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Wave Optics

Wave optics, also known as physical optics, delves into the intricate behavior of light as a wave, revealing phenomena that cannot be explained by geometrical optics alone. It explores interference, diffraction, and polarization—key aspects that shape how light behaves in complex systems. These phenomena are fundamental to technologies in photonics, communications, imaging, and even biomedical devices.

A strong foundation in physics is essential to grasp wave optics, especially its overlap with electricity & magnetism. The study of electromagnetic waves provides the framework to understand how light behaves as a transverse wave, carrying energy and momentum. Applications often draw from related areas such as electrodynamics and electromagnetic induction, which explain wave generation and propagation in materials.

In practical settings, wave optics is crucial in the design of optical instruments, especially those using laser optics for coherent light generation. The precision of interference patterns is applied in spectroscopy, interferometry, and holography. Understanding these effects also aids in innovations within nonlinear optics, where intense light fields alter material properties, and in quantum optics, where light-matter interactions occur at the quantum level.

Wave optics plays a vital role in communication technologies using fiber optics. Here, light pulses travel through narrow cores using principles of interference and wave confinement. This links closely to electronic systems involving electrical circuits and magnetic fields, integrating optics into signal processing and data transmission infrastructure.

Wave behavior is also essential in biological and environmental contexts. In bio-optics, diffraction and scattering affect image clarity and resolution in tissue imaging. Similarly, atmospheric and environmental optics examine how wave phenomena cause halos, mirages, and rainbow formation due to interactions with water droplets and aerosols.

From a theoretical standpoint, wave optics aligns with the quantum view of light. Concepts from modern physics and quantum electrodynamics (QED) help explain photon interference and dual-wave-particle behavior. These ideas also overlap with studies in superconductivity and plasma physics, where wave dynamics are applied in high-energy and fusion environments.

In applied engineering and computational modeling, wave optics is enhanced by techniques from magnetostatics and magnetohydrodynamics (MHD), especially in scenarios involving wave propagation through magnetized fluids or ionized gases.

As optical systems become increasingly complex, insights from visual optics and light and optics are essential for optimizing image formation and correcting aberrations in human and artificial vision systems. Ultimately, wave optics bridges classical and quantum models, playing a pivotal role in expanding our capabilities in science, medicine, and communication.

Wave Optics, depicting diffraction, interference, and polarization of light.
Wave Optics, depicting diffraction, interference, and polarization of light.

Table of Contents

Fundamental Concepts in Wave Optics

Wave Nature of Light

Light exhibits wave-like behavior as a transverse electromagnetic wave, where the electric field (E) and magnetic field (B) oscillate perpendicular to both each other and the direction of propagation. The key characteristics defining light waves include:

  • Wavelength (λ): The distance between two consecutive wave peaks.
  • Frequency (f): The number of oscillations per second.
  • Speed (c): The speed of light in a vacuum, approximately 3 × 10⁸ m/s.

These properties are related through the equation:
c = λf, meaning that as wavelength increases, frequency decreases, and vice versa.

Interference

Interference occurs when two or more coherent light waves overlap, forming a new wave pattern due to superposition. The two primary types of interference are:

  • Constructive Interference: When waves are in phase, their amplitudes combine, producing bright fringes.
  • Destructive Interference: When waves are out of phase by π (180°), they cancel out, forming dark fringes.

One of the most well-known demonstrations of interference is Young’s Double-Slit Experiment, where light passing through two narrow slits creates alternating bright and dark fringes on a screen. The fringe spacing (Δy) is determined by:

Δy = (λD) / d, where:

  • λ = Wavelength of light
  • D = Distance between the slits and the screen
  • d = Distance between the two slits

Diffraction

Diffraction refers to the bending and spreading of light waves when they encounter an obstacle or pass through a narrow opening. It is most prominent when the size of the opening or object is comparable to the wavelength of light.

Single-Slit Diffraction: When light passes through a single narrow slit, it forms a central bright fringe surrounded by alternating dark and bright fringes. The condition for destructive interference (dark fringes) in a single-slit diffraction pattern is:

a sinθ = mλ, where:

  • a = Slit width
  • θ = Angle of diffraction
  • m = Order of the dark fringe (m = ±1, ±2, …)

Polarization

Polarization describes the orientation of the electric field oscillations in a light wave. Natural light is unpolarized, meaning the electric fields oscillate in multiple directions, but it can be manipulated to oscillate in a specific direction.

