Wave Optics (SAQs)

Physics-2 | 3. Wave Optics – SAQs:
Welcome to SAQs in Chapter 3: Wave Optics. This page covers the key FAQs for Short Answer Questions. Answers are provided in simple English, with a Telugu explanation, and formatted according to the exam style. This approach will aid in understanding and help you achieve top marks in your final exams.

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SAQ-1 : Explain Doppler effect in light. Distinguish between redshift and blueshift.

Introduction

The Doppler effect, commonly associated with sound waves, also applies to light waves and the entire electromagnetic spectrum. This effect, also known as the relativistic Doppler effect, is observed when there is relative motion between the source of light and the observer. It results in a change in observed frequency or wavelength of light, leading to phenomena known as “redshift” and “blueshift“.

Understanding Doppler Effect in Light

1. Defining the Doppler Effect for Light: The Doppler effect for light describes the change in wavelength (and thus color) of light due to the relative motion of the source and the observer. Unlike sound, light doesn’t need a medium to travel, so this effect involves changes at the speed of light, making it a relativistic effect.
2. Redshift: When the source of light moves away from the observer, the observed wavelength is longer than the wavelength originally emitted by the source. This elongation shifts the light towards the red end of the spectrum, hence it’s called “redshift“. The formula to calculate the observed wavelength in this case is: $$\lambda_{\text{observed}} = \lambda_{\text{source}} \sqrt{[(1 + v/c) / (1 – v/c)]}$$
3. Blueshift: Conversely, when the source moves towards the observer, the observed wavelength is shorter than the original emitted wavelength. This compression shifts the light towards the blue end of the spectrum, a phenomenon called “blueshift“. The formula to calculate the observed wavelength here is: $$\lambda_{\text{observed}} = \lambda_{\text{source}} \sqrt{[(1 – v/c) / (1 + v/c)]}​$$
4. Note: In both formulas, λ_source​ is the wavelength of light emitted by the source, λ_observed​ is the observed wavelength, v is the velocity of the source, and c is the speed of light.
5. Relative Motion of Observer and Source: The Doppler effect of light isn’t limited to the movement of the source alone. Even if the observer is moving, either away from or towards the source, they will still observe the Doppler effect.

Summary

The Doppler effect of light is a crucial principle in physics and astronomy. It helps in understanding the movement and distance of celestial bodies by observing the redshift or blueshift in the light they emit. Understanding the Doppler effect in light extends our perception beyond the immediate environment, right out into the cosmos.

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SAQ-2 : Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity.

Introduction

The principle of superposition states that when two or more waves overlap in space, the resultant displacement at any point is the algebraic sum of the displacements of the individual waves at that point. Let’s discuss the case where we have two waves with the same amplitude but different phases and derive the resulting intensity.

Step-by-step Solution

1. Given two waves with the same amplitude ‘a’ but with a phase difference ‘ϕ’, their displacements are:
   $$y_1 = a \sin(\omega t)$$
   $$y_2 = a \sin(\omega t + ϕ)$$
2. The resultant displacement, 2y = y_1​ + y₂​, so:
   $$y = a \sin \omega t + a \sin(\omega t + ϕ)$$
3. Expanding this using the trigonometric identity sin(A+B) = sinA cosB + cosA sinB gives:
   $$y = a \sin \omega t [1 + \cos ϕ] + a \cos \omega t \sin ϕ$$
4. Let’s rewrite Eq. (3) in a different form:
   If $$R \cos \theta = a(1 + \cos ϕ)$$
   $$R \sin \theta = a \sin ϕ$$
5. Substituting Eq. (4) and Eq. (5) into Eq. (3) gives:
   $$y = R \sin(\omega t + \theta)$$ where ‘R’ is the resultant amplitude. From Eq. (4) and Eq. (5), by squaring and adding them, we get the resultant intensity ‘I’ as:
   $$I = R^2 = 4a^2 \cos^2(\frac{ϕ}{2})$$
6. Analyzing Eq. (7), we get:
   • Maximum intensity (I_max​) occurs when $$\cos^2(\frac{ϕ}{2}) = 1$$ i.e., ϕ = 2nπ, where n=0, 1, 2, 3,…, leading to ϕ = 0, 2π, 4π, 6π…. Hence, I_max​ = 4a².
   • Minimum intensity (I_min​) occurs when $$\cos^2(\frac{ϕ}{2}) = 0$$ i.e., ϕ = (2n+1)π, where n=0, 1, 2, 3,…, leading to ϕ = π, 3π, 5π, 7π…. Hence, I_min​=0.

