Question:

You have learnt in the text how Huygens principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer:

1. First, recall that Huygens’ Principle states that every point on a wavefront acts as a source of secondary wavelets that travel in all directions.

2. Next, consider a point object placed in front of a plane mirror. The wavelets emitted by the object will reflect off the mirror and travel in a straight line away from it.

3. Since the wavelets are traveling away from the mirror, they will not converge at any point. This means that the image formed by the reflected wavelets will be a virtual image, which is an image that cannot be seen directly but can be seen by looking at the mirror.

4. Finally, since the wavelets are traveling away from the mirror in a straight line, the distance between the object and the virtual image will be the same as the distance between the object and the mirror.

Question:

Let us list some of the factors, which could possibly influence the speed of wave propagation: (i) nature of the source. (ii) direction of propagation. (iii) motion of the source and/or observer. (iv) wavelength. (v) intensity of the wave. On which of these factors, if any, does (a) The speed of light in vacuum (b) The speed of light in a medium (say, glass or water), depend?

Answer:

(a) The speed of light in vacuum does not depend on any of these factors.

(b) The speed of light in a medium (say, glass or water) depends on the wavelength and the intensity of the wave.

Question:

The 6563 A˚ Hα​ line emitted by hydrogen in a star is found to be red-shifted by 15 A˚. Estimate the speed with which the star is receding from the Earth.

Answer:

1. Convert the red-shift of 15 A˚ to km/s by using the formula v = c(Δλ/λ). Here, c is the speed of light (3 x 10^5 km/s) and λ is the wavelength of the 6563 A˚ Hα​ line (6.56 x 10^-7 m).

v = 3 x 10^5 km/s x (15 A˚/ 6.56 x 10^-7 m)

v = 6.5 x 10^5 km/s

2. Calculate the speed with which the star is receding from the Earth by subtracting the speed of light from the calculated velocity.

Speed of receding = 6.5 x 10^5 km/s - 3 x 10^5 km/s

Speed of receding = 3.5 x 10^5 km/s

Question:

Answer the following questions: (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? (b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? (d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily. (e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

Answer:

a) Doubling the width of the slit will increase the size of the central diffraction band and reduce its intensity.

b) Diffraction from each slit will combine to form the interference pattern in a double-slit experiment.

c) The bright spot at the center of the shadow is due to diffraction of light around the circular obstacle.

d) The light and sound waves can bend around obstacles, but the students are unable to see each other because the partition wall is too high for the light to travel over.

e) The ray optics assumption is still used in optical instruments because it provides a good approximation of the behavior of light. Diffraction effects are only significant for very small apertures or obstacles, and can be neglected in most cases.

Question:

Light of wavelength 5000 A˙ falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer:

1. Wavelength of the reflected light: 5000 A˙
2. Frequency of the reflected light: 6.00 x 10^14 Hz
3. Angle of incidence for the reflected ray to be normal to the incident ray: 90°

Question:

Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Answer:

1. Corpuscular theory, also known as the particle theory of light, states that light is composed of tiny particles called corpuscles. This theory predicts that the speed of light in a medium, such as water, should be greater than the speed of light in a vacuum because the corpuscles are slowed down by the medium.

2. However, experimental determination of the speed of light in water has shown that the speed of light in water is actually slower than the speed of light in a vacuum. This result does not confirm the prediction of the corpuscular theory.

3. An alternative picture of light that is consistent with experiment is the wave theory of light, which states that light is composed of waves. This theory predicts that the speed of light in a medium, such as water, should be slower than the speed of light in a vacuum because the waves are slowed down by the medium. This prediction is confirmed by experimental determination of the speed of light in water.

Question:

For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Answer:

Answer:

The Doppler formula for frequency shift in the case of light waves in vacuum is strictly identical for both situations because the speed of light is constant in a vacuum. This is regardless of the motion of either the source or the observer, since the speed of light is the same for both.

In the case of light travelling in a medium, the speed of light will depend on the properties of the medium, such as its refractive index. Therefore, the Doppler formula for frequency shift may not be strictly identical for the two situations, since the speed of light can be different depending on the medium.