Types of Polarization:

  • Linear Polarization: The electric field oscillates in a single plane.
  • Circular Polarization: The electric field rotates in a helical pattern as the wave propagates.
  • Elliptical Polarization: A combination of linear and circular polarization, forming an elliptical trajectory.

Malus’s Law describes how polarized light interacts with a polarizing filter:
I = I₀ cos²θ, where:

  • I₀ = Initial light intensity
  • θ = Angle between the light’s polarization direction and the polarizer’s axis

Huygens’ Principle

Huygens’ Principle states that every point on a wavefront can be considered as a source of secondary wavelets, which spread out in all directions. The new wavefront is formed by the envelope of these secondary wavelets. This principle helps explain:

  • Reflection: Light waves bouncing off surfaces.
  • Refraction: Light bending when passing through different media.
  • Diffraction: The spreading of light waves around obstacles or through narrow openings.

Wave Optics provides a comprehensive understanding of these fundamental phenomena, enabling the development of advanced optical technologies in imaging, communication, and laser applications.

Applications of Wave Optics

  1. Anti-Reflective Coatings: Use interference to minimize reflections on lenses and screens.
  2. Optical Instruments: Diffraction gratings are used in spectrometers to analyze light spectra.
  3. Fiber Optics: Utilize total internal reflection and interference for data transmission.
  4. Polarized Sunglasses: Reduce glare by blocking horizontally polarized light.
  5. Holography: Employs interference and diffraction to create 3D images.
A visually striking depiction of various applications of wave optics, including anti-reflective coatings, diffraction gratings, fiber optics, polarized sunglasses, and holography.
A visually striking depiction of various applications of wave optics, including anti-reflective coatings, diffraction gratings, fiber optics, polarized sunglasses, and holography.

Five Numerical Examples

Example 1: Fringe Spacing in Young’s Double-Slit Experiment

Problem:
Light of wavelength 600 nm passes through two slits separated by 0.2 mm. The screen is 1 m away. Calculate the fringe spacing.

 

Solution:

Δy=λDd\Delta y = \frac{\lambda D}{d}

Δy=600×10−9×10.2×10−3=3×10−3 m=3 mm\Delta y = \frac{600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 3 \times 10^{-3} \, \text{m} = 3 \, \text{mm}

Answer:
The fringe spacing is 3 mm.

Example 2: First-Order Diffraction Angle

Problem:
A slit of width 0.05 mm is illuminated by light of wavelength 500 nm. Find the angle for the first dark fringe.

Solution:

asin⁡θ=mλa \sin \theta = m \lambda

For the first-order dark fringe (m = 1):

0.05×10−3sin⁡θ=500×10−90.05 \times 10^{-3} \sin \theta = 500 \times 10^{-9}
 
sin⁡θ=500×10−90.05×10−3=0.01\sin \theta = \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} = 0.01
 
θ=sin⁡−1(0.01)≈0.57∘\theta = \sin^{-1}(0.01) \approx 0.57^\circ

Answer:
The angle for the first dark fringe is approximately 0.57°.

Example 3: Critical Angle for Total Internal Reflection

Problem:
Find the critical angle for light traveling from glass (n = 1.5) to air n = 1.0)

Solution:

sin⁡θc=n2n1=1.01.5\sin \theta_c = \frac{n_2}{n_1} = \frac{1.0}{1.5}

θc=sin⁡−1(23)≈41.8∘\theta_c = \sin^{-1} \left(\frac{2}{3}\right) \approx 41.8^\circ

Answer:
The critical angle is approximately 41.8°.

Example 4: Intensity of Polarized Light

Problem:
Unpolarized light of intensity 100 W/m² passes through a polarizer set at 30°. What is the transmitted intensity?

Solution:

I=I0cos⁡2θI = I_0 \cos^2 \theta I=100cos⁡230∘=100×(32)2=100×34=75 W/m2I = 100 \cos^2 30^\circ = 100 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 100 \times \frac{3}{4} = 75 \, \text{W/m}^2

Answer:
The transmitted intensity is 75 W/m².

Example 5: Path Difference for Constructive Interference

Problem:
What is the minimum path difference for constructive interference for light of wavelength 600 nm?