Summary

The above derivations show how the resultant amplitude and intensity of two superposed waves with the same amplitude but different phases can be calculated. This analysis helps us understand the behavior of wave superposition and interference in various fields such as acoustics, optics, and quantum mechanics.

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SAQ-3 : Does the principle of conservation of energy hold for interference and diffraction phenomena ? Explain briefly.

Introduction

The principle of conservation of energy states that energy can neither be created nor destroyed; it only transforms from one form to another. This principle is universally applicable, including in the phenomena of light interference and diffraction.

Interference of Light

1. During the interference of light, bright and dark fringes form due to the constructive and destructive superposition of light waves, respectively.
2. All the bright fringes in an interference pattern have maximum and equal intensity, signifying points where light waves add constructively.
3. Conversely, all the dark fringes have minimum intensity and are completely dark, signifying points where light waves cancel each other out destructively.
4. Despite these differences, the total energy in the interference pattern remains constant as energy is equally distributed among the bright and dark fringes, illustrating conservation of energy.

Diffraction of Light

1. Diffraction involves the bending or spreading of light waves around obstacles or through apertures, resulting in bright and dark fringes.
2. Unlike interference, the bright fringes in a diffraction pattern do not have equal intensity. This is due to the different path lengths and varying degrees of constructive interference.
3. Similarly, the dark fringes in a diffraction pattern, though completely dark, have different intensities compared to the bright fringes.
4. However, the total energy remains constant in diffraction as well, as energy is simply redistributed among the bright and dark fringes, once again demonstrating the principle of energy conservation.

Summary

In both light interference and diffraction, the conservation of energy holds true. Although the intensity of the fringes varies in these phenomena due to different degrees of constructive and destructive interference, the total energy remains constant as it’s only redistributed, not created or lost.

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SAQ-4 : How do you determine the resolving power of your eye?

Introduction

The resolving power of an optical system such as the human eye or a telescope is defined as its ability to distinguish or resolve two closely spaced objects. For the human eye, this is often evaluated using a test for visual acuity. A commonly used method to evaluate and correct the resolving power of the human eye in the field of optometry is Retinoscopy.

Retinoscopy

1. Retinoscopy is an objective method used to obtain an estimation of a person’s refractive error and lens prescription.
2. During retinoscopy, an optometrist uses a device known as a retinoscope to shine light into a patient’s eye and observe the reflection (reflex) off the retina.
3. As the spot of light is moved across the pupil, the relative movement of the reflex is observed.
4. Subsequently, the optometrist manually places lenses over the patient’s eye using a temporary frame to “neutralize” the reflex.

Calculation of Resolving Power

1. The actual resolving power of the human eye with 20/20 vision is generally considered to be about one arc minute or 60 arc seconds.
2. This resolving power, however, is not solely dependent on the diameter of the pupil. In reality, it’s more complex and influenced by factors such as the eye’s diffraction limit.
3. The diffraction limit can be calculated using Rayleigh’s criterion, which states that the angular resolution is given by θ = λ/D, where λ is the wavelength of light (approximately 500 nm), and D is the diameter of the eye’s pupil (about 5 mm).
4. Using this formula, we calculate the diffraction-limited angular resolution of the eye to be around 0.008 degrees.
5. Thus, if the eye could resolve images at the diffraction limit, one would be able to distinguish lines in a printed pattern at a distance of about 15 meters.

Summary

The resolving power of the human eye is an important measure of visual acuity and can be measured using retinoscopy. The calculated diffraction limit provides an understanding of the maximum potential resolving power of the eye. However, the actual resolving power is influenced by additional factors, such as the health of the eye’s photoreceptor cells and neural processing in the brain.

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SAQ-5 : Explain the polarization of light by reflection and arrive at Brewster’s law from it.

