| CBSE/JEE Main Focus on Diffraction |
|---|
For JEE Main and CBSE, a strong qualitative understanding of diffraction is essential. You should know:
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| CBSE/JEE Main Focus on Polarization |
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For JEE Main and CBSE, a thorough conceptual understanding of polarization is critical. You should be familiar with:
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| Feature | Fraunhofer Diffraction | Fresnel Diffraction |
|---|---|---|
| Source & Screen Distance | Effectively at infinite distance from the obstacle/aperture. Achieved using lenses to make light parallel. | At finite distances from the obstacle/aperture. |
| Wavefront Shape | Plane wavefronts incident on and emerging from the aperture. | Spherical or cylindrical wavefronts incident on and emerging from the aperture. |
| Pattern | Observed on a screen placed at the focal plane of a converging lens. Pattern consists of equally spaced bright and dark fringes (maxima and minima). | Pattern has complicated features; fringes are not equally spaced. |
| Mathematical Complexity | Simpler to analyze mathematically. | More complex mathematically. |
| JEE Focus | Highly important, especially single-slit diffraction. | Qualitative understanding only. |
| Feature | CBSE Board Exam | JEE Main & Advanced |
|---|---|---|
| Diffraction Concepts | Qualitative understanding of single-slit diffraction. Conditions for maxima/minima. Angular width of central maximum. Simple definitions and diagrams. | In-depth understanding of single-slit diffraction. Quantitative problems on slit width, angular width, positions of minima/maxima. Comparison with Young's double slit interference. Resolving power of optical instruments (based on circular aperture diffraction). |
| Polarization Concepts | Qualitative understanding of unpolarized/polarized light. Methods of polarization (reflection, scattering, absorption). Malus's Law and Brewster's Law statements. Basic definitions. | Detailed understanding of Malus's Law and Brewster's Law, including numerical applications. Identification of polarizing angle. Concepts of polarizers and analyzers. More complex problems involving multiple polaroids or systems. |
| Mathematical Depth | Less emphasis on derivations, more on direct formulas and conceptual reasoning. | Derivations and quantitative problem-solving are crucial. Understanding the underlying physics and applying it to varied scenarios. |
| Problem Solving | Direct formula application. Conceptual questions. | Challenging numerical problems, multi-concept problems, graphical analysis of intensity. |
Mastering concepts like Diffraction and Polarization for JEE and Board exams often involves remembering specific conditions, laws, and characteristic features. Here are some mnemonics and short-cuts to help you recall the key qualitative ideas efficiently.
| Feature | Interference (Double Slit) | Diffraction (Single Slit) |
|---|---|---|
| Fringe Widths | All fringes are of Equal Width. | Central maximum is Twice as wide as others; secondary maxima are narrower. |
| Intensity | All bright fringes have Equal Intensity. | Intensity decreases rapidly for successive maxima away from the center. |
Remember, these mnemonics are tools to aid recall. Always ensure you understand the underlying physical concepts for a strong foundation, especially for JEE Advanced problems.
Master these key qualitative concepts for a strong grasp in JEE & Board Exams!
Understanding wave optics phenomena like diffraction and polarization qualitatively is crucial for both JEE and board exams. It helps build a strong foundation before delving into quantitative aspects.
Imagine sound bending around a corner – you can hear someone even if you can't see them. Diffraction is the exact same phenomenon for light waves. It's the bending or spreading of waves as they pass through an aperture (opening) or around an obstacle.
Diffraction proves that light is a wave. Polarization tells us something more specific: that light is a transverse wave. This means the oscillations (of electric and magnetic fields) are perpendicular to the direction the wave is travelling.
In essence, diffraction shows light's wave nature, while polarization specifically reveals its transverse wave nature. Both are critical for a complete understanding of light.
The phenomena of diffraction and polarization, while seemingly abstract, are fundamental to many everyday observations and advanced technologies. Understanding their qualitative aspects helps appreciate the wave nature of light in various practical scenarios.
Diffraction is the bending of waves around obstacles and the spreading of waves after passing through an aperture. This leads to characteristic interference patterns. Here are some real-world examples:
Polarization refers to the orientation of the oscillations of the electric field vector in an electromagnetic wave. Unpolarized light consists of waves oscillating in all possible planes perpendicular to the direction of propagation. Polarizers selectively transmit light oscillating in a specific plane.
