{ m}$.
Step 2: Calculate part (a).
Distance for $n = 3$ using $\lambda_1$:
$$x = \frac{n\lambda_1 D}{d} = \frac{3 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.17 \times 10^{-3} \text{ m} = 1.17 \text{ mm}$$
Step 3: Set up condition for coinciding fringes (b).
Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$:
$$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2$$
$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5}$$
The minimum integral values are $n_1 = 4$ and $n_2 = 5$.
Step 4: Calculate the actual coinciding distance.
Calculate using $n_1 = 4$ and $\lambda_1$:
$$x_{min} = \frac{4 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.56 \times 10^{-3} \text{ m} = 1.56 \text{ mm}$$
Step 1: Identify Given Values.
$\lambda_1 = 650 \times 10^{-9} \text{ m}$, $\lambda_2 = 520 \times 10^{-9} \text{ m}$
$d = 2 \times 10^{-3} \text{ m}$, $D = 1.2 \text{ m}$.
Step 2: Calculate part (a).
Distance for $n = 3$ using $\lambda_1$:
$$x = \frac{n\lambda_1 D}{d} = \frac{3 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.17 \times 10^{-3} \text{ m} = 1.17 \text{ mm}$$
Step 3: Set up condition for coinciding fringes (b).
Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$:
$$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2$$
$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5}$$
The minimum integral values are $n_1 = 4$ and $n_2 = 5$.
Step 4: Calculate the actual coinciding distance.
Calculate using $n_1 = 4$ and $\lambda_1$:
$$x_{min} = \frac{4 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.56 \times 10^{-3} \text{ m} = 1.56 \text{ mm}$$
Question 10.7: In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Answer:
Step 1: State the angular width formula.
The formula for angular fringe width is $\theta = \frac{\lambda}{d}$.
Step 2: Determine wavelength in water.
When the apparatus is immersed in water, the wavelength changes to $\lambda' = \frac{\lambda}{n}$, while the slit distance $d$ stays constant.
Step 3: Calculate new angular width.
$$\theta' = \frac{\lambda'}{d} = \frac{\lambda}{n \cdot d} = \frac{\theta}{n}$$
Given $\theta = 0.2^\circ$ and $n = 4/3$:
$$\theta' = \frac{0.2^\circ}{4/3} = \frac{0.2 \times 3}{4} = 0.15^\circ$$
The new angular width will be $0.15^\circ$.
Step 1: State the angular width formula.
The formula for angular fringe width is $\theta = \frac{\lambda}{d}$.
Step 2: Determine wavelength in water.
When the apparatus is immersed in water, the wavelength changes to $\lambda' = \frac{\lambda}{n}$, while the slit distance $d$ stays constant.
Step 3: Calculate new angular width.
$$\theta' = \frac{\lambda'}{d} = \frac{\lambda}{n \cdot d} = \frac{\theta}{n}$$
Given $\theta = 0.2^\circ$ and $n = 4/3$:
$$\theta' = \frac{0.2^\circ}{4/3} = \frac{0.2 \times 3}{4} = 0.15^\circ$$
The new angular width will be $0.15^\circ$.
Question 10.8: What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5)
Answer:
Step 1: Apply Brewster's Law.
According to Brewster's Law:
$$\tan(i_p) = n$$
Where $i_p$ is the Brewster polarizing angle and $n = 1.5$.
Step 2: Solve for the angle.
$$i_p = \tan^{-1}(1.5) \approx 56.3^\circ$$
The Brewster angle for this glass transition is $56.3^\circ$.
Step 1: Apply Brewster's Law.
According to Brewster's Law:
$$\tan(i_p) = n$$
Where $i_p$ is the Brewster polarizing angle and $n = 1.5$.
Step 2: Solve for the angle.
$$i_p = \tan^{-1}(1.5) \approx 56.3^\circ$$
The Brewster angle for this glass transition is $56.3^\circ$.
