Wave Optics MCQs with Answers – Part 1 (Class 12 Physics)

Explanation: Wave optics studies light by considering its wave nature. Straight-line ray diagrams explain many image-formation results, but they cannot fully describe how light waves overlap, spread near edges, or have restricted vibrations. Interference needs superposition of waves, diffraction needs bending or spreading of waves, and polarisation needs the transverse character of light. These effects require ideas such as phase, wavelength, and vibration direction. Reflection and ordinary image formation can often be handled by ray optics, but the three effects named in the answer go beyond a purely geometrical model.

Explanation: Ray optics works best when the size of apertures or obstacles is very large compared with the wavelength \( \lambda \) of light. In that case, bending or spreading around edges is very small, so straight rays give a good approximation. When the opening is comparable to \( \lambda \), the wave nature becomes noticeable and the light spreads into regions that a simple ray diagram would mark as shadow. This spreading is the basic idea behind diffraction. The frequency does not become random just because light passes through a small opening, so the failure is due to wave spreading rather than a change of \( \nu \).

Explanation: Alternate bright and dark bands are a signature of interference. Interference is explained by superposition, where two light waves meet and their displacements combine at the same point. Bright regions form when the waves reinforce each other, while dark regions form when they cancel or nearly cancel. A plane mirror image and a lens focusing a beam can often be described by ray optics. The bright-dark band pattern cannot be explained by only drawing independent straight rays because the relative phase of the waves matters.

Explanation: Polarisation means that the vibrations of a transverse wave are restricted to a particular plane or direction. Light can be polarised because its vibrations are transverse to the direction of propagation. Polaroid materials transmit vibrations mainly along one transmission axis and reduce components in the perpendicular direction. Glare from horizontal surfaces is often partially polarised, so suitable sunglasses can reduce it. Longitudinal waves cannot be polarised in this way because their vibrations are along the direction of travel.

Explanation: Colours due to overlapping light waves are connected with interference, because the observed brightness or colour depends on how waves superpose. Spreading of light through a narrow slit is diffraction, which becomes noticeable when the slit width is comparable to the wavelength \( \lambda \). Reduction of glare by a polaroid sheet uses polarisation, because the polaroid selects one vibration direction more strongly than the perpendicular one. The three observations involve different wave aspects, even though all belong to wave optics. The key separation is that interference is about overlap, diffraction is about spreading, and polarisation is about vibration direction.

Explanation: The amplitude \( a \) of a wave measures the maximum displacement of the vibrating quantity from its mean value. For light, the intensity \( I \) is related to the square of the amplitude, so a larger amplitude means a stronger beam when other conditions are unchanged. Colour is mainly associated with frequency \( \nu \), not directly with amplitude. The time period \( T \) and wavelength \( \lambda \) are controlled by frequency and wave speed, not by simply increasing amplitude. This distinction keeps brightness separate from colour in the wave description of light.

Explanation: The speed \(v\) of a wave is the distance travelled by the wave in one second. In one time period \(T\), the wave advances by one wavelength \( \lambda \). Since frequency \( \nu \) is the number of cycles per second, the wave covers \( \nu \) wavelengths in one second. Therefore, the distance covered per second is \( \nu\lambda \), giving \(v=\nu\lambda\). The relation is multiplicative, not additive, because frequency counts how many wavelength-lengths pass per second.

Explanation: Frequency \( \nu \) tells how many complete vibrations occur in one second. Time period \(T\) tells the time taken for one complete vibration. If there are \( \nu \) vibrations in \(1\,\text{s}\), then the time for one vibration is \( \frac{1}{\nu} \). This gives the relation \(T=\frac{1}{\nu}\). The unit check also supports it, because \( \nu \) has unit \( \text{s}^{-1} \), so \( \frac{1}{\nu} \) has unit \( \text{s} \).

Explanation: Phase describes the stage of vibration of a wave at a point and at an instant. Two points are in the same phase if they reach corresponding stages, such as maximum displacement or zero displacement in the same sense, together. Same phase does not require the amplitude to be zero; it is about timing within the cycle. Points in the same medium need not have different speeds just because phase is being compared. Phase is essential in wave optics because interference depends on whether waves arrive in step or out of step.