JEE/CBSE Relevance: While detailed quantitative analysis of these applications is generally beyond the scope, a qualitative understanding of how diffraction and polarization manifest in daily life and technology is important for conceptual clarity. Focus on the core principles illustrated by these examples.
Understanding complex physics concepts like diffraction and polarization often becomes clearer with the help of relatable analogies. These analogies provide qualitative insights, helping you build intuition before diving into mathematical formulations. Both JEE and board exams frequently test conceptual understanding of these phenomena.
Diffraction refers to the bending of waves around obstacles or the spreading of waves after passing through an opening. It is a characteristic property of all types of waves (sound, water, light).
Polarization refers to the direction of oscillation of the electric field vector in an electromagnetic wave (like light). Unpolarized light oscillates in all possible directions perpendicular to the direction of propagation, while polarized light oscillates predominantly in one specific plane.
JEE & Board Exam Focus: While analogies are qualitative, they are excellent for building foundational understanding. For exams, ensure you can describe these phenomena in your own words, identify conditions under which they occur, and distinguish between them. Qualitative descriptions of diffraction patterns and types of polarization (plane, circular, elliptical) are important for both JEE Main and advanced levels, as well as for CBSE board exams.
Before diving into the fascinating phenomena of diffraction and polarization, it's crucial to have a solid understanding of certain fundamental concepts from wave optics. Mastering these prerequisites will make your journey through diffraction and polarization much clearer and more intuitive.
💡 Quick Tip: If any of these concepts feel hazy, revisit them briefly. A strong foundation will simplify the more advanced ideas in diffraction and polarization, allowing you to focus on the new concepts rather than struggling with underlying principles.
Navigating questions on Diffraction and Polarization (qualitative ideas) requires a keen eye for subtle distinctions. Many students fall into common traps due to misconceptions or misinterpretations. This section highlights these pitfalls to help you secure those crucial marks.
By being mindful of these common traps, you can approach questions on diffraction and polarization with greater precision and avoid losing marks on conceptual errors.
This section summarizes the most crucial qualitative concepts of Diffraction and Polarization, essential for JEE Main and board exams. Focus on understanding the phenomena and their basic implications rather than complex derivations.
Keep these fundamental ideas clear. While JEE Main often tests qualitative understanding, a firm grasp of the basic principles is crucial for problem-solving.
Solving problems related to diffraction and polarization primarily involves understanding the underlying phenomena, key relationships, and the conditions under which they occur. While "qualitative ideas" are emphasized, a strong grasp of the relevant formulas is essential to interpret changes and predict outcomes.
Diffraction involves the bending of light waves around obstacles or through apertures. For JEE Main, single-slit diffraction is most common, focusing on the central maximum and positions of minima.
a sinθ = nλ, where n = 1, 2, 3... (for the first, second, third minima, etc., on either side of the central maximum).2θ ≈ 2λ/a (using small angle approximation, sinθ ≈ θ). The linear width on the screen is approximately W = 2D(λ/a).(2λ/a) decreases. This means a wider slit produces a narrower diffraction pattern. (Think: less bending).(2λ/a) increases. This means a longer wavelength light produces a wider diffraction pattern. (Think: more bending).a sinθ = nλ gives the positions of minima. The central maximum spans from -λ/a to +λ/a (angularly).a sinθ = nλ is for single-slit minima.Polarization refers to the restriction of light wave vibrations to a single plane. Problems often involve polaroids, reflection, or scattering.
I₀ passes through an ideal polarizer, the transmitted intensity is I₀/2. The light becomes plane-polarized.I₁ is incident on an analyzer whose transmission axis makes an angle 'θ' with the polarization direction of the incident light, the transmitted intensity is I = I₁ cos²θ.(θ_p) on an interface between two media (refractive index 'n'), the reflected light is completely plane-polarized. The relationship is tan θ_p = n. At this angle, the reflected and refracted rays are perpendicular to each other.cos²90° = 0).cos²θ. It's maximum when axes are parallel (θ=0°) and minimum (zero) when perpendicular (θ=90°).Mastering these qualitative relationships and the conditions under which they apply will significantly aid in solving both conceptual and numerical problems in JEE Main and Board exams. Practice visualizing the wave behavior for a deeper understanding.