Question:

Estimate the distance for which ray optics is good approximation for an aperture of 4mm and wavelength 400nm.

Answer:

Step 1: Calculate the numerical aperture (NA) of the aperture using the formula NA = n*sin(θ), where n is the refractive index of the medium and θ is the half angle of the aperture.

Step 2: Calculate the Rayleigh distance (D) using the formula D = (2*NA)/λ, where λ is the wavelength of the light.

Step 3: The Rayleigh distance is the distance for which ray optics is good approximation. Hence, the estimated distance for which ray optics is good approximation for an aperture of 4mm and wavelength 400nm is D.

Question:

In a double-slit experiment the angular width of a fringe is found to be 0.2o on a screen placed 1m away. The wavelength of light used is 600nm. What will be the angular width of the fringe is the entire experimental apparatus is immersed in water? Take refractive index of water to be 34​.

Answer:

Step 1: Calculate the angular width of the fringe in air. Angular width of the fringe in air = 0.2o

Step 2: Find the refractive index of air. Refractive index of air = 1

Step 3: Calculate the wavelength of light in water. Wavelength of light in water = (600 nm) / (refractive index of water) = (600 nm) / (1.34) = 447.76 nm

Step 4: Calculate the angular width of the fringe in water. Angular width of the fringe in water = (angular width of the fringe in air) x (wavelength of light in air / wavelength of light in water) = (0.2o) x (600 nm / 447.76 nm) = 0.267o

Question:

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Answer:

Step 1: Calculate the distance of the first minimum from the centre of the screen in metres. Distance = 2.5 mm/1000 = 0.0025 m

Step 2: Calculate the angular width of the slit using the formula θ = λ/d, where λ is the wavelength of the light and d is the distance of the first minimum from the centre of the screen. Angular width = λ/d = 500 nm/0.0025 m = 200,000 radians

Step 3: Calculate the width of the slit using the formula w = 2d tan θ, where d is the distance of the first minimum from the centre of the screen and θ is the angular width of the slit. Width of the slit = 2d tan θ = 2 x 0.0025 m x tan 200,000 radians = 0.0014 m = 1.4 mm

Question:

In Youngs double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?

Answer:

Answer:

1. Young’s double-slit experiment uses monochromatic light of wavelength λ.
2. The intensity of light at a point on the screen where path difference is λ is K units.
3. The intensity of light at a point where path difference is λ/3 will be K/3 units.

Question:

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:

(a) The shape of the wavefront in this case is a spherical wave.

(b) The shape of the wavefront in this case is a plane wave.

(c) The shape of the wavefront in this case is a spherical wave.

Question:

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:

a) Reflected light Wavelength: 589 nm Frequency: 5.08 x 10^14 Hz Speed: 3.00 x 10^8 m/s

b) Refracted light Wavelength: 442 nm Frequency: 6.86 x 10^14 Hz Speed: 3.00 x 10^8 m/s

Question:

What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Answer:

Step 1: Understand the terms. The Brewster angle is the angle at which the light is reflected from a surface and the refractive index of glass is 1.5.

Step 2: Calculate the Brewster angle. The Brewster angle for air to glass transition is given by the equation: Brewster angle = arctan(n2/n1), where n1 is the refractive index of air (1.0) and n2 is the refractive index of glass (1.5). Therefore, the Brewster angle for air to glass transition is arctan(1.5/1.0) = 53.1°.

Question:

Answer the following questions: (a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation. (b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Answer:

(a) A possible explanation for the slight shaking of the picture on our TV screen when a low flying aircraft passes overhead is that the aircraft is producing sound waves that are causing the air to vibrate. This vibration is then transmitted to the TV screen, causing it to shake.

(b) The justification of the principle of linear superposition of wave displacement is that it allows us to predict the intensity distribution of diffraction and interference patterns. This is because the principle states that the total displacement of a wave is the sum of the individual displacements of each wave that is present in the pattern. By knowing the individual displacements of each wave, we can then calculate the total displacement and thus predict the intensity distribution.

