RBSE Solutions Class 12 Physics Chapter 12 Nature of Light

RBSE Class 12 Physics Chapter 12 Multiple Choice Type Questions

Question 1. To show the phenomenon of interference we need two sources which emit radiation :
(a) of equal frequency and fixed phase difference
(b) of nearly equal frequency
(c) of equal frequency
(d) of different wavelength
Answer: (a) of equal frequency and fixed phase difference
In simple words: For light waves to show interference, the light sources must have the same frequency and a constant phase difference between them. This is important for the interference pattern to stay steady.

🎯 Exam Tip: Remember the two key conditions for sustained interference: coherence (constant phase difference) and monochromatic light (single wavelength/frequency).

Question 3. In an interference experiment on using light of wavelength 700 nm, third bright fringe received at a point on the screen. At the same point to receive fifth bright fringe the wavelength of light is :
(a) 210 nm
(b) 315 nm
(c) 420 nm
(d) 490 nm
Answer: (c) 420 nm
Given, \( \lambda_1 = 700 \text{ nm} = 700 \times 10^{-9} \text{ m} = 7 \times 10^{-7} \text{ m} \)
\( n_1 = 3 \), \( n_2 = 5 \) (bright fringes)
Let \( X_n \) be the distance of the nth bright fringe from the central bright fringe.
\( X_n = \frac{n \lambda D}{d} \)
Since \( D \), \( X_n \), and \( d \) are constant, we can write:
\( \lambda \propto \frac{1}{n} \)
Therefore, \( \lambda_2 = \frac{n_1 \lambda_1}{n_2} \)
Now, put in the values:
\( \lambda_2 = \frac{3 \times 700}{5} \)
\( \lambda_2 = 420 \text{ nm} \)
In simple words: We know the distance of a bright fringe depends on the wavelength and the order of the fringe. If the position of the fringe stays the same, and we change the order, the wavelength must change in a specific way. Using the given values, the new wavelength comes out to be 420 nm.

🎯 Exam Tip: In interference experiments, the position of a bright fringe is directly proportional to the product of the fringe order and wavelength. If the position is fixed for two different scenarios, the product of order and wavelength remains constant.

Question 4. In Young's double slit experiment if the ratio of the widths of slits are 4 : 9, then the ratio of maximum and minimum frequencies will be:
(a) 196:25
(b) 121:1
(c) 25:1
(d) 1:25
Answer: (c) 25:1
Let \( W_1 \) and \( W_2 \) be the widths of the slits, and \( I_1 \) and \( I_2 \) be the intensities.
We are given \( \frac{W_1}{W_2} = \frac{4}{9} \).
The intensity of light is proportional to the width of the slit, so \( \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{4}{9} \).
The ratio of maximum to minimum intensity is given by:
\( \frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 \)
Divide numerator and denominator by \( \sqrt{I_2} \):
\( \frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{\frac{I_1}{I_2}} + 1}{\sqrt{\frac{I_1}{I_2}} - 1} \right)^2 \)
Substitute \( \frac{I_1}{I_2} = \frac{4}{9} \):
\( \frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{\frac{4}{9}} + 1}{\sqrt{\frac{4}{9}} - 1} \right)^2 \)
\( \frac{I_{max}}{I_{min}} = \left( \frac{\frac{2}{3} + 1}{\frac{2}{3} - 1} \right)^2 \)
\( \frac{I_{max}}{I_{min}} = \left( \frac{\frac{5}{3}}{\frac{-1}{3}} \right)^2 \)
\( \frac{I_{max}}{I_{min}} = (-5)^2 = 25 \)
Therefore, \( I_{max} : I_{min} = 25 : 1 \)
In simple words: The brightness of light (intensity) from each slit depends on its width. Maximum and minimum brightness in an interference pattern are related to the square of the sum and difference of the square roots of these intensities. When we do the math, if the widths are in a 4:9 ratio, the maximum to minimum intensity ratio is 25:1.

🎯 Exam Tip: Remember that intensity is proportional to the square of amplitude and also to slit width. The ratio of maximum to minimum intensity involves the sum and difference of the amplitudes.

Question 5. In Young's double slit experiment, light of two different wavelengths are used. For yellow orange colour (\( \lambda \) = 600 nm) third order bright fringe position coincide with fourth order bright fringe of second colour. The wavelength of second colour will be :
(a) 500 nm
(b) 450 nm
(c) 225 nm
(d) 350 nm
Answer: (b) 450 nm
For a bright fringe, the position \( x_n \) is given by \( x_n = \frac{n \lambda D}{d} \).
Given that the third order bright fringe of the first color coincides with the fourth order bright fringe of the second color.
So, \( x_3 = x_4' \)
\( \frac{3 \lambda_1 D}{d} = \frac{4 \lambda_2 D}{d} \)
Since \( D \) and \( d \) are constant, we can cancel them out:
\( 3 \lambda_1 = 4 \lambda_2 \)
Given \( \lambda_1 = 600 \text{ nm} \).
\( 3 \times 600 \text{ nm} = 4 \lambda_2 \)
\( 1800 \text{ nm} = 4 \lambda_2 \)
\( \lambda_2 = \frac{1800}{4} \text{ nm} \)
\( \lambda_2 = 450 \text{ nm} \)
In simple words: When fringes from two different colors appear at the same spot, it means their path differences are equal. Using the formula for bright fringe positions, and knowing the order and wavelength of one color, we can easily find the wavelength of the second color.

🎯 Exam Tip: When fringe positions coincide, equate their formulas (e.g., \( n_1 \lambda_1 = n_2 \lambda_2 \) for bright fringes) and solve for the unknown variable.

Question 6. In Young's double slit experiment the maximum intensity of light is \( I_{max} \), then the intensity of light of the path difference \( \frac{\lambda}{2} \) will be
(a) \( I_{max} \)
(b) \( \frac{I_{max}}{2} \)
(c) \( \frac{I_{max}}{4} \)
(d) zero
Answer: (d) zero
When the path difference is \( \frac{\lambda}{2} \), destructive interference occurs.
This results in minimum intensity, \( I_{min} \).
Let \( I_1 = I_2 = I_0 \) (assuming equal intensity from each slit for maximum contrast).
Maximum intensity \( I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 2I_0 + 2I_0 = 4I_0 \).
Minimum intensity \( I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = I_0 + I_0 - 2\sqrt{I_0 I_0} = 2I_0 - 2I_0 = 0 \).
So, at a path difference of \( \frac{\lambda}{2} \), the intensity is zero.
In simple words: A path difference of half a wavelength means the light waves cancel each other out completely. This is called destructive interference. When waves perfectly cancel, the light intensity becomes zero at that point.

🎯 Exam Tip: A path difference of \( n\lambda \) (where n is an integer) leads to constructive interference and maximum intensity. A path difference of \( (n + \frac{1}{2})\lambda \) leads to destructive interference and minimum intensity.

Question 7. Choose the correct statement:
In most of the circumstances, the diffraction of sound is more possible compare to diffraction of light:
(a) Medium is necessary for sound propagation
(b) Sound waves are longitudinal while light rays are transverse
(c) Wavelength of light is very less comparable to wavelength of sound
(d) Speed of sound wave order is less than 6 compare to speed of light
Answer: (c) Wavelength of light is very less comparable to wavelength of sound
Diffraction is when waves bend around obstacles. This effect is most noticeable when the size of the obstacle or opening is similar to the wavelength of the wave. Light has a very small wavelength (around \( 10^{-7} \) m), much smaller than everyday objects, so its diffraction is hard to see. Sound waves have larger wavelengths, often comparable to common objects, making their diffraction more obvious, allowing us to hear around corners. Option (b) describes types of waves, which is true, but (c) explains *why* diffraction is more common for sound. The speed difference (d) is also true but doesn't directly explain the diffraction observation. (a) is also true but not the reason for easier diffraction.
In simple words: Waves bend around corners better if their wavelength is similar in size to the obstacle. Light waves have very tiny wavelengths, so they don't bend much around big objects like doors. Sound waves have much bigger wavelengths, similar to common objects, so they bend easily, which is why we can hear people around a corner but not see them.

🎯 Exam Tip: The condition for significant diffraction is that the wavelength of the wave must be comparable to the size of the aperture or obstacle. Sound waves typically have much larger wavelengths than visible light, making their diffraction more observable in everyday scenarios.

Question 9. A light of 5000 Å wavelength is diffracted by a single slit. In diffraction pattern fifth minima formed at 5 mm distance from central maxima. If distance between screen and slit is 1m, then the width of the slit will be :
(a) 0.1 mm
(b) 0.3 mm
(c) 0.5 mm
(d) 0.8 mm
Answer: (c) 0.5 mm
Given:
Wavelength \( \lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m} \)
Order of minima \( n = 5 \)
Distance of fifth minima from central maxima \( X_n = 5 \text{ mm} = 5 \times 10^{-3} \text{ m} \)
Distance between screen and slit \( D = 1 \text{ m} \)
The formula for the position of the nth minimum in a single-slit diffraction pattern is:
\( X_n = \frac{n \lambda D}{a} \)
Where \( a \) is the width of the slit.
We need to find \( a \):
\( a = \frac{n \lambda D}{X_n} \)
Substitute the given values:
\( a = \frac{5 \times (5 \times 10^{-7} \text{ m}) \times (1 \text{ m})}{5 \times 10^{-3} \text{ m}} \)
\( a = \frac{25 \times 10^{-7}}{5 \times 10^{-3}} \text{ m} \)
\( a = 5 \times 10^{-4} \text{ m} \)
Convert to millimeters:
\( a = 5 \times 10^{-4} \times 10^3 \text{ mm} \)
\( a = 0.5 \text{ mm} \)
In simple words: In single-slit diffraction, the position of dark fringes depends on the wavelength of light, the order of the fringe, the distance to the screen, and the width of the slit. By using the formula and the given values for the fifth dark fringe, we can calculate the width of the slit to be 0.5 mm.