For the CBSE board examinations, the emphasis on Diffraction and Polarization is largely qualitative, focusing on fundamental definitions, conceptual understanding, and the application of key principles. While JEE might delve into detailed derivations and quantitative analysis, CBSE prioritizes clarity of concepts and descriptive explanations.
CBSE expects students to understand diffraction as a wave phenomenon distinct from interference. Focus on the following aspects:
This section is crucial for understanding the transverse nature of light waves. CBSE primarily focuses on the methods of producing and detecting polarized light.
Mastering these qualitative descriptions and key formulas will ensure strong performance in the CBSE board exams for this topic.
This section outlines the key concepts and formulas related to Diffraction and Polarization that are frequently tested in the JEE Main examination. A strong conceptual understanding and quick recall of formulas are crucial for success in this high-scoring topic.
Diffraction is the bending of light around obstacles or through small apertures. For observable diffraction, the size of the obstacle/aperture must be comparable to the wavelength of light.
a sin θ = nλ, where 'a' is slit width, 'n' = ±1, ±2, … for first, second, etc., minima.a sin θ = (n + 1/2)λ, where 'n' = ±1, ±2, …. (Less frequently asked in direct problem-solving but essential for conceptual questions).2λ/a, and the linear width on the screen is 2λD/a, where D is the distance to the screen.θ = 1.22λ/D, where 'D' is the diameter of the aperture.Polarization is a phenomenon that demonstrates the transverse nature of light waves. Unpolarized light consists of waves oscillating in all possible planes perpendicular to the direction of propagation, while polarized light has oscillations restricted to a single plane.
i_p), the reflected light is completely plane-polarized, and the reflected and refracted rays are perpendicular to each other.tan i_p = n, where 'n' is the refractive index of the medium.I = I₀ cos²θ, where I₀ is the intensity of the plane-polarized light incident on the analyzer, and θ is the angle between the transmission axes of the polarizer and the analyzer.I_u passes through a single ideal polarizer, its intensity becomes I_u/2.Mastering these qualitative ideas and associated formulas will significantly boost your score in Wave Optics for JEE Main.
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Question: If the width of a single slit used in a diffraction experiment is doubled, how does the width of the central maximum change?
Wrong Thought Process: "If the slit width 'a' doubles, the central maximum should also become wider, perhaps twice as wide."
Question: If the width of a single slit used in a diffraction experiment is doubled, how does the width of the central maximum change?
Correct Thought Process: "The width of the central maximum (W) is inversely proportional to the slit width (a), i.e., W ∝ 1/a. If 'a' is doubled, then W will be halved."
Answer: The width of the central maximum will be halved.
nλ yields bright fringes, and (n+1/2)λ yields dark fringes. However, in single-slit diffraction, a sin θ = nλ gives minima (dark fringes), while approximate secondary maxima occur near a sin θ = (n+1/2)λ. This contrasting nature often leads to misapplication. a sin θ = nλ, where n = ±1, ±2, ±3, ... (n=0 corresponds to the central maximum).a sin θ = (2n + 1)λ/2 or a sin θ = (n + 1/2)λ, where n = ±1, ±2, ±3, ... for the 1st, 2nd, 3rd secondary maxima respectively.a sin θ = (1/2)λ. This is incorrect, as this formula approximately gives the position of the first secondary maximum.a sin θ = 1λ. Similarly, for the first secondary maximum, the approximate formula would be a sin θ = (3/2)λ (for n=1).a sin θ = nλ for minima and then substitute θ or x/D, they might fail to explain *why* this approximation is valid or *how* it qualitatively describes the spread and shape of the diffraction pattern. This leads to a disconnect between the mathematical step and its physical meaning. sin θ ≈ θ merely as a mathematical simplification rather than a statement about the angular spread of the phenomenon.x = Dθ becomes a direct consequence, enabling easy calculation and qualitative analysis of fringe widths.a sin θ = λ, so the width is 2θ." This misses the crucial qualitative step of explaining *why* θ ≈ λ/a is used and what it signifies physically.a sin θ = λ. Since the diffraction angles observed are typically very small, we use the small-angle approximation, sin θ ≈ θ. Therefore, the angular position of the first minimum is approximately θ ≈ λ/a. Qualitatively, this means the angular spread of the diffraction pattern is directly proportional to the ratio of the wavelength (λ) to the slit width (a), explaining why narrower slits or longer wavelengths lead to a wider pattern.Students often use angles in degrees when applying the small angle approximation (sin θ ≈ θ) for diffraction calculations, or when comparing angular widths. This is incorrect because the approximation is valid only when θ is expressed in radians.