Explanation: A perfectly sharp geometrical shadow is expected only when straight-line propagation is a very good approximation. Near a narrow opening, the wavefront does not remain confined to the region predicted by simple rays. Light spreads into nearby regions, making the edge of the bright region less sharp. This behaviour is the beginning of diffraction. The wavelength does not become infinite, and wave speed is not described as negative in this situation.

Explanation: Angular frequency \( \omega \) measures how fast the phase of a wave changes in angular measure. Since phase is measured in radians and time is measured in seconds, the unit of \( \omega \) is \( \text{rad}\,\text{s}^{-1} \). Frequency \( \nu \) is measured in \( \text{Hz} \), which is the same as \( \text{s}^{-1} \), but it counts cycles per second rather than radians per second. Wavelength \( \lambda \) has unit \( \text{m} \), and time period \(T\) has unit \( \text{s} \). The radian in \( \omega \) reminds us that it is an angular rate of phase change.

Explanation: \( \textbf{Given data:} \) \(v=3.0\times10^8\,\text{m s}^{-1}\) and \(\lambda=500\,\text{nm}\). \( \textbf{Required:} \) Frequency \( \nu \). \( \textbf{Wave relation:} \) \[ v=\nu\lambda \] \( \textbf{Reason for using it:} \) The question connects wave speed, wavelength, and frequency for a light wave. \( \textbf{Unit conversion:} \) \[ 500\,\text{nm}=500\times10^{-9}\,\text{m}=5.00\times10^{-7}\,\text{m} \] \( \textbf{Solving for frequency:} \) \[ \nu=\frac{v}{\lambda} \] \( \textbf{Substitution:} \) \[ \nu=\frac{3.0\times10^8}{5.00\times10^{-7}}\,\text{s}^{-1} \] \( \textbf{Power of ten handling:} \) \[ \frac{10^8}{10^{-7}}=10^{15} \] \( \textbf{Numerical simplification:} \) \[ \nu=0.60\times10^{15}\,\text{s}^{-1}=6.0\times10^{14}\,\text{Hz} \] \( \textbf{Final answer:} \) The frequency is \(6.0\times10^{14}\,\text{Hz}\). The main unit step is converting \( \text{nm} \) into \( \text{m} \) before using \(v=\nu\lambda\).

Explanation: Phase \( \phi \) is measured in radians, and phase difference \( \Delta\phi \) is also measured in radians. Frequency \( \nu \) and time period \(T\) are reciprocals in unit type, with \( \nu \) in \( \text{s}^{-1} \) and \(T\) in \( \text{s} \). Wavelength \( \lambda \) is a length, while frequency is a rate of vibration. Wave number \(k\) has unit \( \text{m}^{-1} \), while angular frequency \( \omega \) has unit \( \text{rad}\,\text{s}^{-1} \). The safest way to compare such pairs is to attach each symbol to its physical meaning before comparing units.

Explanation: Wavelength \( \lambda \) is a length, so its SI unit is \( \text{m} \). Angular frequency \( \omega \) measures angular phase change per unit time, so \( \text{rad}\,\text{s}^{-1} \) is suitable. Phase \( \phi \) is measured in radians. Frequency \( \nu \), however, counts vibrations per second, so its unit is \( \text{Hz} \) or \( \text{s}^{-1} \), not \( \text{s} \). The symbol \( \text{s} \) belongs to time period \(T\), which is the reciprocal of frequency.

Explanation: \( \textbf{Given:} \) \(\Delta x=\frac{\lambda}{2}\). \( \textbf{Required:} \) Phase difference \( \Delta\phi \). \( \textbf{Useful relation:} \) \[ \Delta\phi=\frac{2\pi}{\lambda}\Delta x \] \( \textbf{Why this relation applies:} \) One full wavelength \( \lambda \) corresponds to one full phase cycle \(2\pi\,\text{rad}\). \( \textbf{Substitution:} \) \[ \Delta\phi=\frac{2\pi}{\lambda}\times\frac{\lambda}{2} \] \( \textbf{Cancellation:} \) The factor \( \lambda \) cancels from numerator and denominator. \( \textbf{Simplification:} \) \[ \Delta\phi=\pi\,\text{rad} \] \( \textbf{Physical meaning:} \) A path difference of half a wavelength means the two vibrations are opposite in phase. \( \textbf{Option check:} \) \(2\pi\,\text{rad}\) would represent one full wavelength path difference, not half a wavelength. \( \textbf{Final answer:} \) The phase difference is \(\pi\,\text{rad}\).