Solution:

For constructive interference:

ΔL=mλ\Delta L = m \lambda

For the first-order (m = 1):

ΔL=600 nm\Delta L = 600 \, \text{nm}

Answer:
The minimum path difference is 600 nm.

Why Study Wave Optics

Beyond the Ray Approximation

Wave optics, or physical optics, explains phenomena that cannot be described by geometrical optics alone. Students study diffraction, interference, and polarization to understand how light behaves as a wave. This deeper understanding allows them to analyze and design more sophisticated optical systems. It reveals the coherence and superposition principles that underlie light behavior.

Interference and Coherence

Students explore constructive and destructive interference, fringe patterns, and coherence length. These principles are applied in technologies like interferometers, holography, and thin-film coatings. Understanding interference enhances control over light behavior in precision instruments. It is essential for optical metrology and imaging systems.

Diffraction and the Limits of Resolution

Students examine how apertures and obstacles affect light propagation through diffraction. This sets the limits for resolution in microscopes and telescopes. Mastery of these concepts improves the design of high-resolution systems. It also informs the development of techniques to overcome classical limits.

Polarization and Optical Modulation

Wave optics introduces students to polarization states and their manipulation using wave plates and polarizers. They explore how polarization is used in LCD displays, stress analysis, and communication systems. These insights support control of light’s vector properties. They enhance the design of advanced photonic devices.

Bridge to Quantum and Modern Optics

Wave optics provides the theoretical foundation for modern optical science, including quantum optics and photonics. Students who grasp these principles are better prepared for cutting-edge research and development. It unifies classical and quantum perspectives on light. It equips students for future innovations in optical technology.

 

Conclusion

Wave Optics explains many natural and technological phenomena that Geometrical Optics cannot, including interference, diffraction, and polarization. These principles are foundational in optical engineering, communication systems, laser technology, and scientific instrumentation. Through the study of wave behavior, engineers and scientists can design advanced optical devices and systems critical to modern technology and scientific discovery.

Frequently Asked Questions: Wave Optics

1. What is wave optics?

Wave optics (or physical optics) is the branch of optics that explains light behaviour by treating light as a wave. It focuses on phenomena such as interference, diffraction, and polarization, which cannot be fully understood using only straight-line ray models from geometrical optics.

2. How does wave optics differ from geometrical optics?

Geometrical optics treats light as rays that travel in straight lines and change direction at mirrors and lenses, ignoring the wave nature of light. Wave optics, by contrast, uses the ideas of wavefronts, superposition, and coherence to explain how light waves add together, bend around obstacles, and show interference and diffraction patterns.

3. What is Huygens’ principle and what are wavefronts?

Huygens’ principle states that every point on a wavefront acts as a source of secondary spherical wavelets, and the new wavefront is the envelope of these wavelets. A wavefront is a surface joining points of equal phase, such as all points where the light wave has the same displacement in its cycle. This viewpoint helps us understand reflection, refraction, and diffraction in terms of wave propagation.

4. What is interference of light and when does it occur?

Interference of light occurs when two or more coherent light waves overlap and combine through superposition. At points where the waves arrive in phase, they produce constructive interference (bright fringes), and where they arrive out of phase, they produce destructive interference (dark fringes). Stable interference patterns require coherent sources with a fixed phase relationship and similar frequency.

5. How does Young’s double-slit experiment demonstrate wave behaviour of light?

In Young’s double-slit experiment, light passes through two closely spaced slits and the emerging waves overlap on a distant screen. Instead of producing just two bright patches, the overlapping waves form a pattern of alternating bright and dark fringes. This regular fringe pattern arises from constructive and destructive interference and provides strong evidence that light behaves as a wave.

6. What is diffraction and how does it differ from simple shadow casting?

Diffraction is the bending and spreading of waves when they encounter an obstacle or pass through an aperture. In wave optics, light does not simply cast sharp shadows; instead, the edges of openings and obstacles act as new sources of wavelets, producing characteristic intensity patterns. Single-slit diffraction, for example, creates a central bright maximum flanked by weaker side fringes, showing the wave nature of light.

7. What is a diffraction grating and how is it used?

A diffraction grating is an optical element with many closely spaced, parallel slits or grooves. When light passes through or reflects from a grating, the waves from each slit interfere, producing sharp maxima at specific angles determined by wavelength and grating spacing. Diffraction gratings are used in spectrometers to separate light into its component wavelengths for analysis in physics, chemistry, and astronomy.

