GSEB Class 12 Physics Solutions Chapter 10 Wave Optics

GSEB Class 12 Physics Wave Optics Text Book Questions And Answers

Question 1. 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) When light reflects, its wavelength, frequency, and speed do not change. So, for the reflected light, the wavelength is 589 nm, the speed is \(3 \times 10^8 \text{ m/s}\), and the frequency is \( \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.09 \times 10^{14} \text{ Hz} \).
(b) When light refracts (enters water), its frequency stays the same as in air, so \( \nu = 5.09 \times 10^{14} \text{ Hz} \). However, its wavelength and speed change. The new wavelength is \( \lambda' = \frac{\lambda}{n} = \frac{589 \text{ nm}}{1.33} \approx 443 \text{ nm} \), and the new speed is \( V' = \frac{c}{n} = \frac{3 \times 10^8 \text{ m/s}}{1.33} \approx 2.26 \times 10^8 \text{ m/s} \).
In simple words: When light bounces off a surface, its speed, color, and wavelength stay the same. When it enters a new material like water, its speed and wavelength change, but its color (frequency) remains constant.

🎯 Exam Tip: Remember that reflection doesn't change light's properties, while refraction changes its speed and wavelength but not its frequency.

Question 2. 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 wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) Spherical
(b) Plane
(c) Plane
In simple words: Light from a point spreads in spheres. A lens can make these spheres flat. Light from very far away looks like flat sheets.

🎯 Exam Tip: Knowing the standard wavefront shapes for different light sources and optical setups is crucial for understanding wave optics diagrams.

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is \(3.0 \times 10^8 \text{ m/s}^{-1}\))
(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 \(v\) can be found using the formula \(v = \frac{c}{n}\), where \(c\) is the speed of light in vacuum and \(n\) is the refractive index. So, \(v = \frac{3 \times 10^8 \text{ m/s}}{1.5} = 2 \times 10^8 \text{ m/s}\).
(b) No, the speed of light in glass *does* depend on its color. This is because the refractive index of glass is different for different colors. Violet light travels slower than red light in a glass prism, meaning it bends more.
In simple words: Light slows down when it enters glass. Different colors of light travel at different speeds in glass; violet light is slower than red light.

🎯 Exam Tip: Remember that the speed of light in a medium is inversely proportional to the refractive index, and refractive index varies with wavelength (dispersion).

Question 4. In a Young's 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:
Given:
Slit separation, \(d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}\)
Screen distance, \(D = 1.4 \text{ m}\)
Order of bright fringe, \(n = 4\)
Distance of \(4^{th}\) bright fringe from center, \(y_n = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}\)
The formula for the position of the \(n^{th}\) bright fringe in Young's double-slit experiment is:
\(y_n = \frac{n \lambda D}{d}\)
To find the wavelength \( \lambda \), we rearrange the formula:
\( \lambda = \frac{y_n d}{n D} \)
Now, substitute the given values:
\[ \lambda = \frac{(1.2 \times 10^{-2} \text{ m}) \times (0.28 \times 10^{-3} \text{ m})}{4 \times (1.4 \text{ m})} \]
\[ \lambda = \frac{3.36 \times 10^{-6}}{5.6} \]
\[ \lambda = 0.6 \times 10^{-6} \text{ m} = 6 \times 10^{-7} \text{ m} = 600 \text{ nm} \]
The wavelength of light used in the experiment is 600 nm.
In simple words: We used the given values for slit distance, screen distance, and fringe position to calculate the wavelength of light with a specific formula.

🎯 Exam Tip: Ensure correct unit conversions (mm to m, cm to m) and careful substitution into the Young's double-slit fringe position formula.

Question 5. In Young's double-slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen where path difference is \( \lambda \), is K units. What is the intensity of light at a point where path difference is \( \frac{\lambda}{3} \)?
Answer:
Let the intensity from each slit be \(I_{single}\). The total intensity \(I\) at a point due to interference of two waves is given by:
\(I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi\)
Assuming identical sources, \(I_1 = I_2 = I_{single}\). So, the intensity formula simplifies to:
\(I = 2I_{single}(1 + \cos\phi)\)

The phase difference \( \phi \) is related to the path difference \( \Delta x \) by \( \phi = \frac{2\pi}{\lambda} \Delta x \).

When the path difference is \( \lambda \):
\( \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi \)
The intensity \(I\) at this point is:
\(I = 2I_{single}(1 + \cos(2\pi)) = 2I_{single}(1 + 1) = 4I_{single}\)
We are told that this intensity is \(K\) units. So, \(K = 4I_{single}\).

When the path difference is \( \frac{\lambda}{3} \):
\( \phi' = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \)
The intensity \(I'\) at this point is:
\(I' = 2I_{single}(1 + \cos(\frac{2\pi}{3})) = 2I_{single}(1 - \frac{1}{2}) = 2I_{single}(\frac{1}{2}) = I_{single}\)
Since \(K = 4I_{single}\), we know \(I_{single} = \frac{K}{4}\).
Therefore, \(I' = \frac{K}{4}\).
In simple words: When the path difference is one full wavelength, the intensity is K. When the path difference is one-third of a wavelength, the intensity becomes K/4.

🎯 Exam Tip: Remember the relationship between path difference, phase difference, and intensity in interference patterns, especially for common phase values like \(2\pi\) and \(2\pi/3\).

Question 6. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young's double-slit experiment. (Given \(d = 2 \times 10^{-3} \text{ m}\), \(D = 1.2 \text{ m}\))
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer:
(a) Given:
Wavelength, \(\lambda = 650 \text{ nm} = 650 \times 10^{-9} \text{ m}\)
Slit separation, \(d = 2 \times 10^{-3} \text{ m}\)
Screen distance, \(D = 1.2 \text{ m}\)
Order of bright fringe, \(n = 3\)
The distance of the \(n^{th}\) bright fringe from the central maximum is given by:
\(y_n = \frac{n \lambda D}{d}\)
For the third bright fringe (\(n=3\)):
\[ y_3 = \frac{3 \times (650 \times 10^{-9} \text{ m}) \times (1.2 \text{ m})}{2 \times 10^{-3} \text{ m}} \]
\[ y_3 = \frac{2340 \times 10^{-9}}{2 \times 10^{-3}} \text{ m} = 1170 \times 10^{-6} \text{ m} = 1.17 \times 10^{-3} \text{ m} = 1.17 \text{ mm} \]
(b) For bright fringes of two wavelengths (\(\lambda_1\) and \(\lambda_2\)) to coincide at a distance \(y\) from the central maximum, the condition is:
\( n \lambda_1 = m \lambda_2 \)
Given \(\lambda_1 = 650 \text{ nm}\) and \(\lambda_2 = 520 \text{ nm}\).
\( n \times 650 \text{ nm} = m \times 520 \text{ nm} \)
Dividing by 10, we get \(65n = 52m\).
Dividing by 13, we get \(5n = 4m\).
The smallest whole number values that satisfy this are \(n=4\) and \(m=5\).
So, the \(4^{th}\) bright fringe of 650 nm light coincides with the \(5^{th}\) bright fringe of 520 nm light.
The least distance \(y\) from the central maximum where this happens is:
\[ y = \frac{n \lambda_1 D}{d} = \frac{4 \times (650 \times 10^{-9} \text{ m}) \times (1.2 \text{ m})}{2 \times 10^{-3} \text{ m}} \]
\[ y = \frac{3120 \times 10^{-9}}{2 \times 10^{-3}} \text{ m} = 1560 \times 10^{-6} \text{ m} = 1.56 \times 10^{-3} \text{ m} = 1.56 \text{ mm} \]
In simple words: We first calculated the position of a specific bright fringe for one color. Then, we found the closest point where bright fringes from two different colors would align by finding the smallest common multiple of their fringe orders.