Always convert angles to radians when using the small angle approximation (sin θ ≈ θ or tan θ ≈ θ) or when directly using formulas derived from this approximation, such as the angular position of diffraction minima/maxima (θ = nλ/a or θ = (n+1/2)λ/a). Many formulas in wave optics inherently assume angles are in radians for dimensional consistency.
A student is given the angular width of a diffraction pattern as Δθ = 1 degree. To find the linear width (x) on a screen at distance D = 2 m, they use the formula x = D * Δθ.
Incorrect Calculation:
x = 2 m * 1 = 2 m
This is wrong because Δθ = 1 degree was used directly instead of converting it to radians, leading to an incorrect linear width.Given: Angular width of a diffraction pattern Δθ = 1 degree. Distance to screen D = 2 m.
Correct Calculation:
1. Convert Δθ from degrees to radians:
Δθradians = 1 × (π / 180) radians ≈ 0.01745 radians
2. Use the correct angular value in radians for the linear width (x):
x = D * Δθradians
x = 2 m * 0.01745 radians
x ≈ 0.0349 m
This correctly uses radians for the small angle approximation in the formula.2λ/a. This formula clearly shows an inverse relationship between the slit width 'a' and the angular width. Therefore, increasing the slit width 'a' actually decreases the angular (and linear) width of the central maximum, making the diffraction pattern more confined and sharper. Conversely, decreasing 'a' spreads out the pattern significantly. Question: In a single-slit diffraction experiment, if the width of the slit is doubled, how does the width of the central maximum change?
Student's Incorrect Reasoning: "A wider slit means more light passes through, so it will spread out more on the screen, making the central maximum wider."
Conclusion: The central maximum becomes wider.
Question: In a single-slit diffraction experiment, if the width of the slit is doubled, how does the width of the central maximum change?
Correct Reasoning: The width of the central maximum (W) is inversely proportional to the slit width (a) (i.e., W ∝ 1/a). If 'a' is doubled, then 'W' must be halved to maintain the proportionality.
Conclusion: The width of the central maximum will become half its original value.
Students frequently confuse the qualitative features and intensity distribution of a single-slit diffraction pattern with a double-slit interference pattern. This often leads to misapplying principles, especially regarding the width and intensity of central maxima, or the spacing of fringes.
Both phenomena involve wave superposition. Double-slit interference is inherently modulated by single-slit diffraction from each slit. The visual similarities of bright/dark regions obscure their distinct physical origins and governing conditions, leading to confusion.
Assuming that all bright fringes in a single-slit diffraction pattern have roughly the same width and intensity, similar to ideal double-slit interference fringes.
Correctly identifying that the central bright fringe in single-slit diffraction is significantly wider and much more intense than any other bright fringes, whose intensities drop sharply. In contrast, for double-slit, the bright fringes are roughly of equal intensity within the central diffraction maximum.
A student observes a single-slit diffraction pattern and states that all bright fringes (maxima) have the same width and comparable intensity, similar to Young's Double Slit Experiment.
When observing a single-slit diffraction pattern, the central bright fringe is notably wider (twice the width of secondary maxima) and significantly brighter than any other bright fringes. The intensity of the secondary maxima decreases rapidly as you move away from the center. Conversely, in an ideal double-slit interference pattern, all bright fringes are of nearly equal width and intensity.
JEE Advanced Tip: In reality, a double-slit pattern is always modulated by a single-slit diffraction envelope, meaning the intensities of the interference maxima decrease as we move away from the center of the diffraction envelope.
Consider a light of wavelength λ = 500 nm incident on a slit of width d = 0.5 µm.
Incorrect reasoning: Since 500 nm is numerically smaller than 0.5 µm, λ < d.
Consider a light of wavelength λ = 500 nm incident on a slit of width d = 0.5 µm.
Correct conversion:
Students frequently misuse Malus' Law ($I = I_0 cos^2 heta$) by either misidentifying the angle $ heta$ or incorrectly using $I_0$ when unpolarized light is first passed through a polarizer/polaroid. This is a crucial concept for JEE Advanced as questions often involve multiple polaroids.