🎯 Exam Tip: For single-slit diffraction, remember the formula for minima positions: \( X_n = \frac{n \lambda D}{a} \). Pay careful attention to unit conversions (Å to m, mm to m) to avoid errors.

Question 10. A beam of microwaves whose wavelength is 0.052 m is coming towards a rectangular aperture of width 0.35 m. Resultant diffraction pattern is observing on a wall at 8.0 m distance from aperture. The distance between first and second order outer fringes will be :
(a) 1.3 m
(b) 1.8 m
(c) 1.19 m
(d) 2.5 m
Answer: (c) 1.19 m
Given:
Wavelength \( \lambda = 0.052 \text{ m} \)
Aperture width \( a = 0.35 \text{ m} \)
Distance to screen \( D = 8.0 \text{ m} \)
We need to find the distance between the first and second order outer fringes. This refers to the distance between the first and second order maxima.
The angular position for the \( n^{th} \) maximum in single-slit diffraction is given by \( a \sin \theta_n = (2n+1)\frac{\lambda}{2} \). For small angles, \( \sin \theta_n \approx \theta_n \approx \frac{X_n}{D} \).
So, \( \frac{a X_n}{D} = (2n+1)\frac{\lambda}{2} \).
Therefore, \( X_n = \frac{(2n+1)\lambda D}{2a} \).
For the first order maximum (\( n=1 \)):
\( X_1 = \frac{(2(1)+1)\lambda D}{2a} = \frac{3 \lambda D}{2a} \)
For the second order maximum (\( n=2 \)):
\( X_2 = \frac{(2(2)+1)\lambda D}{2a} = \frac{5 \lambda D}{2a} \)
The distance between the first and second order outer fringes (\( \Delta X \)) is \( X_2 - X_1 \):
\( \Delta X = \frac{5 \lambda D}{2a} - \frac{3 \lambda D}{2a} = \frac{2 \lambda D}{2a} = \frac{\lambda D}{a} \)
Substitute the given values:
\( \Delta X = \frac{(0.052 \text{ m}) \times (8.0 \text{ m})}{0.35 \text{ m}} \)
\( \Delta X = \frac{0.416}{0.35} \text{ m} \)
\( \Delta X \approx 1.1885 \text{ m} \approx 1.19 \text{ m} \)
In simple words: For a single slit, the positions of the bright fringes are not evenly spaced. We use a formula to find the distance of each bright fringe from the center. To find the distance between the first and second bright fringes, we calculate each position and then find the difference. This calculation shows the distance is about 1.19 meters.

🎯 Exam Tip: Distinguish between minima and maxima formulas in single-slit diffraction. Minima occur at \( a \sin \theta = n\lambda \) and maxima are approximately at \( a \sin \theta = (n+\frac{1}{2})\lambda \).

Question 12. Two white points are placed on a black paper at 1 mm distance. They are seen by an eye of cornea diameter 3 mm. What will be the maximum distance between them, so they are exactly resolved by eye (wavelength of light = 500 nm)?
(a) 6 m
(b) 3 m
(c) 5 m
(d) 1 m
Answer: (c) 5 m
Given:
Distance between points \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
Diameter of cornea \( D = 3 \text{ mm} = 3 \times 10^{-3} \text{ m} \)
Wavelength \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m} \)
We need to find the maximum distance \( d_{max} \) at which the eye can resolve these two points.
The angular resolution \( \theta_{min} \) of the eye is given by:
\( \theta_{min} = \frac{1.22 \lambda}{D} \)
Also, the angular separation of two points \( \theta \) at a distance \( d_{max} \) from the eye is approximately \( \theta = \frac{d}{d_{max}} \).
For the points to be just resolved, \( \theta = \theta_{min} \).
So, \( \frac{d}{d_{max}} = \frac{1.22 \lambda}{D} \)
Solving for \( d_{max} \):
\( d_{max} = \frac{dD}{1.22 \lambda} \)
Substitute the values:
\( d_{max} = \frac{(1 \times 10^{-3} \text{ m}) \times (3 \times 10^{-3} \text{ m})}{1.22 \times (5 \times 10^{-7} \text{ m})} \)
\( d_{max} = \frac{3 \times 10^{-6}}{6.1 \times 10^{-7}} \text{ m} \)
\( d_{max} = \frac{30 \times 10^{-7}}{6.1 \times 10^{-7}} \text{ m} \)
\( d_{max} \approx \frac{30}{6.1} \text{ m} \approx 4.918 \text{ m} \approx 5 \text{ m} \)
In simple words: The ability of our eyes to distinguish two close objects depends on their angular separation and the diameter of our pupil. Using a special formula for resolution, we can calculate the furthest distance at which two small points, 1 mm apart, can still be seen as separate by an average eye. This distance is about 5 meters.

🎯 Exam Tip: Remember Rayleigh's criterion for resolution, which states that two objects are just resolved when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. The formula \( \theta_{min} = \frac{1.22 \lambda}{D} \) is crucial.

Question 13. Electromagnetic waves are transverse. Evidence of this is given by :
(a) polarisation
(b) interference
(c) reflection
(d) refraction
Answer: (a) polarisation
From Brewster's law, the angle of polarization \( i_p \) is given by \( \tan i_p = n \), where \( n \) is the refractive index. So \( i_p = \tan^{-1}(n) \). This law relates reflection and refraction to polarization. The phenomenon of polarisation occurs only for transverse waves, not for longitudinal waves. In polarization, the vibrations of the wave are restricted to a single plane. If electromagnetic waves were longitudinal, their vibrations would be along the direction of propagation, and thus, they could not be polarized. Interference, reflection, and refraction occur for both transverse and longitudinal waves.
In simple words: Polarisation is a special property where light waves vibrate in only one specific direction. Only waves that move side-to-side (transverse waves) can do this. Waves that push back and forth (longitudinal waves) cannot be polarised. So, seeing light polarise tells us it is a transverse wave.

🎯 Exam Tip: Polarization is the definitive proof of the transverse nature of a wave. Interference and diffraction demonstrate wave nature, but not specifically transverse wave nature.

Question 15. Four polaroids are so placed that the transmission axis of each is inclined at an angle of 30° from the axis of the previous polaroid in the same direction. If unpolarised light beam falls on the first polaroid, then what will be the intensity of light emerging from each polaroid?
(a) 10%
(b) 21%
(c) 50%
(d) 37.5%
Answer: (b) 21%
Let the intensity of the unpolarised light be \( I_0 \).
When unpolarised light passes through the first polaroid, its intensity becomes half:
\( I_1 = \frac{I_0}{2} \)
According to Malus's Law, when polarised light passes through a polaroid, the intensity \( I \) is given by \( I = I_{incident} \cos^2 \theta \), where \( \theta \) is the angle between the transmission axis of the polaroid and the plane of polarisation of the incident light.
For the second polaroid, \( \theta = 30^\circ \):
\( I_2 = I_1 \cos^2 (30^\circ) = \frac{I_0}{2} \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8} \)
For the third polaroid, \( \theta = 30^\circ \) (relative to the second):
\( I_3 = I_2 \cos^2 (30^\circ) = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32} \)
For the fourth polaroid, \( \theta = 30^\circ \) (relative to the third):
\( I_4 = I_3 \cos^2 (30^\circ) = \frac{9I_0}{32} \times \frac{3}{4} = \frac{27I_0}{128} \)
To find the percentage of emerging light compared to the initial unpolarised light:
Percentage \( = \frac{I_4}{I_0} \times 100 = \frac{27I_0/128}{I_0} \times 100 = \frac{27}{128} \times 100 \)
Percentage \( \approx 0.2109 \times 100 \approx 21.09\% \approx 21\% \)
In simple words: When unpolarised light passes through the first filter (polaroid), its brightness is cut in half. Then, each next filter is turned 30 degrees from the one before it. We use a rule called Malus's Law to calculate how much light passes through each filter. After four such filters, about 21% of the original light remains.

🎯 Exam Tip: Remember that unpolarised light becomes half in intensity after passing through the first polaroid. Subsequent polaroids follow Malus's Law: \( I = I_{incident} \cos^2 \theta \), where \( \theta \) is the angle between the transmission axes.

Question 16. Two nicols are so oriented that the angle between their major axis is 60°, then the percentage of emerging out light from the system will be :
(a) 30%
(b) 100%
(c) 12.5%
(d) 37.5%
Answer: (c) 12.5%
Let the intensity of the unpolarised light be \( I_0 \).
When unpolarised light passes through the first Nicol prism (polariser), its intensity becomes half:
\( I_1 = \frac{I_0}{2} \)
This light is now plane-polarised.
This plane-polarised light then passes through the second Nicol prism (analyser), which is oriented at an angle of \( \theta = 60^\circ \) to the first.
According to Malus's Law, the intensity of light emerging from the second Nicol prism \( I_2 \) is:
\( I_2 = I_1 \cos^2 \theta \)
\( I_2 = \frac{I_0}{2} \cos^2 (60^\circ) \)
We know \( \cos (60^\circ) = \frac{1}{2} \), so \( \cos^2 (60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
\( I_2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8} \)
To express this as a percentage of the initial unpolarised light:
Percentage \( = \frac{I_2}{I_0} \times 100 = \frac{I_0/8}{I_0} \times 100 = \frac{1}{8} \times 100 = 12.5\% \)
In simple words: When regular light goes through the first Nicol prism, its brightness becomes half and it gets polarized. When this polarized light then goes through a second Nicol prism that is turned 60 degrees, its brightness drops again following Malus's law. In the end, only 12.5% of the original light intensity passes through both prisms.