🎯 Exam Tip: When dealing with coinciding fringes from multiple wavelengths, always set the fringe position formulas equal (\(n\lambda_1 = m\lambda_2\)) and find the smallest integer values for n and m.

Question 7. In a double-slit experiment, the angular width of a fringe is found to be \(0.2^\circ\) on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be \( \frac{4}{3} \).
Answer:
The angular width of a fringe (\(\theta\)) in a double-slit experiment is given by \( \theta = \frac{\lambda}{d} \) (when expressed in radians), where \( \lambda \) is the wavelength of light and \(d\) is the slit separation.
When the entire experimental setup is placed in water, the wavelength of light changes. The new wavelength \( \lambda' \) in water is related to the wavelength in air \( \lambda \) and the refractive index of water \( n \) by the formula: \( \lambda' = \frac{\lambda}{n} \).
The slit separation \(d\) does not change.
So, the new angular width \( \theta' \) in water will be:
\( \theta' = \frac{\lambda'}{d} = \frac{(\lambda/n)}{d} = \frac{1}{n} \left(\frac{\lambda}{d}\right) \)
Since \( \frac{\lambda}{d} \) is the original angular width \( \theta \), we have:
\( \theta' = \frac{\theta}{n} \)
Given:
Original angular width, \( \theta = 0.2^\circ \)
Refractive index of water, \( n = \frac{4}{3} \)
Now, we calculate the new angular width \( \theta' \):
\[ \theta' = \frac{0.2^\circ}{(4/3)} = \frac{0.2 \times 3}{4} = \frac{0.6}{4} = 0.15^\circ \]
Therefore, the angular width of the fringe when the apparatus is immersed in water will be \(0.15^\circ\).
In simple words: When a double-slit experiment is put in water, the light's wavelength gets shorter. Because of this, the light waves bend less, making the angular width of the fringes narrower.

🎯 Exam Tip: Remember that angular width is inversely proportional to the refractive index of the medium, as wavelength changes but slit separation does not.

Question 8. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5)
Answer:
Brewster's angle (\(i_p\)) is the specific angle of incidence where light reflected from a surface becomes fully polarized. It is linked to the refractive index (\(n\)) of the material by Brewster's law, which states: \( \tan(i_p) = n \).
Given:
Refractive index of glass, \( n = 1.5 \).
Using the formula:
\( \tan(i_p) = 1.5 \)
To find the angle \(i_p\), we take the inverse tangent:
\( i_p = \tan^{-1}(1.5) \)
\( i_p \approx 56.3^\circ \).
So, the Brewster angle for light going from air into glass is about \(56.3^\circ\).
In simple words: Brewster's angle is a special angle where reflected light is completely polarized. For glass, this angle is about 56.3 degrees.

🎯 Exam Tip: Remember Brewster's law \( \tan(i_p) = n \) and its application for calculating the polarization angle.

Question 9. Light of wavelength 5000 Å 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:
First, we convert the wavelength from Ångströms to meters: \( \lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m} \).
The speed of light in a vacuum is \( c = 3 \times 10^8 \text{ m/s} \).

For reflected light:
When light reflects from a surface, its basic properties like wavelength and frequency do not change. So, the reflected light will have the same wavelength of \( 5000 \text{ Å} \) (or \( 5 \times 10^{-7} \text{ m} \)).
The frequency (\( \nu \)) is calculated using the formula \( \nu = \frac{c}{\lambda} \):
\[ \nu = \frac{3 \times 10^8 \text{ m/s}}{5 \times 10^{-7} \text{ m}} = 6 \times 10^{14} \text{ Hz} \]

For the reflected ray to be normal to the incident ray:
This means the angle between the incoming ray and the outgoing reflected ray is \(90^\circ\).
According to the law of reflection, the angle at which light hits the surface (angle of incidence, \(i\)) is equal to the angle at which it bounces off (angle of reflection, \(r\)). So, \(i = r\).
If the incident and reflected rays are at \(90^\circ\) to each other, then \( i + r = 90^\circ \).
Since \(i = r\), we can substitute \(r\) with \(i\):
\( 2i = 90^\circ \)
\( i = \frac{90^\circ}{2} = 45^\circ \)
Therefore, the light must strike the surface at an angle of \(45^\circ\) for the reflected ray to be perpendicular to the incident ray.
In simple words: When light reflects, its color and wavelength stay the same. If the reflected light is at a right angle to the incoming light, the light must have hit the surface at a 45-degree angle.

🎯 Exam Tip: Remember that reflection preserves wavelength and frequency. Also, apply the laws of reflection (angle of incidence equals angle of reflection) carefully when solving geometrical optics problems.

Question 10. Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer:
Ray optics (which treats light as rays) is a good way to understand light behavior when diffraction (light bending around obstacles) is not a big effect. This is true for distances less than or equal to the Fresnel distance (\(Z_F\)).
The formula for the Fresnel distance is:
\( Z_F = \frac{a^2}{\lambda} \)
Where:
\(a\) is the size of the opening (aperture), which is the diameter in this question.
\( \lambda \) is the wavelength of the light.
Given values:
Aperture size, \(a = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}\).
Wavelength, \( \lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} \).
Let's put these values into the formula:
\[ Z_F = \frac{(4 \times 10^{-3} \text{ m})^2}{400 \times 10^{-9} \text{ m}} \]
\[ Z_F = \frac{16 \times 10^{-6} \text{ m}^2}{400 \times 10^{-9} \text{ m}} \]
\[ Z_F = \frac{16 \times 10^{-6}}{0.4 \times 10^{-6}} \text{ m} \]
\[ Z_F = \frac{16}{0.4} \text{ m} = 40 \text{ m} \]
So, ray optics works well for distances up to about 40 meters when using this aperture and wavelength.
In simple words: Ray optics is good for short distances where light behaves like straight lines. The maximum distance for this is called the Fresnel distance, which we calculated to be 40 meters.