Extra Important Questions (Board Style 2026)
Multiple Choice Questions (MCQs)
1. Which of the following phenomena proves that light is a transverse wave?
A) Interference
B) Diffraction
C) Polarization
D) Reflection
Answer: C
Explanation: Only transverse waves can be polarized. Longitudinal waves, like sound, cannot exhibit polarization. (Difficulty Level: Easy)
Explanation: Only transverse waves can be polarized. Longitudinal waves, like sound, cannot exhibit polarization. (Difficulty Level: Easy)
2. If the distance between the source slits and the screen ($D$) is doubled in YDSE, the fringe width ($\beta$) becomes:
A) Halved
B) Doubled
C) Four times
D) Remains unchanged
Answer: B
Explanation: Since $\beta = \frac{\lambda D}{d}$, fringe width is directly proportional to $D$. (Difficulty Level: Easy)
Explanation: Since $\beta = \frac{\lambda D}{d}$, fringe width is directly proportional to $D$. (Difficulty Level: Easy)
3. Two coherent light sources have an intensity ratio of 9:1. The ratio of maximum to minimum intensity in their interference pattern will be:
A) 9:1
B) 3:1
C) 4:1
D) 16:1
Answer: C
Explanation:
Step 1: Apply formula.
$$\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$$
Step 2: Substitute ratio components.
$$\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{9} + \sqrt{1}}{\sqrt{9} - \sqrt{1}}\right)^2 = \left(\frac{3 + 1}{3 - 1}\right)^2 = \left(\frac{4}{2}\right)^2 = 4:1$$
Explanation:
Step 1: Apply formula.
$$\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$$
Step 2: Substitute ratio components.
$$\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{9} + \sqrt{1}}{\sqrt{9} - \sqrt{1}}\right)^2 = \left(\frac{3 + 1}{3 - 1}\right)^2 = \left(\frac{4}{2}\right)^2 = 4:1$$
Short Answer Questions (2-3 Marks)
4. Why can two independent, identical light bulbs never act as coherent sources?
Answer:
Step 1: Understand Light Emission. Light is emitted by individual excited atoms dropping to lower energy states in brief, independent bursts lasting roughly $10^{-8}$ seconds.
Step 2: Recognize Phase Randomness. Because these transitions occur randomly, the phase relationship between two separate bulbs changes millions of times per second. This prevents them from maintaining a constant phase difference over time.
Step 1: Understand Light Emission. Light is emitted by individual excited atoms dropping to lower energy states in brief, independent bursts lasting roughly $10^{-8}$ seconds.
Step 2: Recognize Phase Randomness. Because these transitions occur randomly, the phase relationship between two separate bulbs changes millions of times per second. This prevents them from maintaining a constant phase difference over time.
5. Distinguish between interference and diffraction.
Answer:
* Interference: Results from the superposition of wavelets originating from two distinct coherent wavefront sources (like the two slits in YDSE).
* Diffraction: Results from the superposition of secondary wavelets originating from different parts of the same single wavefront.
* Interference: Results from the superposition of wavelets originating from two distinct coherent wavefront sources (like the two slits in YDSE).
* Diffraction: Results from the superposition of secondary wavelets originating from different parts of the same single wavefront.
6. State Huygens' Principle.
Answer:
Step 1: Source of Secondary Wavelets. Every point on a given primary wavefront behaves as a source for tiny, secondary spherical wavelets that spread out in all directions at the speed of light.
Step 2: The New Wavefront. A surface touching these secondary wavelets tangentially in the forward direction outlines the updated position of the wavefront.
Step 1: Source of Secondary Wavelets. Every point on a given primary wavefront behaves as a source for tiny, secondary spherical wavelets that spread out in all directions at the speed of light.
Step 2: The New Wavefront. A surface touching these secondary wavelets tangentially in the forward direction outlines the updated position of the wavefront.
7. What happens to the interference pattern in YDSE if the monochromatic light source is replaced by a white light source?
Answer:
Step 1: Central Fringe Behavior. The central fringe will appear completely white because all constituent wavelengths arrive in phase at the center.
Step 2: Surrounding Fringes. The surrounding fringes will appear colored, with violet appearing closest to the center and red appearing further out, because fringe width depends on wavelength ($\beta \propto \lambda$).
Step 1: Central Fringe Behavior. The central fringe will appear completely white because all constituent wavelengths arrive in phase at the center.
Step 2: Surrounding Fringes. The surrounding fringes will appear colored, with violet appearing closest to the center and red appearing further out, because fringe width depends on wavelength ($\beta \propto \lambda$).