🎯 Exam Tip: Remember the two key steps: first, unpolarised light passing through a polariser loses half its intensity. Second, subsequent polarisers apply Malus's Law \( I = I_0 \cos^2 \theta \), where \( I_0 \) is the intensity of the incident polarised light and \( \theta \) is the angle between the transmission axes.

RBSE Class 12 Physics Chapter 12 Very Short Answer Type Questions

Question 2. On which physical quantities do the Young's fringe width depend?
Answer: Young's fringe width, often denoted as \( \beta \), depends on the following physical quantities:
(i) Wavelength \( \lambda \) of the light used.
(ii) Distance \( D \) between the slit and the screen.
(iii) Distance \( d \) between the two slits.
The formula for fringe width is \( \beta = \frac{\lambda D}{d} \).
In simple words: The width of the bright and dark bands (fringes) in Young's experiment changes based on three things: how long the light waves are (wavelength), how far the slits are from the screen, and how far apart the two slits are from each other.

🎯 Exam Tip: Clearly state the three factors and, if asked, include the formula for fringe width: \( \beta = \frac{\lambda D}{d} \). This formula helps explain the direct and inverse relationships.

Question 3. Write statement of Huygen's principle of diffraction.
Answer: According to Fresnel's explanation, diffraction is the result of the superposition of secondary waves. These secondary waves start from different parts of the same wavefront that passes through an aperture or around an obstacle. Each point on a wavefront acts as a source of secondary wavelets, and the envelope of these wavelets forms the new wavefront.
In simple words: Diffraction happens because when light waves hit a small opening or object, every point on the part of the wave that gets through acts like a tiny new light source. These tiny new waves then spread out and overlap, creating the bending effect we call diffraction.

🎯 Exam Tip: Focus on the idea that every point on a wavefront becomes a source of secondary wavelets, and diffraction is the result of their interference.

Question 4. What type of wavefront will emerge :
(i) from a point source?
(ii) from a far light source?
Answer:
(i) From a point source, a spherical wavefront will emerge. The waves spread out in all directions from the central point, forming expanding spheres.
(ii) From a far light source, a plane wavefront will emerge. When a spherical wavefront travels a very long distance, a small section of it appears flat, like a plane.
In simple words: (i) If light comes from a tiny spot, the waves spread out like circles in all directions, so we call it a spherical wavefront. (ii) If light comes from a very, very far away source (like the sun), the waves look flat when they reach us, so we call it a plane wavefront.

🎯 Exam Tip: Remember the relationship between source geometry and wavefront shape: point source -> spherical, linear source -> cylindrical, distant source -> plane. This is fundamental to understanding wave propagation.

Question 5. What is the most important condition for interference by two waves?
Answer: The most important condition for sustained interference by two waves is that both sources of light should be coherent. Coherent sources mean they must emit light waves with the same frequency and maintain a constant phase difference between them over time.
In simple words: For two light waves to create a steady pattern of bright and dark bands, they must be "coherent." This means their waves must be exactly the same kind (same color/frequency) and always start their cycles at the same relative time (constant phase difference).

🎯 Exam Tip: Coherence (same frequency and constant phase difference) is critical for observing a stable interference pattern. Without it, the pattern changes rapidly and is not visible.

Question 6. In single slit experiment how does angular separation change when the distance between slit and screen becomes two times?
Answer: The angular separation (\( \theta \)) in a single slit diffraction experiment is inversely proportional to the distance between the slit and the screen (\( D \)), i.e., \( \theta \propto \frac{1}{D} \). Therefore, if the distance \( D \) becomes two times larger, the angular separation \( \theta \) will become half.
In simple words: The angle at which light spreads out after passing through a single slit is related to how far away the screen is. If you double the distance between the slit and the screen, the angle of spread will become half as wide.

🎯 Exam Tip: Understand that angular separation refers to the angle subtended by the diffraction pattern at the slit. Its relationship with D is inverse, while linear separation (fringe width) is directly proportional to D.

Question 7. What should be the order of the aperture or obstacle to observe diffraction with it?
Answer: To observe diffraction effectively, the order of the size of the aperture or obstacle should be comparable to the wavelength of the light (or wave) being used. If the aperture is too large compared to the wavelength, diffraction effects are minimal, and light appears to travel in straight lines.
In simple words: For light to bend around an object or through a hole (diffraction), the object or hole needs to be about the same size as the light's wavelength. If the object or hole is much bigger, the light won't bend much.

🎯 Exam Tip: The key phrase is "comparable to the wavelength." This condition explains why sound diffracts easily but light usually doesn't in everyday situations.

Question 9. Despite the wave nature of light why does it seem to move in straight line.
Answer: Light has a very small wavelength (typically around \( 10^{-7} \) meters). Because the wavelength of light is extremely small compared to the size of most everyday objects and apertures, diffraction effects are usually negligible. This makes light appear to travel in straight lines, following the principles of ray optics.
In simple words: Light waves are incredibly tiny. Because their waves are so small compared to almost everything around us, they don't bend noticeably when they hit objects. This makes light look like it travels in straight lines, even though it is a wave.

🎯 Exam Tip: Connect the small wavelength of light to the condition for observable diffraction. When the obstacle size is much greater than the wavelength, light follows rectilinear propagation.

Question 10. In single slit diffraction, which type of light wave undergo superimpose?
Answer: In single-slit diffraction, the slit itself acts as a source of numerous secondary wavelets, as per Huygens' principle. These wavelets spread out in all directions from different points within the slit and then superimpose (interfere) with each other. This superposition of wavelets from different parts of the same wavefront causes the phenomenon of diffraction, creating a pattern of bright and dark fringes.
In simple words: When light passes through a single narrow slit, every tiny part of the wave inside that slit starts acting like a new light source. These many tiny waves then spread out and overlap, or "superimpose," to create the diffraction pattern.

🎯 Exam Tip: Emphasize that in single-slit diffraction, the interference happens *within* the single wavefront as it passes through the slit, not between two separate sources.

Question 11. What is mathematical form of Malus law?
Answer: The mathematical form of Malus's Law describes the intensity of plane-polarised light that passes through an analyser. If \( I_0 \) is the intensity of the plane-polarised light incident on the analyser, and \( \theta \) is the angle between the transmission axis of the analyser and the plane of polarisation of the incident light, then the intensity \( I \) of the light emerging from the analyser is given by:
\( I = I_0 \cos^2 \theta \)
Here, \( I_0 \) represents the maximum intensity of the transmitted light when the axes are aligned, and \( \theta \) is the angle between the planes of the polariser and analyser.
In simple words: Malus's Law tells us how bright polarised light will be after passing through a second filter (analyser). It says the new brightness is equal to the original brightness multiplied by the square of the cosine of the angle between the two filters.

🎯 Exam Tip: Clearly state the formula \( I = I_0 \cos^2 \theta \) and define each term: \( I \), \( I_0 \), and \( \theta \). Remember that \( I_0 \) is the intensity of *plane-polarised* light incident on the analyser.

RBSE Class 12 Physics Chapter 12 Short Answer Type Questions

Question 1. Describe Huygens' principle for light waves.
Answer: Huygens' principle is a geometrical method used to find the new position of a wavefront at a later time, given its current position. It is based on two main ideas:
1. **Each point on a wavefront acts as a source of secondary wavelets:** When light travels, every point on a wavefront (a surface of constant phase) becomes a new source that sends out tiny, spherical secondary waves, also called wavelets.
2. **The new wavefront is the envelope of these wavelets:** At any later time, the new position of the wavefront is found by drawing a common tangent surface (an envelope) to all these secondary wavelets.
For example, if you drop a stone into calm water, ripples spread out. All points on the edge of a ripple are in the same phase and form a wavefront. According to Huygens, each point on this ripple then acts as a tiny source, creating new small ripples. The outer boundary of all these small ripples forms the next bigger ripple.
This principle helps explain phenomena like reflection, refraction, and diffraction. The wavelets travel with the same speed as the original wave in that medium. While the model initially suggested a "backwave," Huygens later made an ad-hoc assumption that the amplitude of secondary wavelets is zero in the backward direction to align with observations.
In simple words: Huygens' principle says that every point on a wave of light acts like a tiny new light source, sending out its own small waves. To find where the main light wave will be next, you just draw a line (or surface) that touches all these tiny new waves. It's a way to track how light waves spread out over time.

🎯 Exam Tip: Explain the two main postulates of Huygens' principle clearly. Briefly mention how it helps understand wave propagation and phenomena like reflection/refraction, and acknowledge the "backwave" limitation if relevant.