8. What is polarization in wave optics?

Polarization describes the orientation of the electric field oscillations in a transverse light wave. Unpolarized light has electric fields vibrating in many directions perpendicular to the direction of travel, while polarized light has a preferred orientation. Wave optics explains how polarizers, reflection, scattering, and certain crystals can produce and modify polarized light, with applications in sunglasses, displays, stress analysis, and optical communication.

9. What is coherence and why is it important in wave optics?

Coherence refers to the stability of the phase relationship between waves. Temporal coherence describes how long a wave maintains a fixed phase over time, while spatial coherence describes how well-defined the phase is across different points in space. High coherence is essential for clear interference and diffraction patterns; lasers, for example, produce highly coherent light that is ideal for wave optics experiments and applications.

10. How does wave optics explain colours in thin films and CDs?

In thin films, such as soap bubbles or oil on water, reflections from the top and bottom surfaces interfere. Depending on film thickness, viewing angle, and wavelength, certain colours are enhanced while others are suppressed, creating vivid patterns. On compact discs and similar structures, the fine spacing of surface features acts like a diffraction grating, separating white light into coloured patterns explained by interference and diffraction.

11. What are some applications of wave optics in science and technology?

Wave optics underpins technologies such as holography, interferometry, spectroscopy, and many imaging systems. Interferometers measure tiny displacements, refractive index changes, or even gravitational waves. Diffraction gratings and interferometers are central to spectroscopic analysis. Polarization control is used in liquid crystal displays, optical communication, and stress analysis in materials. Wave-based ideas also guide the design of lasers, optical sensors, and modern photonic devices.

12. Why is studying wave optics important for physics and engineering students?

Studying wave optics helps students move beyond simple ray diagrams to understand how light really behaves when structures are comparable to its wavelength. It builds intuition for interference, diffraction, and polarization, which are essential in modern technologies such as lasers, fibre optics, photonics, and imaging systems. Wave optics also provides a natural bridge toward quantum optics and the deeper study of light as both a wave and a stream of photons.

Review Questions and Answers:

1. What is wave optics and how does it differ from geometrical optics?
Answer: Wave optics is the study of light as a wave, focusing on phenomena such as interference, diffraction, and polarization. Unlike geometrical optics, which treats light as rays, wave optics explains how wave properties lead to effects like fringe patterns and spreading of light.

2. How does interference occur in wave optics and what are its observable effects?
Answer: Interference occurs when two or more coherent light waves overlap, resulting in regions of constructive interference (bright fringes) and destructive interference (dark fringes). This phenomenon is clearly observed in experiments like Young’s double-slit, where alternating bright and dark bands are produced.

3. What is diffraction and why is it significant in understanding the behavior of light?
Answer: Diffraction is the bending and spreading of light waves as they pass through a narrow aperture or around an obstacle. It is significant because it demonstrates the wave nature of light and helps explain limits on resolution in optical instruments as well as patterns like the single-slit diffraction envelope.

4. Can you explain the concept of coherence and its importance in wave optics experiments?
Answer: Coherence refers to the fixed phase relationship between waves over time and space. It is essential for producing clear interference patterns, as only coherent light sources, which maintain a constant phase difference, can yield stable and observable interference effects.

5. How does polarization relate to wave optics and what are some common methods to achieve it?
Answer: Polarization is the orientation of the oscillations of a light wave perpendicular to its direction of travel. It can be achieved using polarizers, reflection, or scattering. Polarized light is crucial in various applications, including reducing glare in photography and improving contrast in liquid crystal displays.

6. What role do phase differences play in the interference of light waves?
Answer: Phase differences determine whether the overlapping light waves will interfere constructively or destructively. A phase difference of multiples of 2π results in constructive interference, producing bright regions, while a phase difference of odd multiples of π leads to destructive interference, resulting in dark regions.

7. How can the concept of optical path difference be used to explain interference patterns?
Answer: Optical path difference (OPD) is the difference in the distance traveled by two light waves, adjusted for the medium’s refractive index. When the OPD is an integral multiple of the wavelength, constructive interference occurs, and when it is a half-integral multiple, destructive interference occurs, forming the characteristic fringe patterns.