Long Answer Questions (5 Marks)
8. Using Huygens' Principle, derive the law of reflection ($\angle i = \angle r$) when a plane wavefront hits a reflecting surface.
Answer:
Step 1: Initial Setup. Consider a flat wavefront $AB$ striking a reflecting surface $MN$ at an angle $i$. Let $v$ represent the speed of the wave.
Step 2: Travel Time Calculation. The time it takes for point $B$ to travel to position $C$ on the surface is $t = \frac{BC}{v} \implies BC = vt$.
Step 3: Secondary Wavelet Formation. According to Huygens' principle, point $A$ acts as a secondary source and forms a wavelet of radius $vt$ over this same timeframe $t$.
Step 4: Draw New Wavefront. Draw a tangent line from $C$ to this new sphere at point $D$. This forms the reflected wavefront $CD$, where $AD = vt$.
Step 5: Geometric Proof. Now compare triangles $\triangle ABC$ and $\triangle ADC$:
1. $AC = AC$ (Common side)
2. $BC = AD = vt$
3. $\angle ABC = \angle ADC = 90^\circ$
By the RHS congruence criterion, $\triangle ABC \cong \triangle ADC$. Therefore, their corresponding angles are equal: $\angle BAC = \angle DCA \implies \angle i = \angle r$.
Step 1: Initial Setup. Consider a flat wavefront $AB$ striking a reflecting surface $MN$ at an angle $i$. Let $v$ represent the speed of the wave.
Step 2: Travel Time Calculation. The time it takes for point $B$ to travel to position $C$ on the surface is $t = \frac{BC}{v} \implies BC = vt$.
Step 3: Secondary Wavelet Formation. According to Huygens' principle, point $A$ acts as a secondary source and forms a wavelet of radius $vt$ over this same timeframe $t$.
Step 4: Draw New Wavefront. Draw a tangent line from $C$ to this new sphere at point $D$. This forms the reflected wavefront $CD$, where $AD = vt$.
Step 5: Geometric Proof. Now compare triangles $\triangle ABC$ and $\triangle ADC$:
1. $AC = AC$ (Common side)
2. $BC = AD = vt$
3. $\angle ABC = \angle ADC = 90^\circ$
By the RHS congruence criterion, $\triangle ABC \cong \triangle ADC$. Therefore, their corresponding angles are equal: $\angle BAC = \angle DCA \implies \angle i = \angle r$.
9. Derive an expression for fringe width ($\beta$) in Young's Double Slit Experiment.
Answer:
Step 1: Setup Geometry. Let $S_1$ and $S_2$ be two coherent slits separated by a distance $d$. Let $D$ be the distance from the slits to a screen. Consider a point $P$ on the screen at a distance $x$ from the central origin $O$.
Step 2: Define Path Difference. The path difference between the waves reaching $P$ is $\Delta x = S_2P - S_1P$.
Step 3: Apply Pythagoras. Using the geometry of the setup:
$$(S_2P)^2 = D^2 + \left(x + \frac{d}{2}\right)^2$$
$$(S_1P)^2 = D^2 + \left(x - \frac{d}{2}\right)^2$$
Step 4: Subtract Equations.
$$(S_2P)^2 - (S_1P)^2 = 2xd$$
$$(S_2P - S_1P)(S_2P + S_1P) = 2xd$$
Step 5: Apply Approximation. Since $d \ll D$, we can approximate $S_2P + S_1P \approx 2D$:
$$\Delta x \cdot (2D) = 2xd \implies \Delta x = \frac{xd}{D}$$
Step 6: Condition for Bright Fringe. For a Bright Fringe: Path difference $\Delta x = n\lambda \implies \frac{x_n d}{D} = n\lambda \implies x_n = \frac{n\lambda D}{d}$.
Step 7: Final Fringe Width. Fringe Width ($\beta$) is the distance between two adjacent bright fringes:
$$\beta = x_{n+1} - x_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$
Step 1: Setup Geometry. Let $S_1$ and $S_2$ be two coherent slits separated by a distance $d$. Let $D$ be the distance from the slits to a screen. Consider a point $P$ on the screen at a distance $x$ from the central origin $O$.
Step 2: Define Path Difference. The path difference between the waves reaching $P$ is $\Delta x = S_2P - S_1P$.