Question 2. Write definition of interference of light.
Answer: Interference of light is the phenomenon where two or more light waves combine to form a resultant wave of greater, smaller, or the same amplitude. This results in a pattern of alternating bright and dark regions (fringes) in the medium. This happens when coherent light sources (sources with the same frequency and a constant phase difference) superimpose, causing a redistribution of light energy. When waves combine to increase brightness, it's called constructive interference. When they cancel out, resulting in darkness, it's called destructive interference.
In simple words: Interference of light is when two light waves meet and combine to make a new wave. Sometimes they make the light brighter (constructive interference), and sometimes they cancel each other out, making it darker (destructive interference). This creates patterns of bright and dark lines.

🎯 Exam Tip: The definition should highlight the superposition of waves, the redistribution of energy, and the requirement of coherent sources, leading to observable bright and dark fringes.

Question 4. What do you understand by diffraction of light? Compare between diffraction of sound and diffraction of light.
Answer: Diffraction of light is the phenomenon where light waves bend around obstacles or spread out after passing through small apertures. This bending causes the light to enter regions that would otherwise be shadowed by ray optics, leading to a pattern of bright and dark fringes.
**Comparison of Diffraction of Sound and Light:**
1. **Wavelength:** The most significant difference lies in their wavelengths. Visible light has a very small wavelength (around \( 10^{-7} \) m), while sound waves have much larger wavelengths (typically from centimeters to meters).
2. **Observable Effects:** Because light's wavelength is tiny compared to most everyday objects (like doors, buildings), its diffraction is usually not easily observed, making light appear to travel in straight lines. However, sound waves' wavelengths are comparable to the size of common objects.
3. **Everyday Examples:** This explains why we can easily hear someone talking from around a corner (sound diffracts) but cannot see them (light does not noticeably diffract around the corner). The diffraction effect is more pronounced when the size of the obstacle or aperture is on the order of the wavelength of the waves.
4. **Wave Type:** Sound waves are longitudinal waves (vibrations parallel to propagation), whereas light waves are transverse waves (vibrations perpendicular to propagation). However, diffraction occurs for both types of waves, so this difference doesn't explain the ease of observation.
In simple words: Diffraction is when waves bend around things or spread out after going through a small gap. Sound waves bend easily around corners because their waves are long, similar to the size of everyday objects. Light waves are very short, so they don't bend much around most things, which is why we usually see light traveling in straight lines.

🎯 Exam Tip: Define diffraction first. For comparison, clearly state the difference in wavelengths of sound and light. Emphasize that significant diffraction occurs when the obstacle/aperture size is comparable to the wavelength, and use everyday examples to illustrate the point.

Question 5. Define resolving power of microscope. How does it affect when :
(i) Wavelength of illuminating radiation decreases?
(ii) Diameter of object lens decreases? Give explanation of your answer.
Answer: The resolving power of a microscope tells us how clearly it can separate two very close objects. It is defined as the inverse of the smallest distance between two point objects that can still be seen as separate when viewed through the microscope. This means a higher resolving power helps see smaller details.
The resolving power (R.P.) of a microscope is given by the formula: \( R.P. = \frac{1}{d} = \frac{2 n \sin \theta}{\lambda} \)
(i) As the wavelength \( \lambda \) of the illuminating radiation decreases, the resolving power (R.P.) increases. This is because R.P. is inversely proportional to wavelength.
(ii) When the diameter of the objective lens decreases, the semi-vertical angle \( \theta \) also decreases. Since R.P. is directly proportional to \( \sin \theta \), a decrease in \( \theta \) (and thus \( \sin \theta \)) will cause the resolving power of the microscope to decrease.
In simple words: Resolving power means how well a microscope can show small, separate details. If the light's wavelength gets shorter, the microscope sees more clearly. If the lens is smaller, it sees less clearly.

🎯 Exam Tip: Remember that resolving power is inversely proportional to wavelength but directly proportional to the numerical aperture (which depends on the refractive index 'n' and angle \( \sin \theta \)).

Question 7. Describe structure of Polaroid.
Answer: A Polaroid is an inexpensive and easy way to create plane-polarized light. It is made from a large film held between two glass plates. To make this film, tiny crystals of quinine or sulphate or herpethite are arranged on a sheet of Nitro-cellulose. These crystals are carefully aligned so their optic axes are all parallel to each other. This alignment is crucial for its function.
These crystals have a strong property called dichroism, which means they absorb one of two refracted light rays completely, allowing only the other ray to pass through. This emerging ray is plane-polarized light. Each Polaroid has a specific "Polarizing Direction" that aligns with the crystals. When unpolarized light hits a Polaroid, only the light waves vibrating parallel to this direction pass through, making the light polarized. This helps filter light to reduce glare.
Figure 12.37 shows how the parallel lines within the Polaroid represent its polarizing direction.
In simple words: A Polaroid is like a special filter made of tiny, aligned crystals. It lets only light vibrating in one specific direction pass through, making the light "polarized" and reducing glare.

🎯 Exam Tip: Key features to remember are its cheap production, strong dichroic crystals, and the alignment of optic axes to produce plane-polarized light.

Question 8. What do you understand by double refraction.
Answer: Double refraction, also known as birefringence, happens when a ray of ordinary light enters certain crystals, like calcite or quartz. Instead of passing through as a single ray, it splits into two separate refracted rays inside the crystal. This effect is why objects viewed through these crystals can appear doubled.
One of these split rays behaves normally, following Snell's law of refraction; it's called the 'Ordinary-ray' (O-ray). The other ray does not follow Snell's law and is called the 'Extraordinary-ray' (E-ray). Both the O-ray and E-ray are plane-polarized, but their planes of polarization are perpendicular to each other. By removing one of these two rays, such as with a Nicol prism, pure plane-polarized light can be obtained from the crystal.
A Nicol prism, for instance, is specially designed using a calcite crystal to separate these two rays and allow only the E-ray to emerge as plane-polarized light. Figure 12.36 shows the construction and working of a Nicol prism, illustrating how O-rays are removed through total internal reflection, while Figure 12.38 shows the effect of using two polaroids in parallel and crossed positions to demonstrate polarization.
In simple words: Double refraction is when a single light ray splits into two rays inside special crystals. One ray acts normal (O-ray), and the other acts differently (E-ray). Both are polarized in different directions. Devices like a Nicol prism use this to get pure polarized light.

🎯 Exam Tip: Distinguish clearly between O-ray (follows Snell's law) and E-ray (does not follow Snell's law), and state that both are plane-polarized but in perpendicular planes.

Question 10. Distinguish between Fresnel and Fraunhoffer diffraction.
Answer: Diffraction is when light waves bend around obstacles or spread out after passing through an opening. There are two main types of diffraction:
**Differences between Interference and Diffraction:**
1. Interference happens when light from two different coherent sources combines. Diffraction happens when light from different parts of the same wavefront interacts.
2. In interference, the bright and dark fringes can have different widths, but in diffraction, the fringes always have varying widths.
3. In interference, the darkest fringes are perfectly dark if the light waves have the same amplitude. In diffraction, the darkest fringes are not perfectly dark.
4. All bright fringes in interference have the same intensity, but in diffraction, their intensity varies.
5. Interference patterns show a clear difference between bright and dark fringes. Diffraction patterns have less contrast.
**Types of Diffraction:**
(i) **Fresnel Diffraction:** In this type, the light source and the screen are close to the obstacle causing diffraction. No lenses or mirrors are used to modify the light beam. The wavefronts involved are typically spherical or cylindrical.
(ii) **Fraunhofer Diffraction:** Here, both the light source and the screen are effectively far from the diffracting obstacle. This is usually achieved by placing lenses to make the light parallel before and after the obstacle, creating plane wavefronts. Fraunhofer diffraction patterns are simpler to analyze.
Figure 12.20 illustrates the setup for Fresnel diffraction, showing how the light spreads out from a point source near an aperture.
In simple words: Diffraction means light bending around things. Interference is when light from two sources mixes. Fresnel diffraction happens when the light source and screen are close to the obstacle, making round or curved wavefronts. Fraunhofer diffraction happens when they are far away, making flat wavefronts.

🎯 Exam Tip: For diffraction, remember that Fresnel diffraction involves spherical/cylindrical wavefronts and close sources/screens, while Fraunhofer diffraction involves plane wavefronts and distant sources/screens (often using lenses).

RBSE Class 12 Physics Chapter 12 Long Answer Type Questions

Question 1. Explain the phenomenon of refraction of light on the basis of Huygen's secondary wavelets theory and verify Snell's law of refraction.
Answer: Huygens' principle can be used to explain how light refracts when it passes from one medium to another.
**Refraction at a Plane Surface:**
Imagine a plane wavefront, called AB, hitting a flat surface (PP') that separates two different media, as shown in Figure 12.4. Let the speed of light in the first medium be \( v_1 \) and in the second medium be \( v_2 \). The wavefront AB approaches the surface at an angle 'i', which is the angle of incidence.
When point A of the wavefront touches the surface, it acts as a source for new secondary wavelets in the second medium. As the wavefront continues to move, other points on AB also touch the surface and generate their own wavelets. While the wavefront travels from C to E in the first medium (distance \( v_1 t \)), the wavelet from A travels into the second medium.
To find the new refracted wavefront, we draw a sphere of radius \( v_2 t \) from point A in the second medium. A tangent plane (CE) drawn from point C to this sphere represents the new refracted wavefront.
Using geometry from the triangles formed (e.g., \( \Delta AEC \) and \( \Delta ADC \)), we can show the relationship between the angles and speeds:
\( \sin i = \frac{CE}{AC} = \frac{v_1 t}{AC} \)
\( \sin r = \frac{AE}{AC} = \frac{v_2 t}{AC} \)
Dividing these, we get:
\( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \)
This equation is Snell's Law. If the ray bends towards the normal (meaning \( r < i \)), then \( v_2 \) must be less than \( v_1 \). This implies that light travels slower in the denser medium, which aligns with wave theory and experimental observations. If we consider refractive indices \( n_1 = \frac{c}{v_1} \) and \( n_2 = \frac{c}{v_2} \), where 'c' is the speed of light in vacuum, then Snell's Law can be written as \( n_1 \sin i = n_2 \sin r \). When light enters a denser medium, its wavelength and speed decrease, but its frequency stays the same.
The process can also be applied for refraction from a denser to a rarer medium, as illustrated in Figure 12.5, where the refracted wavefront bends away from the normal.
In simple words: Huygens' principle helps us understand how light bends (refracts). When light passes from one material to another, like air to water, its speed changes. This change in speed causes the light's wavefronts to bend, leading to Snell's Law which links the bending angle to the light's speed in each material.