8. What is a diffraction grating and how does it help in analyzing light spectra?
Answer: A diffraction grating is an optical component with many equally spaced slits that diffract light into several beams. The angle at which light is diffracted depends on its wavelength, allowing the grating to separate light into its component colors and analyze its spectral composition.

9. How does the principle of superposition apply to wave optics?
Answer: The principle of superposition states that when two or more waves overlap, the resultant wave is the sum of the individual waves. This principle underlies the interference and diffraction phenomena observed in wave optics, explaining how different light waves combine to form complex patterns.

10. What practical applications of wave optics can you identify in modern technology?
Answer: Wave optics principles are applied in numerous technologies, including laser systems, fiber-optic communications, holography, and optical sensors. These applications rely on interference, diffraction, and polarization to manipulate and analyze light for improved performance and innovation.

Thought-Provoking Questions and Answers

1. How might emerging quantum optical technologies reshape our understanding of traditional wave optics principles?
Answer: Emerging quantum optical technologies, such as quantum entanglement and quantum computing, could deepen our understanding of wave optics by revealing new aspects of light behavior at the quantum level. These advancements may lead to innovative applications like ultra-secure communications and highly sensitive measurement devices that challenge and expand classical interpretations.

2. In what ways could the integration of wave optics with nanotechnology revolutionize optical device performance?
Answer: Integrating wave optics with nanotechnology could lead to the creation of devices that manipulate light at sub-wavelength scales, enabling unprecedented control over optical properties. This integration may result in ultra-compact sensors, high-resolution imaging systems, and advanced photonic circuits that enhance both performance and efficiency.

3. How can wave optics principles be applied to improve the design of next-generation optical communication systems?
Answer: Wave optics principles can enhance optical communication systems by optimizing the interference and diffraction properties of light within fiber-optic cables. By reducing signal loss and dispersion, engineers can design systems that support higher data rates, improved bandwidth, and more robust signal integrity across long distances.

4. What are the potential challenges in scaling wave optics experiments from controlled laboratory settings to real-world applications?
Answer: Scaling wave optics experiments involves challenges such as maintaining coherence over long distances, managing environmental disturbances, and ensuring the stability of interference patterns. Overcoming these issues requires advanced materials, precise engineering, and innovative techniques to replicate laboratory conditions in practical, real-world environments.

5. Could advancements in wave optics lead to breakthroughs in non-invasive medical imaging techniques?
Answer: Yes, advancements in wave optics could significantly improve non-invasive medical imaging by enhancing resolution and contrast in techniques like optical coherence tomography and holographic imaging. These improvements might allow for earlier detection of diseases and more accurate diagnostic procedures without the need for invasive methods.

6. How might the principles of wave optics contribute to the development of adaptive optics in astronomy?
Answer: Wave optics is crucial for adaptive optics, which corrects distortions caused by Earth’s atmosphere. By understanding interference and diffraction, astronomers can design systems that dynamically adjust optical components to counteract atmospheric turbulence, leading to clearer and more detailed images of celestial bodies.

7. What role does computational modeling play in advancing our understanding of complex wave optics phenomena?
Answer: Computational modeling allows researchers to simulate and analyze intricate wave interactions, such as multiple beam interference and complex diffraction patterns. These models can predict experimental outcomes, optimize system designs, and provide deeper insights into phenomena that are difficult to study experimentally, thereby driving innovation in optical research.

8. How can the study of wave optics contribute to our understanding of natural phenomena like rainbows and auroras?
Answer: Wave optics explains natural phenomena such as rainbows and auroras through principles of refraction, dispersion, and interference. By studying these processes, scientists can uncover the detailed interactions between light and atmospheric particles, enhancing our knowledge of weather patterns, atmospheric composition, and energy transfer in nature.

9. In what ways could advancements in wave optics influence the future development of augmented and virtual reality technologies?
Answer: Advancements in wave optics can lead to more realistic and immersive augmented and virtual reality experiences by improving the display resolution, reducing optical distortions, and enhancing depth perception. Innovative optical components based on diffraction and interference principles could result in lighter, more efficient, and higher-quality head-mounted displays.

10. How might environmental factors such as temperature and atmospheric pressure affect wave optics experiments and applications?
Answer: Environmental factors like temperature and atmospheric pressure can alter the refractive index of materials and the coherence of light sources, affecting interference and diffraction patterns. Understanding these effects is essential for designing robust optical systems that maintain performance under varying environmental conditions, especially in precision applications like astronomy and remote sensing.