Introduction

Polarization of light refers to the phenomenon where light waves oscillate in a particular direction. Polarization by reflection occurs when unpolarized light is reflected at a specific angle from a transparent medium, resulting in light that is predominantly polarized in a plane perpendicular to the plane of incidence. This phenomenon is described by Brewster’s Law.

Polarization by Reflection

1. When light reflects off a surface, the reflected light can become polarized if the angle of incidence is such that the reflected and refracted rays are perpendicular to each other.
2. At this specific angle, known as Brewster’s angle (θ_B​), the light reflected from the surface is perfectly polarized parallel to the surface.
3. The refracted ray and the reflected ray are at 90 degrees to each other.

Derivation of Brewster’s Law

1. According to Snell’s law, n₁​ sin θ_i​ = n_2​ sin θ_r​, where n₁​ and n₂​ are the refractive indices of the two media, and θ_i​ and θ_r​ are the angles of incidence and refraction, respectively.
2. At Brewster’s angle, the angle of reflection (θ_r​) and the angle of refraction add up to 90 degrees $$\theta_r + \theta_{refracted} = 90^\circ$$
3. Using trigonometric identities, we can say that $$\tan \theta_B = \frac{\sin \theta_{refracted}}{\cos \theta_{refracted}}$$
4. Since $$\sin \theta_{refracted} = n_1/n_2 \sin \theta_i$$ (from Snell’s Law) and \cos $$\theta_{refracted} = \cos(90^\circ – \theta_i) = \sin \theta_i$$ we get $$\tan \theta_B = n_2/n_1$$
5. Therefore, Brewster’s Law states that the Brewster’s angle (θ_B​) is such that the tangent of this angle is the ratio of the refractive indices of the two media, or $$\tan \theta_B = n_2/n_1$$

Summary

The polarization of light by reflection is a critical concept in optics, demonstrating how light can be polarized through reflection at a specific angle, known as Brewster’s angle. Brewster’s Law provides a quantitative description of this angle, stating that it is the angle at which light with electric fields perpendicular to the incident plane does not reflect, resulting in polarized reflected light. This principle has numerous applications in photography, glare reduction, and various optical technologies.

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SAQ-6 : Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroid’s.

Introduction

Polaroid sheets are optical filters that allow light waves of a specific polarization to pass through while blocking light waves of other polarizations. When light passes through a Polaroid, it becomes polarized, with the electric field oscillating in a single plane.

Principle of Operation

The principle behind the intensity of transmitted light through a Polaroid sheet is based on Malus’s Law. This law states that the intensity of polarized light transmitted through a polarizing filter varies as the cosine squared (cos⁡2cos2) of the angle between the light’s initial polarization direction and the axis of the filter.

Experiment Setup

In the experiment involving two crossed Polaroid sheets (polarizers), the first and third polarizers are aligned such that their polarization axes are perpendicular to each other. This setup initially allows no light to pass through to the observer. Introducing a third Polaroid sheet between the crossed pair introduces a variable that can change the intensity of the transmitted light.

Effect of Rotating the Middle Polaroid

1. Initial Alignment: When the middle Polaroid’s axis is aligned with either of the outer Polaroids, the intensity of light transmitted is maximized or minimized, depending on the alignment.
2. Rotation: As the middle Polaroid is rotated, the angle between the polarization direction of the light passing through the first Polaroid and the axis of the middle Polaroid changes.
3. Intensity Variation: According to Malus’s Law, the intensity of the light transmitted through the second (middle) Polaroid and observed after the third Polaroid varies as the cosine squared of the rotation angle of the middle Polaroid.
4. Maximum Transmission: The maximum transmission occurs when the middle Polaroid’s axis is at 45 degrees to the axes of the first and third Polaroids. This configuration allows some light to pass through all three Polaroids.

Summary

The intensity of light transmitted through a set of three Polaroid sheets, with the middle one being rotated, demonstrates a fundamental property of polarized light and its interaction with polarizing filters. The experiment vividly illustrates Malus’s Law and the dependence of light intensity on the angle of polarization, emphasizing the polarization angle and the cosine squared relationship as key factors in determining the transmitted light’s intensity.