An unpolarized beam of light of intensity $I_{in}$ is incident on a system of two crossed polaroids. A third polaroid is placed between them such that its transmission axis makes an angle of $30^circ$ with the transmission axis of the first polaroid. A common mistake is to directly use $I_{in}$ as the $I_0$ for the first application of Malus' Law, and incorrectly calculate the angle between the second and third polaroids as $30^circ$ instead of $90^circ - 30^circ = 60^circ$. This leads to an incorrect final intensity.
Consider the scenario described in the wrong example. Let the first polaroid be $P_1$, the intermediate polaroid $P_2$, and the second crossed polaroid $P_3$.
To correctly apply Malus's Law:
Problem: Unpolarized light passes through a polarizer P1, then through an analyzer P2, whose transmission axis is oriented at 30° to P1's axis. What is the final intensity?
Wrong thought: Assuming the final intensity is Iunpolarized ⋅ cos²(30°). This overlooks the initial intensity reduction by the first polarizer.
Problem: Unpolarized light of intensity I₀ is incident on a polarizer P1. The transmitted light then passes through an analyzer P2, whose transmission axis is oriented at 30° with respect to the transmission axis of P1. Calculate the intensity of light transmitted through P2.
Unpolarized light of intensity `I_0` passes through polarizer P1 (transmitted intensity `I_0/2`). This light then passes through analyzer P2, whose axis is at 60° to P1's axis. The final transmitted intensity `I_final` is:
`I_final = (I_0/2) * cos²(60°) = (I_0/2) * (1/4) = I_0/8`
Here, `(I_0/2)` is the incident polarized intensity, and `60°` is the correct angle `θ` between the transmission axes.
a sin θ = λ. For small angles, sin θ ≈ θ, so θ ≈ λ/a. Therefore, the angular width of the central maximum (2θ) is directly proportional to the wavelength (λ) and inversely proportional to the slit width (a). A wider slit or a shorter wavelength leads to a narrower central maximum. Question: What happens to the angular width of the central maximum if the slit width is doubled?
Incorrect Answer: The angular width will also double.
Reasoning Flaw: This assumes a direct relationship between slit width and angular width, which is incorrect.
Question: What happens to the angular width of the central maximum if the slit width is doubled?
Correct Answer: The angular width of the central maximum will halve.
Question: If the wavelength of light used is halved, what happens to the angular width?
Correct Answer: The angular width of the central maximum will also halve.
a sin θ = nλ defines the minima. For the first minima (n=1) and small angles, θ ≈ λ/a.θ ∝ λ (direct) and θ ∝ 1/a (inverse).a sin θ = (1/2)λ, directly borrowing the YDSE dark fringe condition.a sin θ = λ (for n=1). This means the angular position is θ = sin-1(λ/a). This marks the boundary of the central bright maximum.A student states: 'Light passing through a 1-meter wide doorway will show clear diffraction patterns on the wall, similar to interference patterns.' This is incorrect because the doorway width (1 m) is vastly larger than the wavelength of visible light (approx. 500 nm), making diffraction effects imperceptible.
When light passes through a very narrow slit, whose width is, for instance, 1 micrometer (1 µm), and the wavelength of light is 500 nanometers (0.5 µm), then prominent diffraction patterns (central maximum, decreasing intensity fringes) are observed. Here, the slit width is comparable to the wavelength (a ≈ 2λ).
Question: What happens to the width of the central maximum if the slit width 'a' is increased?
Student's Incorrect Answer: "The width of the central maximum will increase because a wider slit allows more light, spreading it out more."
Question: What happens to the width of the central maximum if the slit width 'a' is increased?
Correct Answer: "The width of the central maximum will decrease. The angular half-width is approximately λ/a. As 'a' increases, this angular spread decreases, leading to a narrower central maximum. Conversely, if 'a' decreases, the central maximum becomes wider."
Students frequently interchange the conditions for obtaining diffraction minima and secondary maxima in a single-slit experiment. They might incorrectly state that minima occur when the path difference a sinθ = (2n+1)λ/2 and maxima when a sinθ = nλ, which are the conditions for constructive and destructive interference in double-slit interference, not single-slit diffraction minima.
This confusion often arises from over-generalizing concepts from Young's Double Slit Experiment (YDSE) to single-slit diffraction. In YDSE, maxima occur at d sinθ = nλ and minima at d sinθ = (n+1/2)λ. Students mistakenly apply these forms directly to single-slit diffraction, or misremember the derivation where the condition for minima in diffraction is derived based on dividing the slit into secondary wavelets.