🎯 Exam Tip: The Fresnel distance formula helps determine when to use ray optics versus wave optics. Remember \(Z_F = a^2/\lambda\).

Question 11. The 6563 Å \(H_\alpha\) line emitted by hydrogen in a star is found to be redshifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
Answer:
A redshift means that the wavelength of light from the star has gotten longer. This tells us the star is moving away from us (receding). We use the Doppler effect formula for light to figure out its speed.
The formula for Doppler shift when an object is moving away is:
\( \frac{\Delta\lambda}{\lambda} = \frac{v}{c} \)
Here:
\( \Delta\lambda \) is how much the wavelength changed (the redshift).
\( \lambda \) is the original wavelength of the light the star sent out.
\(v\) is the speed at which the star is moving away from Earth.
\(c\) is the speed of light, which is \(3 \times 10^8 \text{ m/s}\).
Given:
Original wavelength, \( \lambda = 6563 \text{ Å} = 6563 \times 10^{-10} \text{ m} \).
Change in wavelength (redshift), \( \Delta\lambda = 15 \text{ Å} = 15 \times 10^{-10} \text{ m} \).
We need to find \(v\), so we change the formula to:
\( v = c \frac{\Delta\lambda}{\lambda} \)
Now, we put in the numbers:
\[ v = (3 \times 10^8 \text{ m/s}) \times \frac{15 \times 10^{-10} \text{ m}}{6563 \times 10^{-10} \text{ m}} \]
\[ v = (3 \times 10^8 \text{ m/s}) \times \frac{15}{6563} \]
\[ v \approx 6.86 \times 10^5 \text{ m/s} \]
So, the star is moving away from Earth at a speed of about \(6.86 \times 10^5 \text{ m/s}\).
In simple words: When light from a star shifts to redder colors, it means the star is moving away. We use a special formula with the change in light's color to calculate how fast it is moving.

🎯 Exam Tip: Remember that redshift indicates recession (moving away), and blueshift indicates approach (moving closer). Use the correct Doppler shift formula \( \frac{\Delta\lambda}{\lambda} = \frac{v}{c} \) for these calculations, making sure to keep units consistent.

Question 12. 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:
Newton's corpuscular theory suggested that light is made of tiny particles. It explained refraction by saying that these light particles are pulled towards a denser material, like water, when they enter it. This pulling force would make the light particles speed up, so the theory predicted that light travels *faster* in water than in a vacuum (\(v > c\)).

However, this prediction was *not* proven by experiments. Experiments, such as those done by Foucault, clearly showed that light actually travels *slower* in water than in a vacuum (\(v < c\)).

The wave picture of light, particularly Huygens' wave theory, matches what experiments tell us. Wave theory correctly explains that light waves slow down when they move from a less dense medium (like air) to a denser medium (like water), which explains refraction and the measured speed of light.
\[ \frac{c}{\sin i} = \frac{v}{\sin r} \]
\[ v = c \frac{\sin r}{\sin i} = \frac{c}{n} \]
Since \(n > 1\) for water, \(v < c\).
In simple words: The old particle theory of light said light speeds up in water, but experiments showed it slows down. The wave theory correctly explains why light slows down in water.

🎯 Exam Tip: Understand the historical context of light theories. Remember that Newton's corpuscular theory incorrectly predicted \(v_{medium} > c\), while wave theory correctly predicts \(v_{medium} < c\), aligning with experimental evidence.

Question 13. You have learned how Huygens' principle explains the laws of reflection and refraction. Use this principle to directly show that a point object in front of a plane mirror forms a virtual image, and the image is as far behind the mirror as the object is in front.
Answer:
To show how a plane mirror forms an image using Huygens' principle:
Imagine a point object \(O\) placed in front of a plane mirror. Spherical wavefronts spread out from this object.
When these wavefronts hit the mirror, each point on the mirror acts as a secondary source of spherical wavelets (according to Huygens' principle).
These wavelets combine to form the reflected wavefront.
Consider a spherical wavefront from the object \(O\) hitting the mirror. As portions of this wavefront reach the mirror, they reflect.
For any part of the wavefront that has not yet reached the mirror, it would have continued to expand in the absence of the mirror.
The reflected wavefront appears to originate from a virtual point \(I\) behind the mirror.
By geometric construction using Huygens' principle (comparing the path length of incident and reflected wavelets), it can be shown that the virtual image \(I\) is formed at the same distance behind the mirror as the object \(O\) is in front of it, and it is laterally inverted. The reflected wavefronts appear to diverge from this virtual image point.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समतल दर्पण के सामने रखी एक बिंदु वस्तु से परावर्तन को दर्शाता है। वस्तु (O) से गोलाकार तरंगें दर्पण तक पहुँचती हैं। दर्पण से परावर्तित तरंगें दर्पण के पीछे एक आभासी बिंदु (I) से आती हुई प्रतीत होती हैं। तरंगों के परावर्तन और फैलने का यह पैटर्न बताता है कि आभासी प्रतिबिंब दर्पण के पीछे उतनी ही दूरी पर बनता है जितनी वस्तु दर्पण के सामने है।
In simple words: Huygens' principle shows that light waves from an object hitting a mirror bounce off and seem to come from a virtual image behind the mirror, exactly as far away as the object is in front.

🎯 Exam Tip: Be able to draw and explain the reflection of a spherical wavefront from a plane mirror using Huygens' construction to demonstrate image formation.

Question 14. 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) Relative 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, and
(b) the speed of light in a medium (say, glass or water), depend?
Answer:
(a) The speed of light in a vacuum (represented by \(c\)) is a fundamental constant in physics. It does *not* depend on any of the listed factors, including:
- The type of light source.
- The relative movement between the source and the observer (this is a core idea of special relativity).
- The light's wavelength, frequency, or intensity.
It always remains \(3 \times 10^8 \text{ m/s}\) in empty space.

(b) The speed of light in a medium (like glass or water) *does not* depend on:
- The source's nature.
- The direction it travels in (if the medium is the same everywhere and in all directions).
- The movement of the source.
- The intensity of the light (unless it's extremely strong, like a powerful laser, which can cause special effects).
However, the speed of light in a medium *does* depend on these factors:
- The **wavelength** (or color) of the light: This is why different colors bend differently, a process called dispersion.
- The **nature of the medium itself**: Different materials slow light down by different amounts, described by their refractive index.
In simple words: Light always travels at the same super-fast speed in empty space, no matter what. But in materials like water, its speed changes depending on its color and what the material is made of.

🎯 Exam Tip: Clearly distinguish between the speed of light in vacuum (constant) and in a medium (dependent on wavelength and medium properties). This is a key concept in wave optics.

