Class 12 Physics, Chapter 10 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
At t=0 a plane wavefront lies at x=0 and propagates along +x in a uniform medium at 2×10^8 m/s. Where is the forward wavefront after 1 ns?
(d) 20 cm
Each secondary wavelet advances by vt. The forward envelope is displaced by 20 cm, rather than remaining at the fixed source centres.
A plane wavefront travels along +x at 2×10^8 m/s. At t=1 ns it is at x=100 cm. Where was it at t=0?
(c) 80 cm
The envelope advances by 20 cm in the elapsed time. Subtract this from the later position; the old wavelet centres do not coincide with the later envelope.
Monochromatic light of frequency 5×10^14 Hz in air enters a transparent medium of index 2 at a stationary interface. Taking c=3.00×10^8 m/s, what are its refracted frequency and wavelength?
(a) 5 ×10^14 Hz; 300 nm
The frequency remains 5×10^14 Hz. Speed becomes c/2, so wavelength is c/(n f), or 300 nm.
Inside a transparent medium of index 2, monochromatic light has wavelength 300 nm. Taking c=3.00×10^8 m/s, what was its frequency in air before refraction at the stationary interface?
(b) 5 ×10^14 Hz
The medium speed is c/n, so its frequency is c/(nλ_medium), equal to the incident frequency. Treating the medium wavelength as a vacuum wavelength gives the wrong larger frequency.
Two independent sodium lamps illuminate two pinholes. Their light has the same wavelength, but their relative phase fluctuates randomly and rapidly. Why is no sustained interference pattern observed?
(c) A stable pattern requires a fixed relative phase, which these sources lack
The rapid random relative-phase changes wash out the interference term during measurement. Equal colour alone does not establish coherence.
Two same-frequency sources have random relative phase and give no stable fringes. Which additional condition would permit a sustained interference pattern when their parallel-polarised beams overlap?
(b) Their relative phase must be kept constant
A fixed relative phase keeps maxima and minima at fixed positions. Frequency equality alone is insufficient when the phase relation fluctuates.
A uniform thin plane-parallel transparent plate replaces air in one path of a Young experiment. It introduces a nonzero constant phase delay smaller than one full cycle; slit geometry and wavelength outside the plate are unchanged. In the small-angle regime, what happens?
(b) The fringe pattern shifts, while adjacent-bright spacing remains unchanged
A constant added optical path changes the phase offset but not the spatial rate of phase change. The whole pattern moves with the same fringe spacing.
Two coherent waves with parallel polarisation each produce intensity 2 units at a point when acting alone. They arrive in phase. What is the resultant intensity?
(b) 8 units
Equal in-phase field amplitudes add to twice one amplitude. Squaring gives four times the individual intensity.
In an ideal equal-beam Young experiment with in-phase slits, the maximum intensity is 12 units. What is the intensity at path difference λ/2?
(b) 0 units
The phase difference is 2π/2, not π/2. Therefore I=K cos²(π/2), giving the stated fraction of peak intensity.
Two wavelengths 600 nm and 480 nm form Young interference patterns in the same apparatus. In that order, which bright-fringe orders first coincide away from the centre?
(c) 4 and 5
Coincidence requires m1 λ1=m2 λ2. Reduce the wavelength ratio to its coprime factors; the required orders are the opposite factors, not equal orders.
The fringe width of a small-angle Young experiment in air is 440 μm. The complete apparatus is immersed in a uniform transparent medium of refractive index 3/2 without changing its geometry or source frequency. What is the new fringe width?
(c) 880/3 μm
Frequency stays fixed but wavelength becomes λ_air/n. Fringe width therefore becomes beta_air/n, with the geometrical centre unchanged.
The diagram shows bright fringes in a small-angle Young experiment. O is the central maximum and P is the fourth bright maximum on its right. The slits are 0.28 mm apart and the screen is 1.4 m away. What wavelength was used?
(c) 400 nm
The measured distance 8 mm spans four fringe widths, giving beta=2 mm. Then λ=beta d/D=400 nm.
The screen distance in a small-angle Young experiment is multiplied by 3, with slit separation and wavelength unchanged. What happens to angular fringe width and linear fringe width?
(b) Angular unchanged; linear multiplied by 3
Angular width λ/d is independent of D. Linear width λD/d grows with the screen distance.
A monochromatic plane wave illuminates a single slit normally and uniformly. What occurs at zero diffraction angle?
(a) A central maximum because all slit contributions arrive in phase
At θ=0 every part of the slit has equal optical path and reinforces. Pairwise cancellation for minima requires a nonzero integer order.
Which observation shows that light is a transverse wave?
(c) Its intensity can vanish when an analyser is crossed with its linear polarisation
Polarisation selects a transverse vibration direction. Interference and diffraction occur for longitudinal waves as well, so they establish wave behaviour without distinguishing transverse character.
Wrong
Two waves destroy their energy when they make a dark fringe.
Why it feels right. The intensity at that position drops to zero.
Right
Interference redistributes intensity across the pattern.
Worked example