11. What future research directions could expand the practical applications of polarization in wave optics?
Answer: Future research could explore new materials and nanostructures to create advanced polarizers with tunable properties. Such developments might lead to breakthroughs in optical data storage, security systems, and improved imaging techniques, further expanding the range of applications that rely on controlling and manipulating light polarization.

12. How can interdisciplinary collaboration accelerate progress in solving complex challenges in wave optics?
Answer: Interdisciplinary collaboration brings together experts in physics, materials science, engineering, and computer science, fostering a comprehensive approach to complex wave optics challenges. This synergy can lead to innovative experimental techniques, novel theoretical models, and the rapid development of cutting-edge optical devices that address real-world problems across multiple fields.

Numerical Problems and Solutions

1. In a Young’s double-slit experiment, if the distance between the slits is 0.5 mm and the screen is placed 2 m away, calculate the fringe separation for light of wavelength 600 nm.
Solution:
Using the formula for fringe separation, Δy=λDd\Delta y = \frac{\lambda D}{d}, where
λ=600×10−9\lambda = 600 \times 10^{-9}, D=2D = 2 m, and d=0.5×10−3d = 0.5 \times 10^{-3},
Δy=600×10−9×20.5×10−3=1.2×10−60.5×10−3=2.4×10−3\Delta y = \frac{600 \times 10^{-9} \times 2}{0.5 \times 10^{-3}} = \frac{1.2 \times 10^{-6}}{0.5 \times 10^{-3}} = 2.4 \times 10^{-3}.
Thus, the fringe separation is 2.4 mm.

2. A single slit of width 0.1 mm is illuminated by light of wavelength 500 nm. Determine the angular width of the central diffraction maximum.
Solution:
The angular width θ\theta for a single slit is given by θ=2λa\theta = \frac{2\lambda}{a}, where
λ=500×10−9\lambda = 500 \times 10^{-9} m and a=0.1×10−3a = 0.1 \times 10^{-3}.
θ=2×500×10−90.1×10−3=1000×10−90.1×10−3=10−610−4=0.01\theta = \frac{2 \times 500 \times 10^{-9}}{0.1 \times 10^{-3}} = \frac{1000 \times 10^{-9}}{0.1 \times 10^{-3}} = \frac{10^{-6}}{10^{-4}} = 0.01radians.

3. In a diffraction grating experiment, if the grating has 5000 lines per centimeter, what is the grating spacing?
Solution:
First, convert lines per centimeter to lines per meter: 50005000 lines/cm = 5000×100=5000005000 \times 100 = 500000 lines/m.
The grating spacing d=1number of lines per meter=1500000=2×10−6d = \frac{1}{\text{number of lines per meter}} = \frac{1}{500000} = 2 \times 10^{-6} m.

4. In Young’s experiment, if the 3rd bright fringe appears at 4.8 mm from the central maximum, find the wavelength of the light used given that the slit separation is 0.2 mm and the screen is 1.5 m away.
Solution:
For the m-th order fringe, ym=mλDdy_m = \frac{m \lambda D}{d}. For m=3m = 3,
4.8×10−3=3λ×1.50.2×10−34.8 \times 10^{-3} = \frac{3 \lambda \times 1.5}{0.2 \times 10^{-3}}.
Rearrange: λ=4.8×10−3×0.2×10−33×1.5=0.96×10−64.5≈213.33×10−9\lambda = \frac{4.8 \times 10^{-3} \times 0.2 \times 10^{-3}}{3 \times 1.5} = \frac{0.96 \times 10^{-6}}{4.5} \approx 213.33 \times 10^{-9} nm.

5. A beam of unpolarized light of intensity I passes through a polarizer and then through an analyzer oriented at 30° relative to the polarizer. Calculate the transmitted intensity using Malus’s law.
Solution:
First, unpolarized light passing through a polarizer is reduced to I2\frac{I}{2}. Then, the analyzer transmits I2cos⁡2(30°)\frac{I}{2} \cos^2(30°).
Since cos⁡30°=32\cos 30° = \frac{\sqrt{3}}{2},
Itransmitted=I2×(32)2=I2×34=3I8I_{\text{transmitted}} = \frac{I}{2} \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I}{2} \times \frac{3}{4} = \frac{3I}{8}.