For CBSE and JEE, it's crucial to understand the distinct conditions for single-slit diffraction:
a sinθ = nλ, where n = ±1, ±2, ±3, ... (n=0 corresponds to the central maximum).a sinθ = (n + 1/2)λ, where n = ±1, ±2, ±3, .... The central maximum (n=0) is at θ=0 and is significantly wider and brighter.A student identifies the condition for the first diffraction minimum as a sinθ = λ/2.
The correct condition for the first diffraction minimum (dark fringe) is a sinθ = λ (for n=1). Similarly, the first secondary maximum occurs approximately at a sinθ = 3λ/2 (for n=1).
Mastering these foundational concepts is crucial for both theoretical questions and practical applications in exams!
cos²θ dependence is key. A common mistake is to confuse the conditions for maximum and minimum intensity, or to incorrectly apply the angular dependence, leading to an inverted understanding of the transmitted light's behavior. θ (between polarizer and analyzer transmission axes) with the angle of incidence or other irrelevant angles.cos²θ dependence.I = I₀ cos²θ, where I₀ is the intensity of the linearly polarized light incident on the analyzer, and θ is the angle between the transmission axes of the polarizer and the analyzer.I₀/2. This I₀/2 then acts as the I₀ for Malus' Law when incident on the analyzer.θ = 0° or θ = 180° (transmission axes are parallel).θ = 90° or θ = 270° (transmission axes are perpendicular or 'crossed').cos² curve, meaning it decreases from maximum to zero as θ goes from 0° to 90°.cos²(45°) = (1/√2)² = 1/2. So the intensity is half of the maximum possible intensity (I₀/2) when incident on the analyzer, resulting in (I₀/2) * (1/2) = I₀/4 of the original unpolarized light. Thus, 45° gives less intensity than 0°.θ.θ = 0°) and zero transmission (extinction) occurs when axes are perpendicular (θ = 90°).I = I₀ cos²θ mentally and understand the behavior of cos²θ from 0° to 90°.I₀/2 factor after the first polarizer.sinθ ≈ θ for small angles) are valid only when angles are expressed in radians, not degrees.λ = 500 nm passing through a slit of width a = 0.5 mm. The student might directly substitute these values into sinθ = λ/a as 500/0.5, leading to sinθ = 1000, which is physically impossible. This shows a complete disregard for unit consistency.λ = 500 nm = 500 × 10-9 ma = 0.5 mm = 0.5 × 10-3 msinθ = λ/a = (500 × 10-9 m) / (0.5 × 10-3 m) = (500/0.5) × 10-6 = 1000 × 10-6 = 10-3θ is small, θ ≈ sinθ ≈ 10-3 radians. This is a sensible physical value. θp = 56°, for any calculation requiring radians, convert: 56° × (π/180) ≈ 0.977 radians.a sinθ = nλ) with conditions for constructive interference. They also often misunderstand how the angular width of the central maximum is determined, leading to incorrect qualitative predictions. a sinθ = nλ corresponds to destructive interference (minima) in single-slit diffraction, unlike d sinθ = nλ for constructive interference in YDSE. a sinθ = ±nλ (for n = 1, 2, 3,...). The central maximum is bounded by the first minima (n=1), and its angular width is approximately 2λ/a for small angles. Qualitatively, a smaller slit width 'a' or a larger wavelength 'λ' leads to a larger angular spread (wider central maximum). a sinθ = λ, or confuses a sinθ = nλ with conditions for constructive interference.a sinθ = λ. Consequently, the angular width of the central maximum is approximately 2λ/a radians (for small θ), meaning a wider slit 'a' results in a narrower central maximum.a sinθ = nλ gives minima in single-slit diffraction.θ as an absolute angle instead of the relative angle between the incident polarized light's electric field and the analyzer's transmission axis.I = I₀ cos²θ), θ is the angle between the transmission axis of the analyzer and the direction of the electric field vector of the incident polarized light.tan θp = n), the reflected light is completely plane-polarized with its electric field vector oscillating perpendicular to the plane of incidence. (JEE Tip: Remember it's parallel to the reflecting surface).sin²θ mistakenly thinking θ is the angle relative to a perpendicular axis.No summary available yet.
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