Optics

An electromagnetic wave $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$, travelling in air is incident on the x – z plane of a glass slab. The glass slab has a refractive index of 1.5. Which one among the following is the correct direction of propagation of the transmitted electromagnetic wave ?

[amp_mcq option1=”$0.4\hat{i} + 0.2\sqrt{21}\hat{j}$” option2=”$0.4\hat{i} – 0.2\sqrt{21}\hat{j}$” option3=”$0.4\hat{i} + \sqrt{21}\hat{j}$” option4=”$0.4\hat{i} – \sqrt{21}\hat{j}$” correct=”option1″]

Correct Answer: A

– The incident electromagnetic wave is given by $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$. The wave vector $\vec{k}_i$ is the vector part multiplying $\vec{r}=(x,y,z)$ in the exponent (ignoring $k_0$). So, the incident wave vector in air is proportional to $3\hat{i} + 4\hat{j}$. We can write $\vec{k}_i = k_0 (3\hat{i} + 4\hat{j})$.
– The magnitude of the incident wave vector is $|\vec{k}_i| = k_0 \sqrt{3^2 + 4^2} = 5 k_0$. In air, the refractive index $n_1 = 1$, so $|\vec{k}_i| = n_1 (\omega/c) = \omega/c$. This implies $k_0 = \omega/(5c)$.
– The wave is incident on the x-z plane, which is the plane where $y=0$. The normal to this plane is along the $\hat{j}$ direction. Let’s assume the glass slab is in the region $y>0$, and the air is in $y<0$. The incident wave has $k_{iy} = 4k_0 > 0$, so it is travelling towards the interface $y=0$ from the region $y<0$. The normal vector pointing into the glass is $\hat{j}$. - According to Snell's law for wave vectors, the component of the wave vector parallel to the interface is conserved. The interface is the x-z plane, so the components parallel to the plane are the x and z components. - Incident wave vector $\vec{k}_i = 3k_0 \hat{i} + 4k_0 \hat{j} + 0\hat{k}$. The parallel component is $k_{ix} \hat{i} + k_{iz} \hat{k} = 3k_0 \hat{i}$. - The transmitted wave vector in glass is $\vec{k}_t = k_{tx} \hat{i} + k_{ty} \hat{j} + k_{tz} \hat{k}$. - Conservation of parallel component: $k_{tx} = k_{ix} = 3k_0$ and $k_{tz} = k_{iz} = 0$. So $\vec{k}_t = 3k_0 \hat{i} + k_{ty} \hat{j}$. - The magnitude of the transmitted wave vector is $|\vec{k}_t| = n_2 (\omega/c) = n_2 n_1^{-1} |\vec{k}_i| = 1.5 \times 1^{-1} \times 5k_0 = 7.5 k_0$. - $|\vec{k}_t|^2 = k_{tx}^2 + k_{ty}^2 = (3k_0)^2 + k_{ty}^2 = (7.5 k_0)^2$. - $9k_0^2 + k_{ty}^2 = 56.25 k_0^2$. - $k_{ty}^2 = (56.25 - 9) k_0^2 = 47.25 k_0^2 = \frac{189}{4} k_0^2$. - $k_{ty} = \pm \sqrt{\frac{189}{4}} k_0 = \pm \frac{3\sqrt{21}}{2} k_0$. - Since the wave is transmitted from air ($y<0$) into glass ($y>0$), the y-component of the transmitted wave vector must be positive (pointing into the glass region). So $k_{ty} = \frac{3\sqrt{21}}{2} k_0$.
– The transmitted wave vector is $\vec{k}_t = 3k_0 \hat{i} + \frac{3\sqrt{21}}{2} k_0 \hat{j}$. The direction of propagation is given by the unit vector in the direction of $\vec{k}_t$.
– The direction is proportional to $3\hat{i} + \frac{3\sqrt{21}}{2} \hat{j}$.
– Let’s check the options. Option A is $0.4\hat{i} + 0.2\sqrt{21}\hat{j} = \frac{2}{5}\hat{i} + \frac{\sqrt{21}}{5}\hat{j}$.
– Check if $(3, \frac{3\sqrt{21}}{2})$ is proportional to $(\frac{2}{5}, \frac{\sqrt{21}}{5})$. The ratio of x-components is $3 / (2/5) = 15/2$. The ratio of y-components is $(\frac{3\sqrt{21}}{2}) / (\frac{\sqrt{21}}{5}) = \frac{3\sqrt{21}}{2} \times \frac{5}{\sqrt{21}} = \frac{15}{2}$.
– Since the ratios are equal, the vectors are proportional. The direction of propagation of the transmitted wave is proportional to $0.4\hat{i} + 0.2\sqrt{21}\hat{j}$.