Question 15. For sound waves, the Doppler formula for frequency shift is slightly different depending on whether the source is moving and the observer is at rest, or vice-versa. However, for light waves in vacuum, the exact Doppler formulas for these two situations are strictly identical. Explain why this is so. Would the formulas still be strictly identical for light travelling in a medium?
Answer:
**Explanation for light in vacuum:**
The Doppler effect for sound waves depends on how the source and observer are moving *compared to the air (or other medium)* carrying the sound. This means it matters if the source moves towards a still listener, or if the listener moves towards a still source.

But for light in empty space (a vacuum), there's no medium. Einstein's theory of special relativity states that the speed of light in a vacuum is always the same for everyone, no matter how fast they are moving or how fast the light source is moving. This means only the *speed at which the source and observer are moving relative to each other* is important. So, the Doppler formulas for light in a vacuum are the same whether the source is moving or the observer is moving, as long as their relative speed is the same.

**For light travelling in a medium:**
No, the formulas would *not* be exactly the same for light moving through a material like water or glass. When light goes through a medium, that medium acts like a reference point. The speed of light in that medium is specific to that medium. Because of this, the Doppler effect for light in a medium would be more like that for sound waves: the way the source or observer moves *relative to the medium* would make a difference in the formulas.
In simple words: Light in space doesn't care who is moving, just how fast they move apart or together. But light in water acts more like sound in air, where it matters if the source or the listener is moving relative to the water.

🎯 Exam Tip: Understand that the absolute speed of light in vacuum and the absence of an 'ether' are central to why relativistic Doppler effect for light is symmetrical, unlike the classical Doppler effect for sound in a medium.

Question 16. In a double-slit experiment using light of wavelength 600 nm, the angular width of a fringe is found to be \(0.1^\circ\) on a distant screen. What is the spacing between the two slits?
Answer:
In a double-slit experiment, the angular width (\(\Delta\theta\)) of a fringe (how wide it appears in terms of angle) is calculated using the formula:
\( \Delta\theta = \frac{\lambda}{d} \)
(This formula requires \( \Delta\theta \) to be in radians. For very small angles, \( \sin(\Delta\theta) \) is nearly equal to \( \Delta\theta \) in radians.)
Here:
\( \lambda \) is the wavelength of the light used.
\(d\) is the distance between the two slits.
Given values:
Wavelength, \( \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} = 6 \times 10^{-7} \text{ m} \).
Angular width, \( \Delta\theta = 0.1^\circ \).
To use the formula, we can either convert \(0.1^\circ\) to radians or use \( \sin(0.1^\circ) \). The original solution uses the value \(0.0017\) for \( \sin(0.1^\circ) \).
We need to find \(d\), so we rearrange the formula:
\( d = \frac{\lambda}{\Delta\theta_{radians}} \approx \frac{\lambda}{\sin(0.1^\circ)} \)
\[ d = \frac{6 \times 10^{-7} \text{ m}}{0.0017} \]
\[ d \approx 3.529 \times 10^{-4} \text{ m} \]
When rounded to two significant figures, the distance between the two slits is approximately \(3.5 \times 10^{-4} \text{ m}\).
In simple words: We know the light's color and how wide the fringes look. By using a specific formula for angular width, we can calculate the distance between the two tiny openings (slits) that created the fringe pattern.

🎯 Exam Tip: For small angles, remember that \( \sin\theta \approx \tan\theta \approx \theta \) (in radians). This approximation is often used in wave optics problems for angular width calculations.

Question 17. Answer the following questions:
(a) In a single-slit diffraction experiment, if 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, why are the students unable to see each other even though they can converse easily?
(e) Ray optics assumes light travels in straight lines. Diffraction shows this isn't always true. Why is ray optics still so commonly used to understand images in optical instruments? What is the justification?
Answer:
(a) **Effect of doubling slit width in single-slit diffraction:**
- **Size of central band:** The angular width of the central bright band in single-slit diffraction becomes half when the slit width is doubled. This is because angular width is inversely proportional to slit width.
- **Intensity of central band:** The intensity of the central bright band becomes four times stronger when the slit width is doubled. This is because intensity is proportional to the square of the slit width.

(b) **Relation between diffraction and interference in double-slit experiment:**
In a double-slit setup, the pattern we see is a mix of interference (from two slits) and diffraction (from each single slit). The overall brightness of the interference fringes is shaped or "modulated" by the diffraction pattern of each slit. This means the interference fringes will be brightest in the middle and get dimmer as you move away from the center, following the shape of a single-slit diffraction pattern.

(c) **Bright spot at the center of a circular obstacle's shadow (Poisson's Spot):**
When light waves bend around the edges of a very small round object, they meet and combine constructively (add up) exactly at the center of the shadow. This special combining of waves creates a bright spot right in the middle of what should be a dark shadow.

(d) **Why students can converse but not see each other around a wall:**
Both light and sound waves can bend around corners (diffract), but how much they bend depends on their wavelength compared to the size of the obstacle.
- **Sound waves:** Sound waves have relatively long wavelengths (e.g., around 0.3 meters for common speech). A wall that is 7 meters long is not much larger than these wavelengths, so sound waves bend significantly around it, allowing students to hear each other.
- **Light waves:** Light waves have extremely short wavelengths (e.g., around \(10^{-7}\) meters). A 7-meter wall is vastly larger than the wavelength of light. Since the wall is so big compared to light's wavelength, light does not bend much around it, which is why students cannot see each other.

(e) **Justification for using ray optics:**
Ray optics is very useful for explaining things like lenses and mirrors because, in most everyday situations, the size of optical parts (like lenses or apertures) is much, much bigger than the wavelength of light. When this condition is met, the bending effects of diffraction are so small they can be ignored. In such cases, we can simply think of light as traveling in straight lines (rays), which makes calculations and understanding much simpler.
In simple words: When a slit gets wider, the central bright band shrinks but gets much brighter. In two slits, the pattern you see is a mix of bending around each slit and waves combining. A bright spot appears behind small round objects due to waves combining. Sound bends easily around walls because its waves are long, but light doesn't because its waves are very short. Ray optics works for big things because light's tiny waves don't noticeably bend.

🎯 Exam Tip: Remember the conditions for significant diffraction: obstacle size comparable to wavelength. This explains why sound diffracts easily but visible light usually doesn't in everyday scenarios.