🎯 Exam Tip: Clearly draw and label the wavefronts, incident and refracted rays, and angles. The key idea is that each point on a wavefront acts as a source for secondary wavelets, and the new wavefront is the tangent to these wavelets.

Question 2. Describe the reflection of light on the basis of Huygens' wave theory.
Answer: Huygens' wave theory explains reflection by considering light as a series of wavefronts.
**Reflection at a Plane Surface:**
Imagine a plane wavefront, AB, hitting a flat reflecting surface, AY, as shown in Figure 12.3. The wavefront and the reflecting surface are perpendicular to the plane of the paper.
As the wavefront AB reaches the surface, point B touches first. According to Huygens' principle, every point on the wavefront that touches the surface becomes a source of secondary wavelets, which are small spherical waves spreading out.
While the wavefront from A travels to C (a distance of \( ct \), where 'c' is the speed of light in the first medium and 't' is the time), the secondary wavelet starting from B also expands outwards. Since light travels at the same speed in the same medium, this wavelet from B will have expanded into a hemisphere of radius \( ct \).
To find the new reflected wavefront, we draw a tangent plane from point C to the hemisphere originating from B. This tangent plane, CD, represents the new reflected wavefront.
By examining the geometry of the triangles formed, such as \( \Delta ABC \) and \( \Delta DCB \):
- Angle \( \angle BAC = 90^\circ \) (angle of incidence)
- Angle \( \angle CDB = 90^\circ \) (angle of reflection)
- BC is common to both triangles.
- AC (distance traveled by A to C) = BD (radius of wavelet from B) = \( ct \).
Therefore, \( \Delta ABC \) is congruent to \( \Delta DCB \). This congruence proves that the angle of incidence ( \( i \)) is equal to the angle of reflection ( \( r \)).
This derivation from Huygens' principle successfully proves the second law of reflection, which states that the angle of incidence equals the angle of reflection. This also confirms that the incident ray, the reflected ray, and the normal all lie in the same plane.
In simple words: When light hits a mirror, it bounces off. Huygens' theory explains this by saying that each point on a light wave acts like a tiny new light source. When these tiny waves hit the mirror and bounce, they create a new overall wave that reflects the light. This way, the angle at which light hits the mirror is the same as the angle at which it bounces off.

🎯 Exam Tip: Draw a clear diagram showing the incident wavefront, reflected wavefront, and secondary wavelets. The key is to demonstrate how the congruence of triangles proves that the angle of incidence equals the angle of reflection.

Question 3. Describe analytically phenomenon of interference of light and explain constructive and destructive interference.
Answer: **Interference of Two Waves:**
Interference of light is a phenomenon where two or more light waves combine to form a resultant wave of greater, lower, or the same amplitude. This results in an uneven distribution of light energy, creating patterns of bright and dark regions.
When two light waves of the same frequency from coherent sources travel in the same direction and superimpose each other, the resulting intensity of light at different points in the medium varies. Sometimes, the intensity is greater than the sum of individual intensities (constructive interference), and sometimes it is less (destructive interference). This variation in resultant intensity is what we call interference.
**Mathematical Analysis:**
Let's consider two waves with the same frequency (\( \omega = 2\pi\eta \)) given by:
\( y_1 = a \sin(\omega t) \) ...(1)
\( y_2 = b \sin(\omega t + \phi) \) ...(2)
Here, 'a' and 'b' are the amplitudes, and \( \phi \) is the phase difference between the waves. When these waves superimpose, the resultant displacement 'y' is:
\( y = y_1 + y_2 \)
\( y = a \sin(\omega t) + b \sin(\omega t + \phi) \)
Expanding \( b \sin(\omega t + \phi) \) gives \( b(\sin(\omega t) \cos \phi + \cos(\omega t) \sin \phi) \).
So, \( y = a \sin(\omega t) + b \sin(\omega t) \cos \phi + b \cos(\omega t) \sin \phi \)
\( y = (a + b \cos \phi) \sin(\omega t) + (b \sin \phi) \cos(\omega t) \)
Let \( A \cos \theta = a + b \cos \phi \) ...(3)
And \( A \sin \theta = b \sin \phi \) ...(4)
Substituting these, we get:
\( y = A \cos \theta \sin(\omega t) + A \sin \theta \cos(\omega t) \)
\( y = A \sin(\omega t + \theta) \) ...(5)
This is the equation of the resultant wave, where A is its amplitude.
**Resultant Amplitude (A):**
Squaring and adding equations (3) and (4):
\( A^2 \cos^2 \theta + A^2 \sin^2 \theta = (a + b \cos \phi)^2 + (b \sin \phi)^2 \)
\( A^2 (\cos^2 \theta + \sin^2 \theta) = a^2 + 2ab \cos \phi + b^2 \cos^2 \phi + b^2 \sin^2 \phi \)
\( A^2 = a^2 + b^2 + 2ab \cos \phi \)
So, \( A = \sqrt{a^2 + b^2 + 2ab \cos \phi} \) ...(6)
**Resultant Intensity (I):**
Intensity is proportional to the square of the amplitude (\( I \propto A^2 \)). If \( I_1 = Ka^2 \) and \( I_2 = Kb^2 \) for some constant K:
\( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \) ...(10)
**Constructive Interference:**
This occurs when the resultant intensity is maximum. This happens when \( \cos \phi = +1 \), meaning the phase difference \( \phi = 0, 2\pi, 4\pi, \ldots, 2n\pi \) (where \( n = 0, 1, 2, \ldots \)).
In this case, \( A_{max} = \sqrt{a^2 + b^2 + 2ab} = \sqrt{(a+b)^2} = a+b \) ...(7)
And \( I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2 \).
If \( a=b \), then \( I_{max} = 4I_1 \).
**Destructive Interference:**
This occurs when the resultant intensity is minimum. This happens when \( \cos \phi = -1 \), meaning the phase difference \( \phi = \pi, 3\pi, 5\pi, \ldots, (2n-1)\pi \) (where \( n = 1, 2, 3, \ldots \)).
In this case, \( A_{min} = \sqrt{a^2 + b^2 - 2ab} = \sqrt{(a-b)^2} = |a-b| \) ...(8)
And \( I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2 \).
If \( a=b \), then \( I_{min} = 0 \), meaning complete darkness.
These analytical results show how wave superposition leads to varying light intensities, forming the characteristic interference pattern.
In simple words: When two light waves meet, their energies can add up or cancel out. If they add up, it's "constructive interference" (bright spot). If they cancel out, it's "destructive interference" (dark spot). This happens because of their phase difference, which is like how aligned their peaks and valleys are.

🎯 Exam Tip: Remember the conditions for maximum intensity (phase difference \( 2n\pi \)) and minimum intensity (phase difference \( (2n-1)\pi \)), along with the formulas for resultant amplitude and intensity.

Question 4. What is meant by diffraction of light? Why is it easy to see diffraction of sound waves compare to light waves? Differentiate between Fresnel and Fraunhoffer diffraction.
Answer: **Diffraction of Light:**
Diffraction is the phenomenon where light waves bend around the edges of an obstacle or spread out after passing through a small opening. This causes light to enter the geometrical shadow region, which cannot be explained by light traveling only in straight lines. The resulting pattern of bright and dark fringes is called a diffraction pattern. This bending and spreading of light is noticeable when the size of the obstacle or aperture is comparable to the wavelength of the light.
**Why Sound Waves Diffract More Easily Than Light Waves:**
It is much easier to observe the diffraction of sound waves compared to light waves. This is because the extent of diffraction depends on the ratio of the wavelength of the wave to the size of the obstacle or aperture.
- **Wavelength of Sound:** Sound waves have relatively large wavelengths, typically ranging from a few centimeters to several meters.
- **Wavelength of Light:** Light waves have very small wavelengths, in the order of \( 10^{-7} \) meters (hundreds of nanometers).
Common objects and openings in our daily environment, like doorways, windows, or building corners, have sizes comparable to the wavelengths of sound waves. This is why we can hear sound from a room even when we cannot see the source of the sound (e.g., listening to a TV from an adjacent room).
However, these everyday objects are vastly larger than the wavelength of light. For light to diffract noticeably, it would need to pass through extremely tiny apertures or around very small obstacles, which are not common. Therefore, light appears to travel in straight lines in most everyday situations, and its diffraction is less obvious without specialized setups. Diffraction is a fundamental wave characteristic.
**Differentiation between Fresnel and Fraunhofer Diffraction:**
There are two main categories of diffraction:
1. **Fresnel Diffraction:**
- **Source and Screen:** The light source and the observation screen are located at finite (close) distances from the diffracting obstacle or aperture.
- **Wavefronts:** The wavefronts involved are typically spherical or cylindrical.
- **Lenses:** No lenses are used to make the incident or diffracted wavefronts parallel.
- **Pattern:** The diffraction pattern depends on the distance from the source and screen to the obstacle. It can be complex.
- **Example:** As shown in Figure 12.20 (and conceptually in Figure 12.19(b)), light spreads out from a point source near an aperture, creating a diffraction pattern.
2. **Fraunhofer Diffraction:**
- **Source and Screen:** The light source and the observation screen are effectively at infinite distances from the diffracting obstacle or aperture. This is usually achieved by using converging lenses to make the incident light parallel and to focus the diffracted light.
- **Wavefronts:** The wavefronts involved are plane wavefronts.
- **Lenses:** Lenses are typically used to collimate the light before it hits the obstacle and to focus the diffracted light onto the screen.
- **Pattern:** The diffraction pattern is observed in the focal plane of the lens and does not depend on the distance from the source or screen, only on the angle of diffraction.
- **Example:** Diffraction at a single slit or double slit, as observed through a telescope or spectrometer. Figure 12.21 conceptually shows Fraunhofer diffraction, where parallel rays are considered.
In simple words: Diffraction is when waves bend around obstacles. Sound waves bend easily because their wavelength is big, like the size of a door. Light waves have tiny wavelengths, so they only bend around very small things, making their diffraction harder to see. Fresnel diffraction happens when light source and screen are close to the obstacle, while Fraunhofer diffraction happens when they are far away, often using lenses to make the light parallel.