Question:

(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0×10^8ms^−1) (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:

a) The speed of light in glass is 2.0×10^8ms^−1.

b) The speed of light in glass is not independent of the colour of light. Red light travels slower in a glass prism than violet light.

Question:

A beam of light consisting of two wavelengths, 650nm and 520nm, 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 650nm.(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the two slits is 0.28mm and the screen is at a distance of 1.4m from the slits.

Answer:

(a) The distance of the third bright fringe on the screen from the central maximum for wavelength 650nm can be calculated using the equation:

d = (m * λ) / (2 * a)

where m is the order of the fringe, λ is the wavelength of the light, and a is the distance between the two slits.

For m = 3, λ = 650nm, and a = 0.28mm, the distance of the third bright fringe on the screen from the central maximum is:

d = (3 * 650nm) / (2 * 0.28mm)

d = 4.643mm

(b) The least distance from the central maximum where the bright fringes due to both the wavelengths coincide can be calculated using the equation:

d = (m * λ) / (2 * a)

where m is the order of the fringe, λ is the wavelength of the light, and a is the distance between the two slits.

For m = 1, λ1 = 650nm, λ2 = 520nm, and a = 0.28mm, the least distance from the central maximum where the bright fringes due to both the wavelengths coincide is:

d = (1 * (650nm + 520nm)) / (2 * 0.28mm)

d = 3.214mm

Question:

In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.

Answer:

1. The single slit diffraction pattern is a result of interference of waves passing through the slit.

2. When two waves of the same frequency and amplitude meet at a point, they will either constructively or destructively interfere with each other.

3. If the two waves are in phase, they will constructively interfere, resulting in a bright spot in the diffraction pattern.

4. If the two waves are out of phase, they will destructively interfere, resulting in a dark spot in the diffraction pattern.

5. When the slit is divided into two parts, the waves passing through the two parts will be out of phase. This will result in destructive interference, and the intensity of the diffraction pattern at the angle of nλ/a will be zero.

Question:

In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1∘. What is the spacing between the two slits?

Answer:

Given: Wavelength (λ) = 600 nm Angular width of a fringe (θ) = 0.1∘

Step 1: Convert the wavelength from nanometers to meters.

λ = 600 nm = 600 x 10-9 m

Step 2: Calculate the angular width of the fringe using the following equation:

θ = λ/d

Step 3: Solve for the distance (d) between the two slits:

d = λ/θ

d = (600 x 10-9 m)/(0.1∘)

d = 6 x 10-7 m

Question:

Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

Answer:

1. Calculate the distance between the two towers: 40 km.

2. Calculate the distance between the halfway point and one of the towers: 20 km.

3. Calculate the height of the hill halfway between the towers: 50 m.

4. Calculate the angle of diffraction using the formula θ = arcsin(H/D), where H is the height of the hill and D is the distance between the halfway point and one of the towers: θ = arcsin(50/20) = 14.04°.

5. Calculate the longest wavelength of radio waves that can be sent between the towers without appreciable diffraction effects using the formula λ = 2Dsinθ, where D is the distance between the towers and θ is the angle of diffraction: λ = 240sin(14.04°) = 22.76 m.

Therefore, the longest wavelength of radio waves that can be sent between the towers without appreciable diffraction effects is 22.76 m.

Question:

In a Youngs 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:

1. Convert the distance between the central bright fringe and the fourth bright fringe to meters: 1.2 cm = 0.012 m

2. Calculate the difference in the path lengths between the two fringes: 1.4 m - (0.28 mm / 1000) = 1.39972 m

3. Use the formula for Young’s double slit experiment to calculate the wavelength: Wavelength = (Difference in Path Lengths) / (Number of Fringes - 1)

Wavelength = (1.39972 m) / (3 - 1)

Wavelength = 0.46657 m

4. Convert the wavelength from meters to nanometers: 0.46657 m = 466.57 nm

The wavelength of light used in the experiment is 466.57 nm.