6. In a double-slit interference setup, if the fringe separation is 1.2 mm and the screen is 2 m from the slits, determine the slit separation for light of wavelength 550 nm.
Solution:
Using Δy=λDd\Delta y = \frac{\lambda D}{d}, rearrange to find d=λDΔyd = \frac{\lambda D}{\Delta y}.
d=550×10−9×21.2×10−3=1100×10−91.2×10−3≈9.17×10−4d = \frac{550 \times 10^{-9} \times 2}{1.2 \times 10^{-3}} = \frac{1100 \times 10^{-9}}{1.2 \times 10^{-3}} \approx 9.17 \times 10^{-4} mm.

7. Calculate the angular position of the first minimum in the diffraction pattern of a slit with a width of 0.25 mm using light of wavelength 650 nm.
Solution:
For a single slit, the first minimum occurs at sin⁡θ=λa\sin \theta = \frac{\lambda}{a}.
sin⁡θ=650×10−90.25×10−3=0.0026\sin \theta = \frac{650 \times 10^{-9}}{0.25 \times 10^{-3}} = 0.0026.
Thus, θ≈sin⁡−1(0.0026)≈0.15°\theta \approx \sin^{-1}(0.0026) \approx 0.15°.

8. A diffraction grating with 10000 lines per meter is used with light of wavelength 600 nm. What is the angle for the first-order maximum?
Solution:
The grating equation is dsin⁡θ=mλd \sin \theta = m \lambda. For m=1m = 1 and d=110000=1×10−4d = \frac{1}{10000} = 1 \times 10^{-4} m,
sin⁡θ=600×10−91×10−4=0.006\sin \theta = \frac{600 \times 10^{-9}}{1 \times 10^{-4}} = 0.006.
Therefore, θ≈sin⁡−1(0.006)≈0.344°\theta \approx \sin^{-1}(0.006) \approx 0.344°.

9. In a double-slit experiment, if the central maximum is 3 cm wide on the screen and the fringe width is 0.5 cm, how many bright fringes are visible on one side of the central maximum?
Solution:
Half of the central maximum plus subsequent fringes spread symmetrically. If the entire width is 3 cm, then from center to one edge is 1.5 cm.
Number of fringe intervals =1.5 cm0.5 cm=3= \frac{1.5 \text{ cm}}{0.5 \text{ cm}} = 3.
Thus, 3 bright fringes are visible on one side of the central maximum (excluding the central fringe).

10. A light wave undergoes a phase change of 180° upon reflection. If two such reflected waves combine, under what condition will they interfere destructively?
Solution:
Two waves with a 180° phase difference are exactly out of phase. They will interfere destructively if they have equal amplitude and no additional path difference, canceling each other completely.

11. In a setup using a diffraction grating, if the second-order maximum is observed at 20° for light of wavelength 700 nm, calculate the grating spacing.
Solution:
Using the grating equation, dsin⁡θ=mλd \sin \theta = m \lambda with m=2m = 2 and θ=20°\theta = 20°:
d=2×700×10−9sin⁡20°d = \frac{2 \times 700 \times 10^{-9}}{\sin 20°}.
Since sin⁡20°≈0.342\sin 20° \approx 0.342,
d≈1400×10−90.342≈4.09×10−6d \approx \frac{1400 \times 10^{-9}}{0.342} \approx 4.09 \times 10^{-6}.

12. In a single-slit diffraction experiment, if the distance between the first minima on either side of the central maximum is 4 mm and the screen is 1.2 m away, determine the slit width for light of wavelength 550 nm.
Solution:
The angular position of the first minimum is given by sin⁡θ=λa\sin \theta = \frac{\lambda}{a}. For small angles, θ≈yD\theta \approx \frac{y}{D} where y=4 mm2=2y = \frac{4 \text{ mm}}{2} = 2 mm =2×10−3= 2 \times 10^{-3} m.
Thus, θ≈2×10−31.2≈1.67×10−3\theta \approx \frac{2 \times 10^{-3}}{1.2} \approx 1.67 \times 10^{-3} radians.
Then, a=λsin⁡θ≈550×10−91.67×10−3≈3.29×10−4a = \frac{\lambda}{\sin \theta} \approx \frac{550 \times 10^{-9}}{1.67 \times 10^{-3}} \approx 3.29 \times 10^{-4} m, or about 0.329 mm.

Last updated: 21 Dec 2025