Explanation: The wave relation is \(v=\nu\lambda\). If \(v\) is fixed, the product \( \nu\lambda \) must remain constant. Therefore, increasing \( \nu \) by a factor of \(2\) requires \( \lambda \) to decrease by a factor of \(2\). Similarly, doubling \( \lambda \) requires \( \nu \) to become half. Statement III is not acceptable because \( \nu \) and \( \lambda \) are linked by the fixed value of \(v\). The inverse relation applies only after the medium and hence the wave speed have been fixed.

Explanation: \( \textbf{Graph relation needed:} \) Phase difference and path difference are connected by \[ \Delta\phi=\frac{2\pi}{\lambda}\Delta x \] \( \textbf{Graph form:} \) This has the form \(y=mx\). \( \textbf{Vertical variable:} \) Here \(y\) corresponds to \( \Delta\phi \). \( \textbf{Horizontal variable:} \) Here \(x\) corresponds to \( \Delta x \). \( \textbf{Slope identification:} \) \[ m=\frac{\Delta\phi}{\Delta x}=\frac{2\pi}{\lambda} \] \( \textbf{Unit meaning:} \) Since \( \Delta\phi \) is in radians and \( \Delta x \) is in metres, the slope has unit \( \text{rad}\,\text{m}^{-1} \). \( \textbf{Physical reading:} \) A shorter wavelength gives a larger phase change per metre of path difference. \( \textbf{Wrong-option reason:} \) The expression \( \frac{\lambda}{2\pi} \) is the reciprocal of the required slope. \( \textbf{Final answer:} \) The slope is \(\frac{2\pi}{\lambda}\).

Explanation: A separation of one complete wavelength \( \lambda \) corresponds to one full cycle of phase. One full cycle is \(2\pi\,\text{rad}\), which brings the vibration back to the same stage. Therefore, two points separated by \( \lambda \) are in the same phase. A phase difference of \( \pi\,\text{rad} \) would correspond to a separation of \( \frac{\lambda}{2} \). Points on the same monochromatic wave do not acquire different frequencies merely because they are at different positions along the wave.

Explanation: \( \textbf{Given:} \) Amplitude ratio \(a_1:a_2=2:1\). \( \textbf{Required:} \) Intensity ratio \(I_1:I_2\). \( \textbf{Relevant idea:} \) For the same type of wave under comparable conditions, intensity is proportional to the square of amplitude. \[ I\propto a^2 \] \( \textbf{Ratio form:} \) \[ \frac{I_1}{I_2}=\frac{a_1^2}{a_2^2} \] \( \textbf{Substitution of amplitude ratio:} \) \[ \frac{I_1}{I_2}=\frac{2^2}{1^2} \] \( \textbf{Simplification:} \) \[ \frac{I_1}{I_2}=4 \] \( \textbf{Ratio form:} \) \[ I_1:I_2=4:1 \] \( \textbf{Physical meaning:} \) Doubling amplitude does not merely double intensity; it makes intensity four times. \( \textbf{Final answer:} \) The intensity ratio is \(4:1\).

Explanation: Ordinary frequency \( \nu \) counts the number of complete cycles per second. Angular frequency \( \omega \) measures the rate of phase change in radians per second. One complete cycle corresponds to \(2\pi\,\text{rad}\). Therefore, multiplying cycles per second by \(2\pi\,\text{rad}\) per cycle gives \( \omega=2\pi\nu \). The relation \( \omega=\lambda\nu \) would have the dimensions of speed, so it is actually related to \(v=\nu\lambda\), not angular frequency.