Question 18. Two towers on top of two hills are 40 km apart. The direct line joining them passes 50 m above the terrain at its midpoint. What is the longest wavelength of radio waves that can be sent between the towers without appreciable diffraction effects?
Answer:
To find the longest radio wavelength that won't show much bending (diffraction), we look at the difference in path length created by the terrain. "No appreciable diffraction" means the light or radio waves should mostly travel in straight lines, and bending effects should be very small.
Let the total distance between the two towers be \(L = 40 \text{ km} = 40 \times 10^3 \text{ m}\).
The problem states that the direct line of sight between the towers is 50 m above the highest point of the ground (obstruction) at the midpoint. This height is our clearance, \(h = 50 \text{ m}\).
The path difference (\(\Delta x\)) between a direct wave and a wave that just grazes the top of an obstruction at the midpoint can be estimated as:
\( \Delta x \approx \frac{2h^2}{L} \)
Where \(h\) is the height of the obstruction relative to the line of sight, and \(L\) is the total distance.
For "no appreciable diffraction," this path difference should be sufficiently small compared to the wavelength. A common interpretation for the limit of 'appreciable diffraction' in this context is when the path difference is roughly equal to one wavelength (\(\Delta x \approx \lambda\)). This means the direct wave and the wave grazing the obstruction are roughly in phase or just starting to produce a noticeable diffraction pattern.
So, setting \( \lambda \approx \frac{2h^2}{L} \).
Substitute the values:
\( h = 50 \text{ m} \)
\( L = 40 \times 10^3 \text{ m} \)
\[ \lambda \approx \frac{2 \times (50 \text{ m})^2}{40 \times 10^3 \text{ m}} \]
\[ \lambda \approx \frac{2 \times 2500}{40000} \]
\[ \lambda \approx \frac{5000}{40000} \]
\[ \lambda \approx \frac{1}{8} \text{ m} = 0.125 \text{ m} = 12.5 \text{ cm} \]
Therefore, the longest wavelength for radio waves to travel between the towers without significant diffraction effects is approximately 12.5 cm.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो पहाड़ियों पर स्थित टावरों के बीच की सीधी रेखा और मध्य बिंदु पर भूमि की ऊंचाई को दिखाता है। यह दर्शाता है कि तरंगों के संचरण के लिए, बाधा के ऊपर से गुजरने वाली किरणें सीधी रेखा से कैसे विचलित हो सकती हैं, जिससे विवर्तन होता है। यह गणना उस अधिकतम तरंग दैर्ध्य को निर्धारित करने में मदद करती है जिसके लिए विवर्तन प्रभाव नगण्य रहते हैं।
In simple words: For radio waves to travel straight between two towers over hills, the waves must be short enough so they don't bend too much around the ground. We used a formula related to the path difference caused by the hills to find the longest wavelength that will still travel mostly straight.

🎯 Exam Tip: For problems involving radio wave propagation and obstructions, the Fresnel zone concept or path difference calculations are key. Be careful with unit conversions (km to m) and formula interpretation for "appreciable diffraction."

Question 19. 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:
In a single-slit diffraction experiment, dark bands (minima) appear at specific positions. The formula that describes the angular position of the \(n^{th}\) minimum is:
\(a \sin \theta_n = n\lambda\)
Where:
\(a\) is the width of the slit (what we need to find).
\( \theta_n \) is the angle to the \(n^{th}\) minimum from the center.
\(n\) is the order of the minimum (for the first minimum, \(n=1\)).
\( \lambda \) is the wavelength of the light.
The linear distance (\(y_n\)) of a minimum from the center on a screen placed \(D\) distance away is \( y_n = D \tan \theta_n \).
For very small angles (which is usually the case in diffraction), \( \sin \theta_n \approx \tan \theta_n \approx \theta_n \) (when \( \theta_n \) is in radians).
So, we can use the simplified formula for the position of the first minimum (\(n=1\)):
\( y_1 = \frac{\lambda D}{a} \)
We want to find \(a\), so we rearrange the formula:
\( a = \frac{\lambda D}{y_1} \)
Given values:
Wavelength, \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \).
Distance to screen, \(D = 1 \text{ m}\).
Distance of the first minimum from the center, \(y_1 = 2.5 \text{ mm} = 2.5 \times 10^{-3} \text{ m}\).
Now, substitute these values into the formula:
\[ a = \frac{(500 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{2.5 \times 10^{-3} \text{ m}} \]
\[ a = \frac{500 \times 10^{-9}}{2.5 \times 10^{-3}} \text{ m} \]
\[ a = 200 \times 10^{-6} \text{ m} = 0.2 \times 10^{-3} \text{ m} = 0.2 \text{ mm} \]
Thus, the width of the slit is 0.2 mm.
In simple words: We know the light's color, how far the screen is, and where the first dark spot appears. Using a specific formula for single-slit diffraction, we can calculate the width of the narrow opening that the light passed through.

🎯 Exam Tip: Remember the single-slit diffraction formula for minima \(a \sin \theta = n\lambda\) and its small-angle approximation \(y = \frac{n\lambda D}{a}\). Ensure correct unit conversions.

Question 20. 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) The principle of linear superposition of wave displacement is fundamental to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
Answer:
(a) **Explanation for TV picture shaking:**
The slight shaking on your TV screen when an aircraft flies over is due to **interference of radio signals**. Your TV antenna receives two versions of the broadcast signal:
(i) One signal comes directly from the TV station.
(ii) Another, weaker signal bounces off the airplane and then reaches your antenna.
These two signals travel different distances to reach your TV. As the plane moves, this path difference changes, causing the signals to sometimes add up (constructive interference) and sometimes cancel out (destructive interference). This changing signal strength makes the TV picture unsteady.

(b) **Justification of the principle of linear superposition:**
The principle of linear superposition says that when multiple waves meet, the total displacement (or effect) at any point is simply the sum of the individual displacements from each wave. This principle is true because the mathematical equations that describe wave motion are "linear."
If one wave (\(y_1\)) and another wave (\(y_2\)) are both solutions to a linear wave equation, then combining them in any linear way (like \(y_1 + y_2\)) will also be a solution. This means that waves don't change each other permanently; they just overlap and then continue on their way. This principle works well for most waves with small amplitudes. However, for very strong waves, like intense laser light, the wave equations can become non-linear, and superposition may not perfectly apply.
In simple words: TV screens shake because the TV gets two signals: one direct and one bounced off the plane, which interfere. The idea that waves simply add up (superposition) works because the math rules for waves are simple and linear.

🎯 Exam Tip: Interference and superposition are key concepts in wave phenomena. Understand that superposition is a consequence of the linearity of wave equations, with exceptions for very high amplitudes.

