Every Book Back multiple-choice question from Wave Optics (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
- A. red
- B. yellow
- C. green
- D. violetCorrect
Explanation. Apparent shift in position is directly proportional to the refractive index of the medium for a given thickness. Since violet light has the highest refractive index in glass, it undergoes the largest shift and appears most raised.
Q2
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, \(\lambda\) = 500 nm]
- A. 1 m
- B. 5 mCorrect
- C. 3 m
- D. 6 m
Explanation. The maximum distance is determined by the Rayleigh criterion for resolving power. By equating the angular separation of the dots to the minimum resolvable angle of the eye, the distance is calculated based on pupil diameter and wavelength.
Q3
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
- A. 2DCorrect
- B. D/2
- C. \(\sqrt{2}\)D
- D. D/\(\sqrt{2}\)
Explanation. Fringe width is directly proportional to the screen distance and inversely proportional to the slit separation. If the separation is doubled, the distance to the screen must also be doubled to keep the fringe spacing constant.
Q4
Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are
- A. 5I and I
- B. 5I and 3I
- C. 9I and ICorrect
- D. 9I and 3I
Explanation. Maximum and minimum intensities result from constructive and destructive interference of amplitudes. Taking the square roots of the given intensities to find amplitudes, their sum squared gives 9I and their difference squared gives I.
Q5
When light is incident on a soap film of thickness 5\(\times\)10\(^{–5}\) cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be,
- A. 1.22
- B. 1.33Correct
- C. 1.51
- D. 1.83
Explanation. The wavelength of maximum reflection is determined by the constructive interference condition in a thin film. Using the film thickness and the given wavelength, the refractive index is calculated considering the phase shifts involved in visible light reflection.
Q6
First diffraction minimum due to a single slit of width 1.0\(\times\)10\(^{–5}\) cm is at 30\(^{o}\). Then wavelength of light used is,
- A. 400 Å
- B. 500 ÅCorrect
- C. 600 Å
- D. 700 Å
Explanation. The position of the first minimum in single-slit diffraction is where the slit width multiplied by the sine of the diffraction angle equals the wavelength. This relationship allows for the direct calculation of the incident light wavelength.
Q7
A ray of light strikes a glass plate at an angle 60\(^{o}\). If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
- A. \(\sqrt{3}\)Correct
- B. \(\frac{\sqrt{3}}{2}\)
- C. \(\frac{3}{2}\)
- D. 2
Explanation. When reflected and refracted rays are perpendicular, the angle of incidence corresponds to the Brewster angle. According to Brewster's law, the refractive index is the tangent of this angle, which identifies the index for the glass plate.
Q8
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,
- A. get shifted downwards
- B. get shifted upwardsCorrect
- C. will remain the same
- D. data insufficient to conclude
Explanation. Adding a glass plate increases the optical path length for light passing through that slit. The interference pattern, including the central maximum, shifts toward the covered slit to compensate for the introduced path difference.
Q9
Light transmitted by Nicol prism is,
- A. partially polarised
- B. unpolarised
- C. plane polarisedCorrect
- D. elliptically polarised
Explanation. A Nicol prism utilizes double refraction and total internal reflection to block the ordinary ray. Only the extraordinary ray is transmitted through the prism, resulting in output light that is completely plane polarized.
Q10
The transverse nature of light is shown in,
- A. interference
- B. diffraction
- C. scattering
- D. polarisationCorrect
Explanation. Polarization is a phenomenon unique to transverse waves. By restricting wave vibrations to a single plane perpendicular to the direction of travel, polarization demonstrates the transverse nature of electromagnetic light waves.
About these Wave Optics questions
These are the Book Back multiple-choice questions for Wave Optics from the Tamil Nadu State Board (Samacheer Kalvi) 12th Standard Physics syllabus. Each question shows the correct option and an original, step-by-step explanation so you understand the method, not just the answer. Use the answer key above to jump to any question, then take the practice test to check yourself under exam-like conditions.
Frequently asked questions
How many MCQs are there in Wave Optics?
This chapter has 10 book-back multiple-choice questions, each with the correct answer and a step-by-step explanation.
Are these 12th Standard Physics MCQs free to practise online?
Yes. Every question, answer and explanation here is free, and you can also take them as a timed practice test.
Where can I find the Wave Optics book-back answers?
The correct option for each question is highlighted on this page with a worked explanation, plus a quick answer-key summary at the top.