🎯 Exam Tip: The key difference in observable diffraction lies in the wavelength-to-obstacle size ratio. Clearly define both types of diffraction and list their distinguishing features, especially regarding source/screen distance and wavefront shape.

Question 5. Explain the phenomenon of diffraction of light at a single slit.
Answer: **Fraunhofer Diffraction due to a Single Slit:**
When a plane wavefront of monochromatic light (light of a single color) passes through a very narrow single slit, it spreads out, and a diffraction pattern is observed on a screen placed far away. This spreading of light is called Fraunhofer diffraction.
As shown in Figure 12.22, a source (S) of monochromatic light is placed at the focus of a convex lens (L1). This creates a parallel beam of light, forming a plane wavefront (WW'), which then strikes a narrow rectangular slit (AB) of width 'a'.
According to Huygens' principle, every point within the slit (AB) acts as a source of secondary wavelets. These wavelets start in the same phase but travel in all directions after passing through the slit. A second convex lens (L2) is used to focus these diffracted wavelets onto a screen placed in its focal plane.
1. **Central Bright Fringe (Principal Maximum):** All secondary wavelets traveling straight through the slit (at an angle \( \theta = 0 \)) are focused at the central point (O) on the screen. Since they all travel the same path distance to O (or path difference is zero), they arrive in phase and interfere constructively, producing a very bright and wide central maximum.
2. **Minima (Dark Fringes):** Dark fringes (minima) occur when the secondary wavelets from different parts of the slit interfere destructively. This happens when the path difference between the wavelets from the extreme ends of the slit (A and B) is an integer multiple of the wavelength \( \lambda \).
- The path difference is given by \( BN = a \sin \theta \).
- For the first dark fringe, \( a \sin \theta_1 = \lambda \)
- For the second dark fringe, \( a \sin \theta_2 = 2\lambda \)
- In general, for the n-th dark fringe, the condition is:
\( a \sin \theta_n = n\lambda \) (where \( n = 1, 2, 3, \ldots \)) ...(1)
- The angular positions of minima are given by \( \sin \theta_n = \frac{n\lambda}{a} \).
3. **Secondary Maxima (Bright Fringes):** Bright fringes (secondary maxima) occur between the minima. These are less intense and narrower than the central maximum. They occur when the path difference is an odd multiple of \( \frac{\lambda}{2} \).
- For the first secondary maximum, \( a \sin \theta'_1 = \frac{3\lambda}{2} \)
- For the second secondary maximum, \( a \sin \theta'_2 = \frac{5\lambda}{2} \)
- In general, for the n-th secondary maximum, the condition is:
\( a \sin \theta'_n = (2n + 1)\frac{\lambda}{2} \) (where \( n = 1, 2, 3, \ldots \)) ...(2)
- The angular positions of secondary maxima are given by \( \sin \theta'_n = \frac{(2n+1)\lambda}{2a} \).
The intensity of these secondary maxima decreases significantly as 'n' increases. The diffraction pattern thus consists of a bright central maximum flanked by alternating dark and progressively fainter bright fringes.
In simple words: When light shines through a tiny slit, it doesn't just make a single bright line. Instead, it spreads out and creates a pattern of bright and dark bands on a screen. The middle band is the brightest, and the others get dimmer. This happens because the light waves from different parts of the slit interfere with each other, sometimes adding up and sometimes canceling out.

🎯 Exam Tip: Remember that the central maximum is twice as wide as the secondary maxima, and its intensity is much greater. Also, know the conditions for both minima ( \( a \sin \theta = n\lambda \) ) and secondary maxima ( \( a \sin \theta = (2n+1)\frac{\lambda}{2} \) ).

Question 6. What is polarisation? Explain polarisation with the help of electric vector. Clarify why is it a characteristic of transverse waves.
Answer: **Polarisation:**
Polarisation is the phenomenon where the vibrations of the electric field vector (and magnetic field vector) of a light wave are restricted to a single plane or a specific orientation perpendicular to the direction of wave propagation. Unpolarized light has electric field vibrations in all possible directions perpendicular to its propagation. Polarized light has these vibrations confined to one specific direction.
**Explanation with Electric Vector:**
According to Maxwell's electromagnetic wave theory, light waves are electromagnetic waves. They consist of oscillating electric (\( \vec{E} \)) and magnetic (\( \vec{B} \)) field vectors that are perpendicular to each other and also perpendicular to the direction of wave propagation. In Figure 12.31, the electric field vector (\( \vec{E} \)) is responsible for all optical effects and is often called the 'light vector'.
In unpolarized light, the electric vector vibrates randomly in all possible planes perpendicular to the direction of propagation. When light undergoes polarization, these vibrations are restricted. For example, in plane-polarized light, the electric vector vibrates only in one specific plane containing the direction of propagation. This plane is called the 'plane of vibration'. The plane perpendicular to the plane of vibration, containing the direction of propagation, is called the 'plane of polarization'.
**Why Polarisation is a Characteristic of Transverse Waves:**
Polarisation is a unique characteristic of transverse waves. A transverse wave is one where the particles of the medium (or the field vectors, in the case of electromagnetic waves) vibrate perpendicular to the direction the wave travels.
- **Transverse Waves:** Since the electric field vibrations in light waves occur perpendicular to the direction of propagation, these vibrations have multiple possible directions in that perpendicular plane (e.g., up-down, left-right, diagonal). Polarisation simply restricts these vibrations to one specific direction. This is possible because there are multiple possible directions of vibration perpendicular to the wave's motion.
- **Longitudinal Waves:** In contrast, a longitudinal wave is one where the particles of the medium vibrate parallel to the direction of wave propagation. For example, sound waves are longitudinal. There is only one direction for vibrations parallel to propagation. Therefore, there's no way to restrict these vibrations to a 'plane' or 'orientation' other than the direction of motion itself. You cannot isolate a particular plane of vibration for a longitudinal wave.
**Experimental Proof (Tourmaline Experiment):**
The transverse nature of light can be experimentally proven using tourmaline crystals (Figure 12.30). If unpolarized light passes through a tourmaline crystal (N1), the emergent light is plane-polarized. If a second tourmaline crystal (N2), called the analyser, is placed after N1, the intensity of light emerging from N2 changes as N2 is rotated. When the optic axes of N1 and N2 are parallel, maximum light passes through. When they are perpendicular (crossed position), almost no light passes through, and darkness is observed. This variation in intensity with rotation demonstrates that the light waves have vibrations restricted to a specific plane, confirming their transverse nature. If light were longitudinal, rotating the analyser would not affect the intensity, as vibrations would always be along the direction of propagation. This dependence on the inclination of the axes is only possible if the waves are transverse.
In simple words: Polarisation is when light waves vibrate only in one specific direction, not all directions. Imagine shaking a rope up-and-down only, instead of all over the place. Light can do this because it's a "transverse" wave, meaning its vibrations are sideways to its travel direction, giving it different "sides" to vibrate on. Sound waves, which vibrate back-and-forth along their path, cannot be polarized because they don't have these "sides".

🎯 Exam Tip: Emphasize that polarisation restricts transverse vibrations to a single plane. Clearly explain why this is not possible for longitudinal waves due to their vibration direction. The tourmaline crystal experiment is a classic way to demonstrate this principle.

Explanation of Tourmaline-Experiment:

According to Maxwell’s electromagnetic wave theory, light waves are electromagnetic waves, not mechanical waves. For these waves to travel, a theoretical medium called ether was thought to be needed. In these waves, two field vectors – the electric field vector \( \vec{E} \) and the magnetic field vector \( \vec{B} \) – vibrate perpendicular to each other and also perpendicular to the direction the electromagnetic wave travels, as shown in Figure 12.31. The electric field vector \( \vec{E} \) is responsible for all light effects, so it is also called the 'light vector'. Because both field vectors are perpendicular to the direction of travel, light waves are transverse in nature. Light has different properties based on its transverse nature, which are explored through experiments.