Question 21. In deriving the single-slit diffraction pattern, it is stated that the intensity is zero at angular positions given by \( \sin \theta = \frac{n\lambda}{a} \) (for \(n = \pm 1, \pm 2, \ldots\)). Justify this by suitably dividing the slit to demonstrate cancellation.
Answer:
To understand why light intensity becomes zero at certain angles in single-slit diffraction, we can use Huygens' principle by dividing the slit into smaller sections.
Let's consider the first dark spot (minimum) where the angle \( \theta \) satisfies \( \sin \theta = \frac{\lambda}{a} \), with \(a\) being the slit width and \( \lambda \) the wavelength.
We divide the slit of width \(a\) into two equal parts, each with a width of \( \frac{a}{2} \).
Now, imagine a light wave coming from the very top of the first half of the slit and another wave from the very top of the second half (which is the exact middle of the whole slit). The extra distance one wave travels compared to the other (path difference) in the direction of \( \theta \) is \( \frac{a}{2} \sin \theta \).
If we substitute \( \sin \theta = \frac{\lambda}{a} \) into this path difference, we get \( \frac{a}{2} \left(\frac{\lambda}{a}\right) = \frac{\lambda}{2} \).
A path difference of \( \frac{\lambda}{2} \) means these two waves are perfectly out of phase (one is at its peak when the other is at its trough). Therefore, they cancel each other out completely.
We can do this for every tiny wave source in the first half of the slit; there will always be a matching tiny wave source in the second half that cancels it out. This results in a total intensity of zero in that specific direction. This method can be expanded for all other dark spots by dividing the slit into \(2n\) equal parts, where \(n\) is the order of the minimum.
In simple words: When light passes through a narrow opening, it bends. At certain angles, if we imagine the opening split into tiny sections, the light waves from one section perfectly cancel out the light waves from another section, making the screen completely dark at that spot.

🎯 Exam Tip: The division of the slit into two (or \(2n\)) halves is a crucial method to justify the positions of minima in single-slit diffraction. Understand the concept of pairwise destructive interference.

GSEB Class 12 Physics Wave Optics Additional Important Questions And Answers

Question 1. What is the path difference between two waves when the crest of one wave falls on the crest of the other?
Answer:
When the crest of one wave perfectly meets the crest of another wave, they are said to be "in phase." This leads to stronger waves (constructive interference). The path difference (\( \delta \)) between two waves that meet crest-on-crest must be a whole number multiple of the wavelength (\( \lambda \)).
So, the path difference is \( \delta = n\lambda \), where \(n\) can be \(0, \pm 1, \pm 2, \ldots \).
In simple words: When wave peaks meet perfectly, they are in step. The distance difference between them is a whole number of wavelengths.

🎯 Exam Tip: Remember the conditions for constructive interference: path difference is \(n\lambda\) and phase difference is \(2n\pi\).

Question 2. What is the path difference between two waves when the crest of one wave falls on the trough of the other?
Answer:
When the crest (highest point) of one wave meets the trough (lowest point) of another wave, they are completely "out of phase." This causes them to cancel each other out (destructive interference). The path difference (\( \delta \)) for this to happen must be an odd number of half-wavelengths (\( \lambda/2 \)).
So, the path difference is \( \delta = (2n + 1)\frac{\lambda}{2} \), where \(n\) can be \(0, \pm 1, \pm 2, \ldots \).
In simple words: When a wave peak meets a wave valley, they are out of step. The distance difference between them is an odd number of half-wavelengths.

🎯 Exam Tip: Remember the conditions for destructive interference: path difference is \((2n+1)\frac{\lambda}{2}\) and phase difference is \((2n+1)\pi\).

Question 3. What is constructive interference?
Answer:
Constructive interference happens when two or more light waves (or other types of waves) meet and overlap in such a way that their peaks (crests) line up with other peaks, and their valleys (troughs) line up with other valleys. When waves combine like this, they are "in phase," and their strengths (amplitudes) add up. This makes the resulting wave much stronger, and for light, this means the intensity becomes maximum, creating bright spots in an interference pattern. The specific condition for constructive interference is when the path difference between the waves is a whole number of wavelengths (\( n\lambda \)), or when their phase difference is a whole number of \(2\pi\) radians (\( 2n\pi \)), where \(n\) can be \(0, \pm 1, \pm 2, \ldots \).
In simple words: Constructive interference is when waves meet and add up, making a stronger wave or a brighter light spot. This happens when their peaks and valleys line up perfectly.

🎯 Exam Tip: Focus on the alignment of crests/troughs and the resulting increase in amplitude and intensity for constructive interference. Remember the path and phase difference conditions.

Question 4. In Young's double-slit experiment, why are the light waves considered coherent?
Answer:
In Young's double-slit experiment, the light waves are coherent because this is essential for seeing a steady and clear interference pattern. Coherent light sources produce waves that always have the same phase relationship and the same frequency (which means the same wavelength). If the light wasn't coherent, the peaks and valleys from the two slits would meet randomly, causing the interference pattern to flicker too fast to be seen, making it look like just a bright blur instead of distinct lines.
In simple words: Light waves in Young's experiment must be coherent (stay in step and have the same color) to create a clear, stable interference pattern; otherwise, the pattern would disappear.

🎯 Exam Tip: Remember that "coherent sources" are critical for observing sustained interference patterns, meaning they must maintain a constant phase difference and have identical frequencies.

Question 5. In a double-slit experiment, if one of the slits is closed, what happens?
Answer:
If one of the slits in a double-slit experiment is closed, the setup stops being a two-source interference system. Instead, light will only pass through the single open slit. What you see on the screen then is not the distinct pattern of bright and dark lines (interference fringes) anymore. Instead, you would observe a single-slit diffraction pattern, which for white light, would appear as a broad, generally white patch of light in the center, possibly with some faint color separation at its edges, without the sharp, alternating bands.
In simple words: If one slit is closed, the clear striped pattern (interference) disappears, and you only see a wide, somewhat blurry bright area (diffraction) on the screen.

🎯 Exam Tip: Understand the difference between double-slit interference and single-slit diffraction. Closing one slit eliminates the two-source interference effect, leaving only single-slit diffraction.

Question 6. When both slits are opened in a double-slit experiment, we can see an interference pattern. Why?
Answer:
When both slits are open in a double-slit experiment, two sets of light waves spread out from them. These waves are coherent, meaning they stay in step with each other. As these waves travel and overlap on the screen, they combine (superpose). Where the peaks of the waves meet peaks, they add up to make a brighter light (constructive interference). Where the peaks meet valleys, they cancel each other out, making a dark area (destructive interference). This regular pattern of adding and canceling creates the visible interference pattern.
In simple words: When both slits are open, light waves from each slit meet and combine. Sometimes they make light brighter, and sometimes they cancel each other out, creating a striped interference pattern.

🎯 Exam Tip: The core reason for interference patterns is the superposition of coherent waves from two sources, leading to both constructive and destructive interference.

Question 7. If two coherent sources are used, what is the resultant amplitude of the electric field and intensity of light for constructive interference?
Answer:
When two coherent sources produce constructive interference, their light waves combine perfectly. If each source has an electric field amplitude of \(E\):
- **Resultant amplitude of electric field:** The amplitudes add up directly, so the combined amplitude is \(E + E = 2E\).
- **Resultant intensity of light:** The brightness (intensity) of light is related to the square of its amplitude. If the intensity from one source is proportional to \(E^2\), then for constructive interference, the combined intensity will be proportional to \((2E)^2 = 4E^2\). This means the light is four times brighter than from a single source.
In simple words: When light waves add up perfectly (constructive interference), their electric field strength doubles, and the brightness (intensity) becomes four times stronger.