Step 3: Apply Pythagoras. Using the geometry of the setup:
$$(S_2P)^2 = D^2 + \left(x + \frac{d}{2}\right)^2$$
$$(S_1P)^2 = D^2 + \left(x - \frac{d}{2}\right)^2$$
Step 4: Subtract Equations.
$$(S_2P)^2 - (S_1P)^2 = 2xd$$
$$(S_2P - S_1P)(S_2P + S_1P) = 2xd$$
Step 5: Apply Approximation. Since $d \ll D$, we can approximate $S_2P + S_1P \approx 2D$:
$$\Delta x \cdot (2D) = 2xd \implies \Delta x = \frac{xd}{D}$$
Step 6: Condition for Bright Fringe. For a Bright Fringe: Path difference $\Delta x = n\lambda \implies \frac{x_n d}{D} = n\lambda \implies x_n = \frac{n\lambda D}{d}$.
Step 7: Final Fringe Width. Fringe Width ($\beta$) is the distance between two adjacent bright fringes:
$$\beta = x_{n+1} - x_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$
10. Describe diffraction at a single slit and find the conditions for minima and maxima.
Answer:
Step 1: Setup. A parallel beam of light passes through a narrow slit of width $a$.
Step 2: Central Maximum. Secondary wavelets traveling straight forward focus at the center $O$ to form a bright central maximum.
Step 3: General Path Difference. Wavelets traveling at an angle $\theta$ have a path difference of $a \sin\theta$ between the edges of the slit.
Step 4: Minima Condition. Occur when $a \sin\theta = n\lambda$. Dividing the slit into even parts causes the wavelets to cancel each other out in pairs.
Step 5: Secondary Maxima Condition. Occur when $a \sin\theta = (2n+1)\frac{\lambda}{2}$. Dividing the slit into odd parts leaves an unpaired section that creates a weaker bright fringe.
Step 1: Setup. A parallel beam of light passes through a narrow slit of width $a$.
Step 2: Central Maximum. Secondary wavelets traveling straight forward focus at the center $O$ to form a bright central maximum.
Step 3: General Path Difference. Wavelets traveling at an angle $\theta$ have a path difference of $a \sin\theta$ between the edges of the slit.
Step 4: Minima Condition. Occur when $a \sin\theta = n\lambda$. Dividing the slit into even parts causes the wavelets to cancel each other out in pairs.
Step 5: Secondary Maxima Condition. Occur when $a \sin\theta = (2n+1)\frac{\lambda}{2}$. Dividing the slit into odd parts leaves an unpaired section that creates a weaker bright fringe.
Case-Based Questions
11-13. Read the passage and answer questions:
Young's Double Slit Experiment fundamentally altered our understanding of light by providing clear evidence of its wave nature. By passing light through two closely spaced micro-slits, alternating bands of bright and dark regions appear on a distant screen. This outcome occurs because the waves periodically reinforce or cancel each other out across the screen.
11. What happens to the fringe width if the entire YDSE setup is submerged in a liquid of refractive index 1.5?
Answer: The fringe width decreases. Since $\lambda' = \frac{\lambda}{1.5}$, the width shrinks to $\beta' = \frac{\beta}{1.5}$.
12. If one of the two slits is covered completely, what pattern will you see on the screen?
Answer: The regular interference pattern will disappear and be replaced by a wider single-slit diffraction pattern.
13. Why must the distance between the two slits ($d$) be kept extremely small?
Answer: Because fringe width is inversely proportional to slit separation ($\beta \propto \frac{1}{d}$). If $d$ is too large, the fringes will blur together and become impossible to resolve.
Answer: The fringe width decreases. Since $\lambda' = \frac{\lambda}{1.5}$, the width shrinks to $\beta' = \frac{\beta}{1.5}$.
12. If one of the two slits is covered completely, what pattern will you see on the screen?
Answer: The regular interference pattern will disappear and be replaced by a wider single-slit diffraction pattern.
13. Why must the distance between the two slits ($d$) be kept extremely small?
Answer: Because fringe width is inversely proportional to slit separation ($\beta \propto \frac{1}{d}$). If $d$ is too large, the fringes will blur together and become impossible to resolve.