We can explain the experiment done with tourmaline crystals using the Electromagnetic Wave Theory. Since light waves are transverse, their vibrations are spread evenly around the direction of propagation. This is why when normal light passes through a tourmaline crystal, its intensity doesn't change when the crystal is rotated. This is because the light vector always finds a parallel path to the crystal's optic axis. It's important to remember that only vibrations parallel to the optic axis come out of the crystal. So, the vibrations in the light that comes out are only in one plane, which is called 'plane polarized light'. This lack of symmetry in the vibrations of the emerging light is caused by the crystal, and this whole effect is called the polarization of light.

When the light coming from the first crystal \( N_1 \) hits the second crystal \( N_2 \), and both crystals' optic axes are parallel, all the vibrations from \( N_1 \) pass through \( N_2 \), making the light intensity maximum. If \( N_2 \) is rotated, only the part of the vibrations aligned with \( N_2 \)'s optic axis passes through, causing the intensity to change. The intensity from \( N_2 \) becomes minimum (almost zero) when the axes of \( N_1 \) and \( N_2 \) are perpendicular to each other. In this position, the crystals are said to be 'crossed'. This shows that the intensity of the light that comes out depends on how the axes are tilted. Similar results are seen when \( N_1 \) is rotated and \( N_2 \) stays still.

(2) Polarisation by Scattering

(3) Polarisation by Double Refraction

Polarisation by Double Refraction:

Some crystals, like calcite and quartz, cause an ordinary light ray to split into parts when it hits them, as seen in Figure 12.35. This effect is known as double refraction. One of these refracted rays follows the normal laws of refraction and is called the 'Ordinary-ray' or 'O-ray'. The other ray does not follow these laws and is called the 'Extraordinary-ray' or 'E-ray'. Both of these rays are plane polarized, with their planes perpendicular to each other. In the O-ray, the vibrations are in a plane perpendicular to the plane of incidence, and in the E-ray, the vibrations are in the plane of incidence. By removing one of these two rays, plane polarized light can be obtained from the crystal.

Question 8. How can you obtain plane polarised light using reflection? What is Brewster's law? Prove that when an unpolarised light is incident on a plane glass surface, then reflected and refracted rays are 90° to each other.
Answer:
Production of Plane Polarised Light
(1) Polarisation by Reflection: In 1808, French engineer Malus discovered that when ordinary light reflects off a transparent surface (like glass), it becomes partially plane polarized. In 1811, scientist 'Brewster' studied this in detail and found that the amount of polarized light in the reflected ray changes with the angle of incidence. When the angle of incidence is adjusted to a specific value, the reflected light becomes completely plane polarized. This special angle is known as the polarizing angle or Brewster's angle. This is a very useful way to produce polarized light. The reflected and refracted rays are always at right angles to each other at this specific angle.
Brewster's Law states that when light is incident at the polarizing angle, the refractive index (n) of the medium is equal to the tangent of the polarizing angle (\( i_p \)). So, \( n = \tan i_p \).
Let's prove that reflected and refracted rays are at right angles to each other when light is incident at Brewster's angle:
According to Snell's law:
\( \frac{\sin i_p}{\sin r} = n \) ... (2)
From Brewster's law:
\( n = \tan i_p = \frac{\sin i_p}{\cos i_p} \) ... (3)
Comparing equations (2) and (3):
\( \frac{\sin i_p}{\sin r} = \frac{\sin i_p}{\cos i_p} \)
\( \implies \sin r = \cos i_p \)
We know that \( \cos i_p = \sin (90^\circ - i_p) \). So,
\( \implies \sin r = \sin (90^\circ - i_p) \)
\( \implies r = 90^\circ - i_p \)
\( \implies i_p + r = 90^\circ \) ... (4)
Now, from Figure 12.33, the angle between the reflected ray and the refracted ray is \( \theta \). The sum of angles on a straight line is \( 180^\circ \). So, \( i_p + \theta + r = 180^\circ \).
Substitute \( i_p + r = 90^\circ \) into this equation:
\( \implies 90^\circ + \theta = 180^\circ \)
\( \implies \theta = 180^\circ - 90^\circ = 90^\circ \) ... (5)
This proves that the reflected and refracted rays are at right angles (\( 90^\circ \)) to each other when light is incident at the polarizing angle. This phenomenon highlights the transverse nature of light waves.
In simple words: Plane polarized light can be made by reflecting unpolarized light at a special angle called Brewster's angle. At this angle, the reflected light is completely polarized, and the reflected ray is exactly \( 90^\circ \) from the refracted ray. This rule helps us understand light's wave nature.

🎯 Exam Tip: Remember Brewster's law \( n = \tan i_p \) and the condition \( i_p + r = 90^\circ \) for reflected and refracted rays to be perpendicular. Clearly stating these points will fetch full marks.

Question 9. Give definition of 'plane of vibration' and 'plane of polarisation'. Define Malus law and explain the parallel and cross condition of Polaroid.
Answer:
Plane of Vibration: The plane that contains both the direction of vibration of the electric field vector and the direction in which the wave is traveling is called the plane of vibration.
For example, if a light wave moves along the X-axis and its electric field vibrates parallel to the Y-axis, then the XY-plane is the plane of vibration.
Plane of Polarisation: The plane that is perpendicular to the plane of vibration, but still contains the direction of wave propagation, is called the plane of polarization. In the example above, if the XY-plane is the plane of vibration, then the XZ-plane (or EFGH in the figure) is the plane of polarization.
Malus's Law: This law describes the intensity of light passing through a polarizer. If \( I_0 \) is the intensity of plane polarized light incident on an analyser (second polarizer) and \( \theta \) is the angle between the transmission axes of the polarizer and the analyser, then the intensity of the transmitted light \( I \) is given by: \( I = I_0 \cos^2 \theta \). This means the light intensity decreases as the angle between the polarizers increases.
Polaroid Conditions (Parallel and Crossed):
Polaroids are thin films placed between glass plates, made of tiny crystals aligned to allow only light vibrating in a specific direction to pass through, thus polarizing it.
Parallel Condition: When two polaroids are placed with their transmission axes parallel to each other (\( \theta = 0^\circ \)), the intensity of light emerging from the second polaroid (analyser) is maximum. This is because \( \cos^2 0^\circ = 1 \), so \( I = I_0 \). In this case, most of the light from the first polaroid passes through the second one. This is shown in Figure 12.30(a) and 12.38(a).
Crossed Condition: When two polaroids are placed with their transmission axes perpendicular to each other (\( \theta = 90^\circ \)), the intensity of light emerging from the second polaroid is minimum (ideally zero). This is because \( \cos^2 90^\circ = 0 \), so \( I = 0 \). In this arrangement, the second polaroid blocks all the light that passed through the first one. This is shown in Figure 12.30(b) and 12.38(b). These observations clearly demonstrate that the light emerging from the first polaroid is plane polarized.
In simple words: The 'plane of vibration' is where light waves wiggle, and the 'plane of polarization' is a plane at right angles to that. Malus's law tells us how much light passes through a second filter based on its angle. If two filters are lined up (parallel), you see full light. If they are crossed (perpendicular), you see no light.

🎯 Exam Tip: When defining plane of vibration and polarization, specify the orientation relative to wave propagation. For Malus's Law and polaroid conditions, the \( \cos^2 \theta \) relationship is key, and be sure to state the angles for maximum and minimum intensity.

RBSE Class 12 Physics Chapter 12 Numerical Questions

Question 1. Two waves of same shape have amplitude ratio 2 : 1. In interference region find out the ratio of maximum and minimum intensities.
Answer:
Given, amplitude ratio \( \frac{a_1}{a_2} = \frac{2}{1} \)
The ratio of maximum to minimum amplitude is:
\( \frac{A_{max}}{A_{min}} = \frac{a_1 + a_2}{a_1 - a_2} \)
Divide numerator and denominator by \( a_2 \):
\( \frac{A_{max}}{A_{min}} = \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} \)
So, \( A_{max} : A_{min} = 3 : 1 \)
Intensity is proportional to the square of amplitude: \( I \propto A^2 \)
\( \frac{I_{max}}{I_{min}} = \left(\frac{A_{max}}{A_{min}}\right)^2 = \left(\frac{3}{1}\right)^2 = \frac{9}{1} \)
Thus, the ratio of maximum to minimum intensities is \( 9 : 1 \). This relationship is fundamental for understanding interference patterns.
In simple words: If two waves have amplitudes in a 2 to 1 ratio, then their maximum combined intensity will be 9 times their minimum combined intensity. This happens because intensity is related to the square of how big the wave is.

🎯 Exam Tip: Remember the relationship between amplitude ratio and intensity ratio: \( \frac{I_{max}}{I_{min}} = \left(\frac{A_{max} + A_{min}}{A_{max} - A_{min}}\right)^2 \). Also, clearly show the steps for calculating the amplitude ratio first.