🎯 Exam Tip: Remember that for coherent sources, amplitudes add linearly in constructive interference (e.g., \(E+E = 2E\)), but intensity (which is proportional to amplitude squared) increases quadratically (e.g., \( (2E)^2 = 4E^2 \)).

Question 8. For destructive interference, what is the value of the resultant amplitude of the electric field and intensity of light?
Answer:
When two coherent sources produce destructive interference, their light waves meet perfectly out of step (peaks meet valleys). If each source has an electric field amplitude of \(E\):
- **Resultant amplitude of electric field:** The amplitudes cancel each other out. If they are equal, the combined amplitude is \(E - E = 0\).
- **Resultant intensity of light:** Since brightness (intensity) is proportional to the square of the amplitude, if the combined amplitude is zero, then the combined intensity is also zero. This creates a completely dark spot.
In simple words: When light waves cancel each other out (destructive interference), their electric field strength becomes zero, and there is no light (zero intensity).

🎯 Exam Tip: For destructive interference, equal amplitudes lead to complete cancellation, resulting in zero resultant amplitude and zero intensity.

Question 9. In an interference experiment, if one of the slits is closed, what will be your observation?
Answer:
If one of the slits in an interference experiment is blocked, the experiment no longer shows two-source interference. Instead, light only goes through one slit. The pattern seen on the screen will then be a **diffraction of light** pattern from a single slit. This pattern has a very wide and bright center, with much dimmer and narrower bright and dark bands on either side.
In simple words: Closing one slit stops interference, and you only see the light bending (diffracting) through the single open slit, creating a different kind of pattern.

🎯 Exam Tip: Understand that interference requires at least two coherent sources, while diffraction occurs when light passes through a single aperture or bends around an obstacle.

Question 10. Compare the intensities of maxima and minima in terms of:
(a) Bandwidth
(b) Intensity ratio
Answer:
(a) **Comparison in terms of Bandwidth (Fringe Width):**
In interference patterns, like those from Young's double-slit experiment, the "bandwidth" usually refers to the width of the bright (maxima) or dark (minima) fringes. Generally, the width of all bright fringes and all dark fringes is the same, given by the formula \( \beta = \frac{\lambda D}{d} \). So, both the maxima and minima have the same bandwidth.
However, their intensities are very different: maxima are bright, and minima are dark (ideally completely dark).

(b) **Comparison in terms of Intensity ratio:**
For interference patterns made by two waves with amplitudes \(a_1\) and \(a_2\):
- **Maximum intensity (\(I_{max}\)):** This happens during constructive interference when the waves add up. The total amplitude is \(A_{max} = a_1 + a_2\). So, the maximum intensity is proportional to \( (a_1 + a_2)^2 \).
- **Minimum intensity (\(I_{min}\)):** This happens during destructive interference when the waves partially or fully cancel out. The total amplitude is \(A_{min} = |a_1 - a_2|\). So, the minimum intensity is proportional to \( (a_1 - a_2)^2 \).
The ratio of maximum to minimum intensity is given by:
\[ \frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} \]
If the amplitudes of the two waves are exactly equal (\(a_1 = a_2 = a\)), then:
\(I_{max} \propto (a+a)^2 = (2a)^2 = 4a^2\).
\(I_{min} \propto (a-a)^2 = 0\).
In this special case, the ratio \( \frac{I_{max}}{I_{min}} \) becomes infinitely large, meaning the dark fringes are perfectly black.
In simple words: The width of bright and dark fringes is usually the same. But bright fringes are much brighter than dark fringes. How much brighter depends on how different the strengths (amplitudes) of the two waves are. If the wave strengths are equal, dark fringes are completely black.

🎯 Exam Tip: Remember the intensity formulas \(I_{max} \propto (a_1+a_2)^2\) and \(I_{min} \propto (a_1-a_2)^2\). A common scenario is \(a_1=a_2\), leading to \(I_{min}=0\) and perfect contrast.

 

Question 11. Give an example of light interference in our daily life.
Answer: The beautiful colors seen on a thin oil film spread over water are an example of light interference.
In simple words: When oil spreads on water, it forms a thin layer. Sunlight reflecting off the top and bottom of this thin oil layer interferes, creating the pretty colors we see.

🎯 Exam Tip: Remember to provide a clear, real-world example like oil on water or soap bubbles to illustrate interference phenomena effectively for full marks.

 

Question 12. Why does the picture on a TV screen shake slightly when a low-flying aircraft passes overhead?
Answer: This happens because the direct TV signal received by the antenna interferes with a weaker signal reflected by the passing aircraft.
In simple words: Your TV antenna gets two signals: one directly from the broadcast tower and another bounced off the airplane. These two signals clash, causing the picture to shake.

🎯 Exam Tip: Focus on explaining the "interference" of two signals (direct and reflected) as the core reason for the TV picture disturbance.

 

Question 13. The diffraction pattern becomes invisible when the slit is very wide. Explain why.
Answer: To see diffraction clearly, the width or size of the slit must be similar to the wavelength of the light used. If the slit becomes much wider than the light's wavelength, the diffraction pattern will not be visible.
In simple words: Diffraction happens best when the opening is about the same size as the light's wave. If the opening is too big, the light waves pass straight through without bending much, so you don't see the pattern.

🎯 Exam Tip: Highlight the relationship between slit width and wavelength as the critical factor for observing diffraction patterns. When the slit is much larger than the wavelength, diffraction effects are negligible.

 

Question 14. Diffraction is common with sound but not with light. Why is this so?
Answer: Diffraction occurs significantly when the size of an obstacle or aperture is comparable to the wavelength of the waves. Sound waves have a long wavelength, so everyday objects and openings are often similar in size to sound wavelengths, making sound diffraction common. Light waves, however, have very short wavelengths. It's difficult to find obstacles or slits in everyday life that are small enough to cause noticeable light diffraction.
In simple words: Sound waves are long, so they easily bend around common objects like doors. Light waves are very short, so they need tiny openings or obstacles to bend, which we don't usually find, making light diffraction harder to see.

🎯 Exam Tip: The key to this explanation is comparing the wavelengths of sound and light with the typical sizes of obstacles or apertures encountered in daily life. Sound's longer wavelength makes its diffraction more evident.