Assertion-Reason Questions
(Options: A) Both A & R are true, R is correct explanation. B) Both are true, R is not correct explanation. C) A is true, R is false. D) A is false, R is true.)
14. Assertion (A): No interference pattern is detected when two separate sodium lamps illuminate a pair of adjacent slits.
Reason (R): Phase variations between independent light sources occur rapidly and randomly.
Answer: A
Explanation: The lamps are independent sources, meaning they are incoherent. Rapid phase shifts blur the interference pattern into a uniform background before it can be resolved.
Explanation: The lamps are independent sources, meaning they are incoherent. Rapid phase shifts blur the interference pattern into a uniform background before it can be resolved.
15. Assertion (A): A single-slit diffraction pattern features a central bright maximum that is twice as wide as the secondary maxima.
Reason (R): The first minimum on either side occurs at an angular position defined by $\theta = \pm\frac{\lambda}{a}$.
Answer: A
Explanation: The central maximum spans from $-\frac{\lambda}{a}$ to $+\frac{\lambda}{a}$, giving it an angular width of $\frac{2\lambda}{a}$. This is exactly twice the width ($\frac{\lambda}{a}$) of any subsequent secondary maximum.
Explanation: The central maximum spans from $-\frac{\lambda}{a}$ to $+\frac{\lambda}{a}$, giving it an angular width of $\frac{2\lambda}{a}$. This is exactly twice the width ($\frac{\lambda}{a}$) of any subsequent secondary maximum.
Common Mistakes Students Make
- Mixing up Intersecting Conditions: Students often swap the conditions for bright and dark spots between interference and diffraction. Remember that in YDSE, $a\sin\theta = n\lambda$ gives a bright fringe, whereas in single-slit diffraction, it defines a minimum (dark) spot.
- Neglecting Phase and Path Differences: Remember that $\Delta x = n\lambda$ represents a path difference, not a phase angle! Use $\phi = \frac{2\pi}{\lambda}\Delta x$ to convert between the two.
- Messy Wavefront Sketches: When asked to draw wavefront refraction, ensure your incoming and outgoing rays stay perpendicular to their respective wavefront lines.
Exam Preparation Tips
- Master the Derivations: Focus your revision on proving the laws of reflection using Huygens' principle, along with the YDSE fringe width derivation. These are highly likely to appear as 5-mark questions.
- Keep Calculations Neat: YDSE calculations involve tiny values like nanometers ($10^{-9}$) and millimeters ($10^{-3}$). Group your coefficients and powers of 10 separately to avoid making simple arithmetic errors.
- Sketch Diagrams with a Ruler: Ensure your ray paths and wavefront positions are drawn cleanly with a pencil and straightedge. Clear labeling can earn you partial marks even if you make a calculation error later on.
FAQ Section
Is Wave Optics very difficult compared to Ray Optics?
It feels more abstract at first because you have to think of light as waves rather than straight rays. However, it requires far less geometry to master, and the numerical questions follow predictable formulas.
What is the most important derivation in CBSE Class 12 Wave Optics?
Proving the laws of reflection/refraction using Huygens' Principle and deriving the YDSE fringe width formula are the two most crucial derivations for boards.
What are coherent sources, and why are they necessary?
Coherent sources maintain a constant phase difference over time. Without them, the interference pattern shifts too rapidly for our eyes or screens to resolve, resulting in a flat, uniform blur.
How does fringe width change if we use blue light instead of red light?
Since blue light has a shorter wavelength than red light ($\lambda_{blue} < \lambda_{red}$), the fringe width will shrink ($\beta \propto \lambda$) and the pattern will appear more tightly packed.
Can we observe diffraction with sound waves easily?
Yes! Sound waves have long wavelengths (ranging from centimeters to meters), allowing them to bend easily around everyday obstacles like doors and walls. Light waves have tiny wavelengths, so they only exhibit noticeable diffraction when passing through microscopic slits.
Conclusion: Mastering Wave Optics comes down to understanding two core ideas: how wavefronts travel and how overlapping waves combine. Practice the core derivations by hand, keep your unit conversions accurate during numerical problems, and review past exam questions regularly. Save this guide, keep studying consistently, and you will tackle your 2026 board exams with complete confidence!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics