CBSE Class 12 Physics Ray and Wave Optics Assignment Set 02

Question. Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is
(a) 10 : 8
(b) 9 : 1
(c) 4 : 1
(d) 2 : 1

Answer: C

Question. Four independent waves are expressed as
(i) y₁ = a₁ sin~t (ii) y₂ = a₂ sin 2ωt
(iii) y₃ = a₃ cos^~t and (iv) y₄ = a₄ sin (ωt + π/3)
The interference is possible between
(a) (i) and (iii)
(b) (i) and (iv)
(c) (iii) and (iv)
(d) not possible at all

Answer: D

Question. Consider sunlight incident on a slit of width 10⁴ A. The image seen through the slit shall
(a) be a fine sharp slit white in colour at the center.
(b) a bright slit white at the center diffusing to zero intensities at the edges.
(c) a bright slit white at the center diffusing to regions of different colours.
(d) only be a diffused slit white in colour.

Answer: A

Question. In a Young’s double-slit experiment the fringe width is found to be 0.4 mm. If the whole apparatus is dipped in water of refractive index 4/3, without disturbing the arrangement, the new fringe width will be
(a) 0.30 mm
(b) 0.40 mm
(c) 0.53 mm
(d) 0.2 mm

Answer: A

Question. In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength m of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to
(a) 1.8 mm
(b) 1.9 mm
(c) 2.1 mm
(d) 1.7 mm

Answer: B

Question. In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case
(a) there shall be alternate interference patterns of red and blue.
(b) there shall be an interference pattern for red distinct from that for blue.
(c) there shall be no interference fringes.
(d) there shall be an interference pattern for red mixing with one for blue.

Answer: C

Question. In a Young’s double-slit experiment, the source S and two slits A and B are horizontal, with slit A above slit B. The fringes are observed on a vertical screen K. The optical path length from S to B is increased very slightly (by introducing a transparent material of higher refractive index) and optical path length from S to A is not changed. As a result, the fringe system on K moves
(a) vertically downwards slightly
(b) vertically upwards slightly
(c) horizontally slightly to the left
(d) horizontally slightly to the right

Answer: A

Question. In Young’s double-slit experiment, the distance between the slit sources and the screen is 1 m. If the distance between the slits is 2 mm and the wavelength of light used is 600 nm, the fringe width is
(a) 3 mm
(b) 0.3 mm
(c) 6 mm
(d) 0.6 mm

Answer: B

Question. The Young’s double-slit experiment is performed with blue and green lights of wavelengths 4360 Å and 5460 Å respectively. If x is the distance of 4th maxima from the central one, then
(a) (x)_blue = (x)_green
(b) (x)_blue > (x)_green
(c) (x)_blue < (x)_green
(d) (x)_blue / (x)_green = 5460 / 4360

Answer: C

Question. The angular resolution of a 10 cm diameter telescope at a-wavelength 500 nm is of the order of
(a) 10^–4 rad
(b) 10^–6 rad
(c) 10^–3 rad
(d) 10⁷ rad

Answer: B

Question. A telescope has an objective lens of 10 cm diameter and is situated at a distance of 1 km from two objects. The minimum distance between these objects that can be resolved by the telescope, when the mean wavelength of light is 5000 Å is of the order of
(a) 5 mm
(b) 5 cm
(c) 2.5 m
(d) 5 m

Answer: A

Question. Which of the following phenomenon cannot take place with longitudinal waves (e.g., sound waves)?
(a) reflection
(b) interference
(c) diffraction
(d) polarisation

Answer: D

Question. Unpolarised light is incident from air on a plane surface of a material of refractive index n. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
(a) Reflected light is polarised with its electric vector parallel to the plane of incidence
(b) Reflected light is polarised with its electric vector perpendicular to the plane of incidence
(c) i = sin^–1 (1/n)
(d) i = tan^–1 (1/n)

Answer: A

Question. An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
(a) small focal length and large diameter
(b) large focal length and small diameter
(c) large focal length and large diameter
(d) small focal length and small diameter

Answer: B

Question. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewsters angle φ. If n represents the refractive index of glass with respect to air, then the angle between the reflected and refracted rays is
(a) 90 + φ
(b) sin^–1 (n cosΦ)
(c) 90°
(d) 90° - sin 1(sin Φ / n)

Answer: B

Question. Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(a) the size decreases
(b) the intensity increases
(c) the size increases
(d) the intensity decreases

Answer: A, B

Question. For light diverging from a point source
(a) the wavefront is spherical.
(b) the intensity decreases in proportion to the distance squared.
(c) the wavefront is parabolic.
(d) the intensity at the wavefront does not depend on the distance.

Answer: A, B

Fill in the Blanks

Question. A beam of light is incident normally upon a polariser and the intensity of emergent beam is IO.
The intensity of the emergent beam is found to be unchanged when the polariser is rotated about an axis perpendicular to the pass axis. Incident beam is _________________ in nature.
Answer: unpolarised

Question. The value of Brewster angle depends on the nature of the transparent refracting medium and the _________________ of light used.
Answer: wavelength

Question. In Young’s double slit experiment, the fringe width is given by _________________.
Answer: b = Dλ/d

Question. The phase difference between two waves in _________________ interference is given as an even multiple of p.
Answer: constructive

Question. Fringe width is different as separation between two consecutive _________________ or _________________.
Answer: maxima, minima

Question. _________________ of light occurs when size of the obstacle of aperture is comparable to the wavelength of light.
Answer: Diffraction

Question. Continuous locus of oscillation with constant phase is called as _________________.
Answer: wave-front

Question. In interference and _________________, the light energy is redistributed, increases in one region and decreases in other.
Answer: diffraction

Question. At polarising angle the refracted and reflected rays are _________________ to each other.
Answer: perpendicular

Question. The tangent of angle of polarization as light ray travels from air to glass is equal to the refractive index. This law is called as _________________.
Answer: Brewster’s law

WAVE OPTICS
 

Short Answer Type Questions
 

Question. State the conditions which must be satisfied for two light sources to be coherent.
Answer: For two light sources to be coherent, they must satisfy these criteria:
(i) Both sources must emit light of the exact same frequency (or wavelength).
(ii) The phase difference between the waves emitted by them must remain constant over time.
In simple words: To be coherent, two light waves must have the same frequency and keep a steady phase relationship with each other without changing over time.

Exam Tip: Always mention both "same frequency/wavelength" and "constant phase difference" as these are the primary keywords that examiners look for when grading this definition.

Question. Two independent light sources cannot act as coherent sources. Why?
Answer: It is impossible for two separate light sources to be coherent. Light emission occurs when individual excited atoms de-excite and return to their ground state. Since even a tiny source consists of billions of such atoms, they transition independently. Consequently, it is impossible for them to produce light waves that maintain a constant phase relationship.
In simple words: Different light bulbs or candles cannot be coherent because their atoms emit light in random bursts, making it impossible to keep their light waves perfectly in step with one another.

Exam Tip: When explaining this, highlight the random and independent emission of light by individual atoms when returning to the ground state.

Question. No interference pattern is detected when two coherent sources are infinitely close to one another. Why?
Answer: The fringe width in an interference pattern is determined by the formula: \[ \beta = \frac{\lambda D}{d} \] This relation shows that the fringe width is inversely proportional to the slit separation, or \( \beta \propto \frac{1}{d} \). As the distance between the two sources \( d \) approaches zero (\( d \to 0 \)), the fringe width \( \beta \) approaches infinity (\( \beta \to \infty \)). Consequently, a single bright or dark fringe expands to cover the entire observation screen, making it impossible to detect any distinct interference pattern.
In simple words: If the two slits are extremely close together, the interference stripes become so wide that just one stripe fills the whole screen, making the pattern invisible.

Exam Tip: State the mathematical formula for fringe width and explicitly show the limit where \( d \to 0 \) causes \( \beta \to \infty \) to score full marks on this question.

Question. If the path difference produced due to interference of light coming out of two slits for yellow colour of light at a point on the screen be \( \frac{3\lambda}{2} \), what will be the colour of the fringe at the point? Give reason also.
Answer: The specified path difference of \( \frac{3\lambda}{2} \) corresponds to the condition for destructive interference (minimum intensity) of the yellow wavelength. Therefore, if only yellow light is used, a dark band will appear at that location on the screen. However, if white light is employed, all of its spectral components will be visible at that point except for the yellow light, which gets cancelled out.
In simple words: The path difference cancels out the yellow light. So, yellow light leaves a dark spot, while white light leaves a spot with all colors except yellow.

Exam Tip: For questions involving path differences of odd multiples of half-wavelengths, always specify that it results in destructive interference and minimum intensity.

Question. What happens to the interference pattern if the phase difference between the two sources varies continuously?
Answer: If the phase relationship between the sources changes continuously, the locations of the bright and dark bands will shift at an extremely fast rate. Because the human eye cannot resolve such rapid changes, we perceive only a steady, uniform distribution of light across the screen, causing the interference pattern to vanish.
In simple words: If the phase changes constantly, the stripes dance around so fast that they blur together, and we only see a solid, flat glow on the screen.

Exam Tip: Mention the rapid shifting of fringe positions and the limitation of human visual persistence (persistence of vision) which leads to the observation of uniform illumination.

Question. Radiowaves diffract pronouncedly around the buildings, while light waves, which are e.m. waves do not why?
Answer: Significant wave diffraction only occurs when the wavelength of the wave is comparable to the physical dimensions of the obstructing object. Radio waves, especially short-wavelength radio waves, have wavelengths that are roughly the same scale as buildings and other everyday barriers, allowing them to bend around these obstacles easily. In contrast, the wavelength of visible light is extremely small compared to such large structures, meaning it does not experience noticeable diffraction around them.
In simple words: Waves only bend around things that are close to their own size. Radio waves are large enough to bend around buildings, but light waves are far too tiny to do so.

Exam Tip: The key marking point here is stating the condition for diffraction: the wavelength must be of the order of the size of the obstacle (\( \lambda \approx a \)).

Question. Coloured spectrum is seen, when we look through a muslin cloth. Why.
Answer: A muslin cloth is woven from extremely thin threads, creating a network of tiny slits. When white light travels through these narrow openings, it undergoes diffraction, which produces a multi-colored pattern. The central peak remains white because all wavelengths reinforce each other at the center, whereas the outer secondary maxima are colored because their positions vary with wavelength. For coarser fabrics, the gaps between the threads are much wider, which makes the diffraction effects weak and prevents the formation of any observable color spectrum.
In simple words: The fine threads in muslin cloth act like tiny slits that bend different colors of light by different amounts, spreading them into a rainbow. Coarser cloth has wider gaps, so it does not bend the light enough.

Exam Tip: Differentiate between the central maximum (which is white because there is no path difference for any wavelength) and the secondary maxima (which are colored because their positions depend on wavelength).

Question. How is a wavefront different from a ray? Draw the geometrical shape of the wavefronts when (i) light diverges from a point source, and (ii) light emerges out of convex lens when a point source is placed at its focus.
Answer: A wavefront represents a continuous locus of points that share the exact same phase of vibration. On the other hand, a ray is a directional line drawn normal to the wavefront, indicating where the wave energy is propagating.
(i) When light spreads outward from a localized point source, the resulting wavefronts are spherical.
(ii) When a point source is positioned at the focal point of a convex lens, the refracted light emerges as a parallel beam, forming plane wavefronts.
In simple words: A wavefront is the surface of a wave where all points are in sync, while a ray is simply an arrow showing the direction the wave is traveling. A point source makes round wavefronts, but after passing through a lens, they flatten into straight planes.

Exam Tip: Always draw rays perpendicular (at 90 degrees) to the wavefronts. Use dashed lines for wavefronts and solid lines with arrows for rays to make your diagrams neat and easy to understand for the examiner.

Question. In a young's double slit experiment, the position of the first fringe coincides with S1 and S2 respectively. What is the wavelength of light?
Answer: Based on the experimental geometry, the first-order bright bands, \( B_1 \) and \( B_2 \), located symmetrically on either side of the central point \( O \), are aligned directly with the slits \( S_1 \) and \( S_2 \).
Consequently, the distance from the center to the first bright fringe is exactly half of the slit separation: \[ \beta = \frac{d}{2} \] Using the standard fringe width equation: \[ \beta = \frac{D\lambda}{d} \] We can equate these two expressions: \[ \frac{d}{2} = \frac{D\lambda}{d} \]
which leads to: \[ \lambda = \frac{d^2}{2D} \]
In simple words: Since the first bright bands align with the slits, the fringe width is half of the slit separation. Setting this equal to the standard formula gives the wavelength as \( \frac{d^2}{2D} \).

Exam Tip: Be sure to show each algebraic step clearly when deriving this relation, beginning with the substitution \( \beta = \frac{d}{2} \) into the standard formula.

Question. Draw the diagram showing intensity distribution of light on the screen for diffraction of light at a single slit. How is the width of central maxima affected on increasing the (i) Wavelength of light used (ii) width of the slit/ What happens to the width of the central maxima if the whole apparatus is immersed in water and why?
Answer: The angular width of the central maximum in single-slit diffraction is expressed by the formula: \[ \beta_0 = \frac{2D\lambda}{d} \] where \( D \) is the distance to the screen, \( \lambda \) is the wavelength of the light, and \( d \) is the width of the slit.
(i) If the wavelength \( \lambda \) is increased, the width of the central maximum directly increases because \( \beta_0 \propto \lambda \).
(ii) If the slit width \( d \) is increased, the width of the central maximum decreases since \( \beta_0 \propto \frac{1}{d} \).
When the entire apparatus is submerged in water, the refractive index of water reduces the wavelength of light (\( \lambda_w = \frac{\lambda}{n} \)). Because of this decrease in wavelength, the width of the central maximum also decreases.
In simple words: The central bright stripe of diffracted light spreads wider when the wavelength increases or the slit gets narrower. In water, light waves shrink, so the central stripe gets narrower too.

Exam Tip: Always draw the intensity of secondary maxima decreasing sharply on either side of the central peak, and label the positions of minima as \( \pm \frac{\lambda}{d} \), \( \pm \frac{2\lambda}{d} \) to gain full marks.

Question. What two main changes in diffraction pattern of single slit will you observe when the monochromatic source of light is replaced by a source of white light?
Answer: Replacing a monochromatic light source with a white light source produces two major alterations in the diffraction pattern:
(i) Each diffraction band splits into a colored spectrum. Since band width is proportional to the wavelength (\( \beta \propto \lambda \)), the red components (having longer wavelengths) appear wider than the violet components (having shorter wavelengths).
(ii) At higher diffraction orders, the angular dispersion becomes much more pronounced, leading to the overlapping of different spectral colors and causing the pattern to blur.
In simple words: Instead of single-color bands, you will see rainbow-like stripes with red on the outside and violet on the inside. Further out, these colors will overlap and mix together.

Exam Tip: Remember to specify that the central maximum remains white because all the constituent colors travel without path difference to the center and recombine.

Question. Explain with reason, how the resolving power of a compound microscope will change when (i) frequency of the incident light on the objective lens is increased. (ii) focal length of the objective lens is increased, and (iii) aperture of the objective lens is increased.
Answer: The resolving power (R.P.) of a compound microscope is given by the relation: \[ \text{R.P.} = \frac{2 \mu \sin\theta}{\lambda} = \frac{2 \mu \sin\theta \cdot \nu}{c} \] where \( \mu \) is the refractive index of the medium, \( \theta \) is the semi-vertical angle of the cone of light, \( \lambda \) is the wavelength, \( \nu \) is the frequency, and \( c \) is the speed of light.
(i) **Increase in Frequency (\( \nu \)):** Since the resolving power is directly proportional to frequency (\( \text{R.P.} \propto \nu \)), raising the frequency of the light will increase the microscope's resolving power.
(ii) **Increase in Focal Length:** The formula for resolving power does not contain the focal length of the objective lens. Therefore, changing the focal length does not affect the resolving power.
(iii) **Increase in Aperture:** When the aperture of the lens is widened, the semi-vertical angle \( \theta \) increases. Because \( \text{R.P.} \propto \sin\theta \), this enlargement of the angle increases the resolving power of the instrument.
In simple words: Using higher frequency light or a wider lens helps a microscope see fine details better. However, changing how long the lens is does not make any difference to its resolution.

Exam Tip: Always state the mathematical expression for the resolving power of a microscope first, as this helps you justify each of your three answers logically.

Question. The critical angle between a given transparent medium and air is denoted by \( i_c \). A ray of light in air medium enters this transparent medium at an angle of incidence equal to the polarising angle \( i_p \). Show that the angle of refraction \( r_p \) is given by \( r_p = \tan^{-1}(\sin i_c) \).
Answer: From Brewster's Law, when light hits a transparent medium at the polarizing angle \( i_p \), the refractive index \( \mu \) is: \[ \mu = \tan i_p \] At this polarizing angle, the reflected and refracted rays are perpendicular to each other, giving: \[ i_p + r_p = 90^\circ \implies i_p = 90^\circ - r_p \] Substituting this into the Brewster's expression: \[ \mu = \tan(90^\circ - r_p) = \cot r_p = \frac{1}{\tan r_p} \quad \text{--- (i)} \] Furthermore, if \( i_c \) represents the critical angle for the medium, we have the standard relation: \[ \mu = \frac{1}{\sin i_c} \quad \text{--- (ii)} \] By comparing equation (i) and equation (ii), we find: \[ \frac{1}{\tan r_p} = \frac{1}{\sin i_c} \implies \tan r_p = \sin i_c \] Taking the inverse tangent of both sides yields: \[ r_p = \tan^{-1}(\sin i_c) \]
In simple words: Using Brewster's law and the definition of critical angle, we find that both tell us about the material's bending power. Setting them equal to each other shows that the tangent of the refracting angle equals the sine of the critical angle.

Exam Tip: Ensure you state that the reflected and refracted rays are perpendicular at the polarizing angle, as this is the fundamental physical assumption behind \( i_p + r_p = 90^\circ \).

Question. Two Sources of Intensity I and 4I are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase difference (i) zero (ii) π/2 (iii) π.
Answer: The combined intensity \( I_R \) at any point where two light waves overlap with a phase difference \( \phi \) is determined by: \[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi \] Given that the individual source intensities are \( I_1 = I \) and \( I_2 = 4I \), we can substitute these values into the equation: \[ I_R = I + 4I + 2\sqrt{I \cdot 4I} \cos\phi \] \[ I_R = 5I + 4I \cos\phi \] Now, we calculate the resultant intensity for each specified phase difference:
(i) **When \( \phi = 0 \):** \[ I_R = 5I + 4I \cos(0) = 5I + 4I(1) = 9I \]
(ii) **When \( \phi = \frac{\pi}{2} \):** \[ I_R = 5I + 4I \cos\left(\frac{\pi}{2}\right) = 5I + 4I(0) = 5I \]
(iii) **When \( \phi = \pi \):** \[ I_R = 5I + 4I \cos(\pi) = 5I + 4I(-1) = I \]
In simple words: By using the wave intensity formula with our two starting values, we find that the final brightness is \( 9I \) when the waves are perfectly in step, \( 5I \) when they are partly offset, and \( I \) when they are opposite.

Exam Tip: Be careful with the term \( 2\sqrt{I_1 I_2} \). Substituting \( I \) and \( 4I \) gives \( 2\sqrt{4I^2} = 4I \), not \( 2I \). Simplifying this term correctly is a common place where students lose points.

Question. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance D from the slits. If the screen is moved 5 x 10-2 m towards the slits, the change in fringe width is 3 x 10-5 m. If the distance between the slit is 10-3 m. Calculate the wavelength of the light used.
Answer: Let the initial and final fringe widths be designated as \( \beta \) and \( \beta' \) respectively: \[ \beta = \frac{D\lambda}{d} \quad \text{and} \quad \beta' = \frac{D'\lambda}{d} \] The change in fringe width is given by: \[ \beta - \beta' = \frac{(D - D')\lambda}{d} \] Solving for the wavelength of light \( \lambda \): \[ \lambda = \frac{(\beta - \beta') d}{D - D'} \] From the problem description, we have: - Screen displacement, \( D - D' = 5 \times 10^{-2} \text{ m} \) - Fringe width difference, \( \beta - \beta' = 3 \times 10^{-5} \text{ m} \) - Separation between slits, \( d = 10^{-3} \text{ m} \) Substituting these values into our expression: \[ \lambda = \frac{(3 \times 10^{-5} \text{ m}) \times (10^{-3} \text{ m})}{5 \times 10^{-2} \text{ m}} \] \[ \lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} \text{ m} = 6 \times 10^{-7} \text{ m} = 6000\text{ \AA} \]
In simple words: By measuring how much the stripes shrink when the screen is moved closer, we can use the fringe formula to calculate that the wavelength of the light is \( 6 \times 10^{-7} \text{ m} \), which equals \( 6000\text{ \AA} \).

Exam Tip: Always write down the units (meters and Angstroms) clearly at the end of your calculation. Forgetting units can result in minor point deductions.

Question. A narrow monochromatic beam of light of intensity I is incident on a glass plate. Another identical glass plate is kept close to the first one and parallel to it. Each plate reflects 25% of the incident light and transmits the remaining. Calculate the ratio of minimum and maximum intensity in the interference pattern formed by the two beams obtained after reflection from each plate.
Answer: Let the original intensity of the incident light beam be \( I \). Since each glass plate reflects \( 25\% \) (or \( \frac{1}{4} \)) of the incident light and transmits \( 75\% \) (or \( \frac{3}{4} \)), we trace the intensities of the interfering beams: First, the beam reflected off the front surface of the first plate is \( I_2 \): \[ I_2 = \frac{25}{100} I = \frac{1}{4} I \] The portion of the beam transmitted through this first plate is \( I_3 \): \[ I_3 = \frac{75}{100} I = \frac{3}{4} I \] This transmitted light now hits the second glass plate. The portion reflected by the second plate is \( I_4 \): \[ I_4 = \frac{25}{100} I_3 = \frac{1}{4} \times \frac{3}{4} I = \frac{3}{16} I \] This reflected light must pass back through the first plate to interfere. The portion transmitted back through the first plate is \( I_5 \): \[ I_5 = \frac{75}{100} I_4 = \frac{3}{4} \times \frac{3}{16} I = \frac{9}{64} I \] The interference pattern is formed by the superposition of beam 2 and beam 5. The amplitude ratio \( r \) of these interfering beams is: \[ r = \sqrt{\frac{I_2}{I_5}} = \sqrt{\frac{\frac{1}{4} I}{\frac{9}{64} I}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \] The ratio of minimum intensity to maximum intensity in the resulting interference pattern is: \[ \frac{I_{\min}}{I_{\max}} = \left( \frac{r - 1}{r + 1} \right)^2 = \left( \frac{\frac{4}{3} - 1}{\frac{4}{3} + 1} \right)^2 = \left( \frac{\frac{1}{3}}{\frac{7}{3}} \right)^2 = \left( \frac{1}{7} \right)^2 = \frac{1}{49} \] Thus, the ratio of minimum to maximum intensity is 1:49.
In simple words: By tracing the light as it bounces and passes through both glass plates, we find the intensities of the two interfering rays. Their amplitude ratio is \( \frac{4}{3} \), which gives an intensity ratio of 1:49.

Exam Tip: Always trace each step of reflection and transmission carefully, using fractions rather than decimals to make simplifying the square root of the intensity ratio much easier.