 

Question 15. Answer the following questions:
(a) When we look through a muslin cloth, we can see a colored spectrum. How will you explain this?
(b) Observe the shadow of your book when it is held a few centimeters above a table with a lamp several centimeters above the book. Why is the shadow of the book fuzzy at the edges?
(c) If you observe a distant street light between two fingers pinched together, you can visualize alternate bright and dark fringes. What is this due to? Explain.
Answer:
(a) The muslin cloth's fine fibers act like tiny slits. Light diffracts through these slits, spreading into different colors to form a spectrum.
(b) When light hits the edges of the book, it bends slightly due to diffraction. This bending creates a less sharp boundary for the shadow, making its edges appear fuzzy instead of perfectly clear.
(c) When you pinch your fingers, you create a narrow slit. Light from the street lamp diffracts through this tiny opening. Since the slit's size is similar to the light's wavelength, it creates an interference pattern of alternating bright and dark fringes.
In simple words: (a) Muslin cloth fibers act like tiny gaps that split white light into its colors. (b) Light bends a little around the book's edges, so the shadow isn't perfectly sharp. (c) The small gap between your fingers acts like a tiny opening, causing the light from the street lamp to spread out and create light and dark bands.

🎯 Exam Tip: For (a) and (c), emphasize diffraction through narrow openings (fibers/fingers). For (b), focus on diffraction at the edges of the book causing the fuzziness, leading to a blurred image rather than a sharp pattern.

 

Question 16. What longitudinal waves can be polarized? OR Light waves can be polarized while sound waves can't. Why?
Answer: Only transverse waves can be polarized. Longitudinal waves cannot be polarized. This is because sound waves are longitudinal, meaning their vibrations are along the direction of wave travel. Transverse waves, like light, vibrate perpendicular to the direction of travel, allowing their vibrations to be restricted to a single plane (polarized).
In simple words: Polarization is like filtering waves so they only wiggle in one direction. Light waves wiggle sideways, so you can filter them. Sound waves push and pull forwards and backwards, so they can't be filtered in the same way.

🎯 Exam Tip: The crucial distinction is that polarization is a property unique to transverse waves, where vibrations occur perpendicular to the wave's direction of motion. Longitudinal waves, with vibrations parallel to motion, cannot be polarized.

 

Question 17. Will ultrasonic waves show any polarization? Give a reason for your answer.
Answer: No, ultrasonic waves will not show polarization. The reason is that ultrasonic waves are a type of sound wave, and sound waves are longitudinal waves. As established, only transverse waves can be polarized.
In simple words: Ultrasonic waves are just high-frequency sound. Since sound waves are push-and-pull waves (longitudinal), they cannot be polarized like light waves.

🎯 Exam Tip: Reinforce the concept that polarization applies exclusively to transverse waves. Since ultrasonic waves are longitudinal, they cannot exhibit polarization, similar to audible sound waves.

 

Question 18. How would the angular separation of interference bands in Young's double-slit experiment change when the distance of separation between the slits and the screen is doubled?
Answer: The angular separation of fringes in Young's double-slit experiment is given by \( \beta = \frac{\lambda}{d} \), where \( \lambda \) is the wavelength and \( d \) is the slit separation. The linear fringe width is \( \beta' = \frac{\lambda D}{d} \), where \( D \) is the distance to the screen. The question asks about *angular separation*, which is \( \theta = \frac{\beta'}{D} = \frac{\lambda}{d} \). If the distance of separation between the slits \( d \) is doubled, the angular separation will become half. However, the question states "distance of separation between the slits *and the screen* is doubled". This phrasing is ambiguous. Let's assume it means *both* \( d \) and \( D \) are doubled. If \( D \) (distance to screen) is doubled from \( D_1 \) to \( D_2 = 2D_1 \), the angular separation \( \theta = \frac{\lambda}{d} \) remains unchanged because angular separation is independent of \( D \). If \( d \) (slit separation) is doubled from \( d_1 \) to \( d_2 = 2d_1 \), then the new angular separation \( \theta' = \frac{\lambda}{2d_1} = \frac{1}{2} \theta \). The previous OCR solution had an equation for \( \beta_1 \) and \( \beta_2 \) (linear fringe width) and concluded \( \beta_1 = \beta_2 \) when \( D \) and \( d \) are both doubled. Let's re-evaluate based on angular width definition. Angular width \( \theta = \frac{\lambda}{d} \). If *only* \( D \) is doubled, \( \theta \) does not change. If *only* \( d \) is doubled, \( \theta \) becomes \( \frac{1}{2} \theta \). The question is "distance of separation between the slits and the screen is doubled". This likely refers to \( D \) (screen distance). If it refers to \( d \) (slit separation) also, then the problem is compounded. Let's assume the question implicitly refers to the *linear* separation when it mentions "interference bands" and how it changes when *both* D and d are mentioned as "doubled". Let the original linear fringe width be \( \beta_1 = \frac{D_1 \lambda}{d_1} \). If both the distance of separation between the slits \( d \) and the screen distance \( D \) are doubled: \( D_2 = 2D_1 \) \( d_2 = 2d_1 \) New fringe width \( \beta_2 = \frac{D_2 \lambda}{d_2} = \frac{(2D_1) \lambda}{(2d_1)} = \frac{D_1 \lambda}{d_1} = \beta_1 \). So, the linear fringe width remains the same. Therefore, the angular separation \( \theta = \frac{\beta}{D} \). Original angular separation \( \theta_1 = \frac{\beta_1}{D_1} \). New angular separation \( \theta_2 = \frac{\beta_2}{D_2} = \frac{\beta_1}{2D_1} = \frac{1}{2} \theta_1 \). Thus, the angular separation becomes half.
(i) Original angular separation: \( \theta_1 = \frac{\lambda}{d_1} \)
(ii) Original linear fringe width: \( \beta_1 = \frac{D_1 \lambda}{d_1} \)
If the distance between slits \( d \) and the screen distance \( D \) are both doubled:
New slit separation: \( d_2 = 2d_1 \)
New screen distance: \( D_2 = 2D_1 \)
New linear fringe width: \( \beta_2 = \frac{D_2 \lambda}{d_2} = \frac{(2D_1) \lambda}{(2d_1)} = \frac{D_1 \lambda}{d_1} = \beta_1 \)
New angular separation: \( \theta_2 = \frac{\beta_2}{D_2} = \frac{\beta_1}{2D_1} = \frac{1}{2} \left( \frac{\beta_1}{D_1} \right) = \frac{1}{2} \theta_1 \)
Therefore, the angular separation would become half of its original value.
In simple words: The question asks about angular separation. Angular separation depends on the wavelength of light and the distance between the slits. It does not depend on how far the screen is. If both the slit distance and screen distance are doubled, the angular separation will become half.

🎯 Exam Tip: Clearly distinguish between linear fringe width and angular separation. Angular separation \( (\theta = \lambda/d) \) is independent of screen distance \( D \). However, if both \( D \) and \( d \) are doubled, the linear width remains the same, but the angular width is halved because it's \( \beta/D \).