When an electromagnetic wave passes from one medium to another, its frequency remains constant, but its wavelength and speed change, which in turn changes the magnitude of the wave vector ($|\vec{k}| = 2\pi/\lambda = \omega/v = n\omega/c$). Snell’s law for refraction of light is a consequence of the conservation of the wave vector component parallel to the interface.

Which one of the following experiments explains the wave nature of electromagnetic radiation ?

[amp_mcq option1=”Black-body radiation” option2=”Photoelectric effect” option3=”Rutherford’s scattering experiment” option4=”Diffraction of light” correct=”option4″]

Diffraction of light is a phenomenon that demonstrates the wave nature of electromagnetic radiation.

Diffraction occurs when waves bend around obstacles or spread out through openings. This bending and spreading are characteristic wave behaviors.

Black-body radiation (explained by Planck) and the photoelectric effect (explained by Einstein) are phenomena that provided evidence for the particle nature of light (photons). Rutherford’s scattering experiment demonstrated the nuclear model of the atom, not the wave nature of light. Other phenomena demonstrating the wave nature of light include interference and polarization.

The refractive index of a prism made of flint glass is

[amp_mcq option1=”the same for all wavelengths in white light” option2=”higher for red light than for violet light” option3=”higher for violet light than for red light” option4=”highest for green and yellow lights and lowest for violet and red lights” correct=”option3″]

The refractive index of a material varies with the wavelength of light. For most transparent materials, including flint glass, the refractive index is higher for shorter wavelengths (like violet light) than for longer wavelengths (like red light). This phenomenon is known as dispersion.

– Shorter wavelengths are bent more than longer wavelengths when passing through a prism. This is why white light is dispersed into its constituent colours.
– Higher refractive index means light bends more, which corresponds to shorter wavelengths (violet end of the spectrum).
– Lower refractive index means light bends less, which corresponds to longer wavelengths (red end of the spectrum).

The relationship between refractive index (n) and wavelength (λ) is described by Cauchy’s equation, which states that for a given material, n is approximately proportional to A + B/λ², where A and B are constants. As wavelength (λ) increases, the refractive index (n) decreases. Violet light has a shorter wavelength than red light.

On a very hot day, we often see shimmering wavy lines near the ground. It is due to

[amp_mcq option1=”dispersion of light” option2=”refraction of light” option3=”reflection of light” option4=”total internal reflection of light” correct=”option2″]

The shimmering wavy lines observed near the ground on a hot day are caused by the refraction of light as it passes through layers of air with varying temperatures and densities, leading to variations in the refractive index.

On hot days, the ground heats the air just above it, creating a significant temperature gradient. Hot air is less dense than cooler air and has a slightly lower refractive index. Light rays passing through these layers of air with different refractive indices are continuously bent (refracted). As convection causes these pockets of hot and cool air to move and mix, the light rays are constantly being bent in different directions, making distant objects appear distorted, blurred, and wavy or shimmering. This is a form of atmospheric refraction.

This phenomenon is closely related to the formation of mirages, which are more pronounced examples of atmospheric refraction occurring when light is strongly bent as it passes from cooler, denser air into warmer, less dense air near a hot surface. The shimmering effect is a less extreme, constantly changing manifestation of the same principle of refraction due to thermal gradients in the atmosphere.

The appearance of a rainbow in the sky after a rain shower is due to

[amp_mcq option1=”diffraction and refraction of light in water droplets” option2=”total internal reflection of light in water droplets only” option3=”refraction of light in water droplets only” option4=”both total internal reflection and refraction of light in water droplets” correct=”option4″]

The formation of a rainbow is due to the combined effects of refraction and total internal reflection (or simply internal reflection) of sunlight as it passes through spherical rain droplets.

When sunlight enters a rain droplet, it is refracted (bends) and dispersed (separated into different colors because different wavelengths refract at slightly different angles). This dispersed light then travels to the back inner surface of the droplet, where it undergoes internal reflection. Finally, as the light exits the droplet, it is refracted again, further separating the colors and sending them towards the observer’s eye at specific angles depending on the wavelength, creating the arc of the rainbow. For the primary rainbow, the reflection at the back surface is usually total internal reflection, but even if not strictly TIR for all angles, it is an internal reflection that directs the light back towards the observer.

A double rainbow occurs due to light undergoing two internal reflections within the droplet. Diffraction effects can lead to phenomena like supernumerary bows, which are faint bands seen just inside the primary rainbow, but the primary and secondary rainbows themselves are primarily explained by the principles of refraction and internal reflection.

Which one of the following statements is true about the appearance of colour of the Sun in the sky?

[amp_mcq option1=”At sunset (or sunrise), sunlight travels more distance in the atmosphere and higher frequency radiations scatter away resulting into red sunset (or sunrise).” option2=”At sunset (or sunrise), sunlight travels least distance in the atmosphere and higher frequency radiations scatter away resulting into red sunset (or sunrise).” option3=”At noon, sunlight travels least distance in the atmosphere and relatively less amount of sunlight is scattered and therefore the Sun appears reddish.” option4=”At noon, sunlight travels least distance in the atmosphere and larger amount of sunlight is scattered and therefore the Sun appears reddish.” correct=”option1″]

At sunset or sunrise, sunlight travels a longer path through the Earth’s atmosphere. This longer path means more scattering of sunlight occurs. According to Rayleigh scattering, shorter wavelengths (blue and violet light) are scattered much more effectively than longer wavelengths (red and orange light). As the blue and violet light is scattered away from the line of sight, the light that reaches the observer’s eyes is richer in longer wavelengths, making the Sun appear reddish.

The color of the sky and the Sun’s appearance depend on how sunlight is scattered by molecules in the atmosphere. Rayleigh scattering dictates that scattering is inversely proportional to the fourth power of the wavelength (scattering ∝ 1/λ⁴). This means shorter wavelengths scatter significantly more than longer ones. The path length of sunlight through the atmosphere is the key factor explaining the difference between the Sun’s appearance at noon (shortest path, less scattering) and at sunrise/sunset (longest path, more scattering of short wavelengths).

At noon, when the Sun is high in the sky, the path through the atmosphere is shortest. Less scattering occurs, and although blue light is still scattered away (making the sky blue), enough of the other wavelengths remain in the direct beam that the Sun appears white or slightly yellow.

Which one of the following waves has the longest wavelength?

[amp_mcq option1=”Visible light” option2=”Ultraviolet radiation” option3=”Infra-red radiation” option4=”X-rays” correct=”option3″]

Electromagnetic waves are ordered by their frequency and wavelength in the electromagnetic spectrum. From longest wavelength to shortest (and lowest frequency to highest), the order is typically: Radio waves, Microwaves, Infra-red radiation, Visible light, Ultra-violet radiation, X-rays, Gamma rays. Among the given options, Infra-red radiation has the longest wavelength.

Wavelength and frequency are inversely proportional for electromagnetic waves traveling at the speed of light (c = fλ). Longer wavelength corresponds to lower frequency, and shorter wavelength corresponds to higher frequency.

Visible light occupies a narrow band in the spectrum. Infra-red radiation is just below visible light in frequency (longer wavelength), while Ultra-violet radiation and X-rays are above visible light in frequency (shorter wavelengths).

A ray of light strikes a glass slab (from air). Which one of the following is NOT correct?

[amp_mcq option1=”The speed of light in glass is less than that in air” option2=”The frequency of light in glass is same as that in air” option3=”The wavelength of light in glass is less than that in air” option4=”The frequency of light in glass is less than that in air” correct=”option4″]

When light passes from one medium (air) to another medium (glass), its speed decreases, and its wavelength decreases. However, the frequency of the light wave is determined by the source and remains constant as it passes from one medium to another. Therefore, the statement “The frequency of light in glass is less than that in air” is incorrect.

The speed of light (v), frequency (f), and wavelength (λ) are related by the equation v = fλ. When light enters a denser medium like glass from air, its speed (v) decreases, and since the frequency (f) remains constant, the wavelength (λ) must decrease proportionally (λ = v/f).

The change in speed and wavelength causes the bending of light, known as refraction. The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.

The scattering of a beam of light by colloidal particles is termed as:

[amp_mcq option1=”Thomson effect” option2=”Raman effect” option3=”Tyndall effect” option4=”Compton effect” correct=”option3″]

The scattering of a beam of light by colloidal particles is termed as the Tyndall effect.

The Tyndall effect is the scattering of light as a light beam passes through a colloid. The individual suspension particles scatter and reflect light, making the beam visible. This effect is exhibited by colloids and fine suspensions, but not by true solutions.

The Thomson effect and Compton effect relate to the interaction of radiation (light or X-rays) with electrons and charged particles, respectively, but not specifically the scattering by colloidal particles. The Raman effect involves inelastic scattering of light by molecules, resulting in a change in frequency.