Question 2. In an interference experiment, two sources of light are used of intensities 7 and 47. Find out the intensities at those points where the phase difference of two waves superimposing are (a) zero (b) \( \frac{\pi}{2} \) (c) \( \pi \).
Answer:
Given, \( I_1 = I \) and \( I_2 = 4I \). We need to find the resultant intensity \( I_R \).
The general formula for resultant intensity is:
\( I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \)
Substituting \( I_1 = I \) and \( I_2 = 4I \):
\( I_R = I + 4I + 2\sqrt{I \cdot 4I} \cos \phi \)
\( I_R = 5I + 2\sqrt{4I^2} \cos \phi \)
\( I_R = 5I + 2(2I) \cos \phi \)
\( I_R = 5I + 4I \cos \phi \)

(a) When phase difference \( \phi = 0 \) (for constructive interference):
\( I_R = 5I + 4I \cos 0 \)
\( I_R = 5I + 4I (1) \)
\( I_R = 9I \)

(b) When phase difference \( \phi = \frac{\pi}{2} \):
\( I_R = 5I + 4I \cos \left(\frac{\pi}{2}\right) \)
\( I_R = 5I + 4I (0) \)
\( I_R = 5I \)

(c) When phase difference \( \phi = \pi \) (for destructive interference):
\( I_R = 5I + 4I \cos \pi \)
\( I_R = 5I + 4I (-1) \)
\( I_R = 5I - 4I \)
\( I_R = I \)
These results show how the phase difference critically impacts the observed light intensity in interference patterns.
In simple words: When two light waves meet, how bright the light gets depends on their phase difference. If they are perfectly in sync (\( \phi = 0 \)), the light is brightest (9I). If they are a quarter-wave out of sync (\( \phi = \frac{\pi}{2} \)), the brightness is 5I. If they are perfectly out of sync (\( \phi = \pi \)), the light is dimmest (I).

🎯 Exam Tip: Always use the general formula for resultant intensity, \( I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \). Remember the cosine values for \( 0, \frac{\pi}{2}, \) and \( \pi \). Ensure consistency if the question and given values differ, using the values employed in the solution steps.

Question 4. The light of wavelength 5500 Å is falling normally on an aperture of width 22 × 10⁻⁵cm. Find the angular positions between two minima on both side of central maxima.
Answer:
Given values:
Wavelength \( \lambda = 5500 \, \text{Å} = 5500 \times 10^{-10} \, \text{m} = 55 \times 10^{-8} \, \text{m} \)
Aperture width \( a = 22 \times 10^{-5} \, \text{cm} = 22 \times 10^{-7} \, \text{m} \)
For minima in a single-slit diffraction pattern, the condition is \( a \sin \theta = n\lambda \), where \( n = 1, 2, 3, \ldots \).
So, \( \sin \theta = \frac{n\lambda}{a} \)
For small angles, \( \sin \theta \approx \theta \). So, \( \theta = \frac{n\lambda}{a} \).

For the first minima on either side of the central maximum (\( n = 1 \)):
\( \theta_1 = \frac{1 \cdot \lambda}{a} \)
\( \theta_1 = \frac{55 \times 10^{-8} \, \text{m}}{22 \times 10^{-7} \, \text{m}} \)
\( \theta_1 = \frac{55}{220} = \frac{1}{4} = 0.25 \, \text{radians} \)

For the second minima on either side of the central maximum (\( n = 2 \)):
\( \theta_2 = \frac{2 \cdot \lambda}{a} \)
\( \theta_2 = 2 \times 0.25 = 0.50 \, \text{radians} \)
These angular positions define where the dark fringes appear relative to the central bright fringe. Knowing these angles helps predict the entire diffraction pattern.
In simple words: We want to find the angles where the dark lines (minima) appear when light passes through a small opening. Using the light's wavelength and the opening's width, we calculate that the first dark line appears at an angle of 0.25 radians from the center, and the second dark line appears at 0.50 radians.

🎯 Exam Tip: Remember the condition for minima in single-slit diffraction: \( a \sin \theta = n\lambda \). For small angles, use \( \theta \approx \frac{n\lambda}{a} \) (in radians). Pay close attention to unit conversions (Å to m, cm to m).

Question 5. Two polaroids are so placed, such that the emerging light from them has maximum intensity. If one polaroid is rotated through 30°, from 90° with respect to other, then find out the ratio of part of the emerging light compared with light of maximum intensity.
Answer:
When two polaroids are placed for maximum intensity, their transmission axes are parallel (\( \theta = 0^\circ \)). If unpolarized light of intensity \( I_0 \) falls on the first polaroid, the intensity emerging from it is \( I_1 = \frac{I_0}{2} \). Since the polaroids are parallel for maximum intensity, the emerging intensity from the second polaroid would also be \( I_{max} = \frac{I_0}{2} \).
The problem states that initially the emerging light has maximum intensity, implying the axes are parallel, and then one polaroid is rotated. We use Malus's Law, \( I = I_{max} \cos^2 \theta \), where \( I_{max} \) is the initial maximum intensity passing through the parallel polaroids, and \( \theta \) is the angle between their axes after rotation.
The problem states: "If one polaroid is rotated through 30°, from 90° with respect to other". This phrasing is a bit ambiguous. It could mean: 1. The initial angle was \( 90^\circ \) (crossed), and then one was rotated \( 30^\circ \), making the new angle \( 90^\circ - 30^\circ = 60^\circ \) or \( 90^\circ + 30^\circ = 120^\circ \). 2. The initial angle was \( 0^\circ \) (parallel for maximum intensity), and it's rotated \( 30^\circ \), making the angle \( 30^\circ \). The "from 90° with respect to other" part might be a distractor or misphrased. Given the phrase "emerging light from them has maximum intensity" initially, it's safer to assume the initial angle between the polaroids was \( 0^\circ \) (parallel). If one polaroid is then rotated by \( 30^\circ \), the new angle \( \theta \) between their transmission axes becomes \( 30^\circ \).
According to Malus's Law, the new intensity \( I \) is:
\( I = I_{max} \cos^2 (30^\circ) \)
\( I = I_{max} \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( I = I_{max} \left(\frac{3}{4}\right) \)
The ratio of the emerging light to the maximum intensity is:
\( \frac{I}{I_{max}} = \frac{3}{4} \)
So, the emerging light intensity will be \( \frac{3}{4} \) of the maximum intensity. This calculation shows how rotating a polarizer affects the light's intensity.
In simple words: When two light filters are set up to let the most light through, they are parallel. If one filter is then turned by 30 degrees, the amount of light that passes through will be three-fourths of the brightest light they could let through.

🎯 Exam Tip: When dealing with rotation of polaroids, always clarify the initial and final angles between their transmission axes. Malus's Law \( I = I_0 \cos^2 \theta \) is the key. Make sure to square the cosine value correctly.

Question 6. When Sun is at 37° angle from horizon, then reflected light from water surface is totally polarised. Find refractive index of water.
Answer:
When reflected light is completely polarized, the angle of incidence is the polarizing angle \( i_p \).
The angle of the Sun from the horizon is \( 37^\circ \). This means the angle of incidence \( i_p \) with the water surface (which is usually horizontal) is the angle from the normal.
The angle of incidence is measured from the normal to the surface.
Since the Sun is \( 37^\circ \) from the horizon, the angle it makes with the *surface* is \( 37^\circ \). The normal is perpendicular to the surface. So the angle of incidence \( i_p \) is \( 90^\circ - 37^\circ = 53^\circ \).
According to Brewster's law:
\( n = \tan i_p \)
\( n = \tan (53^\circ) \)
Using the given approximation, \( \tan 53^\circ \approx \frac{4}{3} \approx 1.33 \).
So, the refractive index of water is approximately \( 1.33 \). This is a common value for the refractive index of water.
In simple words: When sunlight hits water at an angle of 53 degrees, the reflected light becomes fully polarized. Brewster's law states that the refractive index of water is the tangent of this angle. So, the refractive index of water is about 1.33.

🎯 Exam Tip: Be careful to distinguish between the angle from the horizon and the angle of incidence. The angle of incidence \( i_p \) is always measured with respect to the normal to the surface, so \( i_p = 90^\circ - \text{angle from horizon} \). Then apply Brewster's law \( n = \tan i_p \).

Question 7. Polarising directions of two polarising plates are parallel and intensity of light is maximum. How much one plate be rotated that intensity of emerging out light becomes one fourth?
Answer:
Initially, the polarizing plates are parallel, and the emerging light has maximum intensity, \( I_{max} \).
We want the intensity of the emerging light to become one-fourth of the maximum intensity, i.e., \( I = \frac{I_{max}}{4} \).
According to Malus's Law, the intensity of light transmitted through an analyser is given by:
\( I = I_{max} \cos^2 \theta \)
Where \( \theta \) is the angle between the transmission axes of the two polaroids.
Substitute the desired intensity:
\( \frac{I_{max}}{4} = I_{max} \cos^2 \theta \)
Divide both sides by \( I_{max} \):
\( \frac{1}{4} = \cos^2 \theta \)
Take the square root of both sides:
\( \cos \theta = \pm \sqrt{\frac{1}{4}} \)
\( \cos \theta = \pm \frac{1}{2} \)
Since the rotation is from an initial parallel position, we consider the positive angle for simplicity. The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is \( 60^\circ \).
So, one plate must be rotated by \( 60^\circ \) for the intensity to become one-fourth of the maximum. This demonstrates the angular dependence of light intensity through polarizers.
In simple words: If two light filters are parallel and let through maximum light, you need to turn one of them by 60 degrees. After this turn, only one-fourth of the original maximum light will pass through.

🎯 Exam Tip: Remember Malus's law \( I = I_0 \cos^2 \theta \). When solving for \( \theta \), ensure you take the square root first to find \( \cos \theta \), and then determine the angle. Always express the angle in degrees or radians as specified.

RBSE Solutions for Class 12 Physics