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General ScienceHardAsked in: Rajasthan · 09 Sept, 2025

The headlights of a distant car are 1.4 m apart. If the diameter of the pupil of the eye is 3 mm, what is the maximum distance at which the two headlights can be resolved? Consider the headlights as point sources of wavelength 500 nm.

  1. A.3448 m
  2. B.68.96 m
  3. C.34.48 m
  4. D.6896 m

Correct answer

D. 6896 m

Explanation

The correct answer is D, 6896 m. By the Rayleigh criterion the smallest angle the eye can resolve is 1.22 times the wavelength divided by the diameter of the pupil, that is 1.22 multiplied by 500 nanometres and divided by 3 millimetres, which comes to about 2.03 times ten to the power minus four radian. The two headlights are just resolved when the angle they subtend equals this value, so the distance equals their separation of 1.4 metre divided by that angle, giving nearly 6900 metres. Option A is wrong because 3448 metres is half the correct distance and follows from using twice the angle. Option C is wrong because 34.48 metres is smaller by a factor of two hundred and corresponds to mistaking millimetres for metres. Option B is wrong for the same reason of a misplaced power of ten; at only 69 metres the two lamps would be seen separately with ease.

Read the full article: Interference, Diffraction and Polarisation of Light Notes

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Q1.General ScienceAsked in: Rajasthan · RSMSSB Lab Assistant 2018 Official PaperMedium

Brewster's law is (where the symbol mu stands for the refractive index and i_p for the polarising angle)

  1. A.i_p = 90 degrees
  2. B.mu = tan i_p
  3. C.i_p minus r = tan mu
  4. D.i_p = tan mu
Show answer
Correct answer: B. mu = tan i_p

Brewster found that when unpolarised light falls on a transparent surface at one particular angle of incidence, called the polarising angle, the reflected light is completely plane polarised, and at that angle the refractive index of the surface equals the tangent of the polarising angle.

Q2.General ScienceAsked in: Uttar Pradesh · UPPSC GIC Lecturer Physics 2021Hard

When a thin glass plate of refractive index 1.5 is introduced in one of the arms of a Michelson interferometer using light of wavelength 5890 angstrom, there is a shift of 10 fringes. The thickness of the plate is

  1. A.5.89 m
  2. B.5.89 times ten to the power minus three metre
  3. C.5.89 times ten to the power minus six metre
  4. D.5.89 times ten to the power minus four metre
Show answer
Correct answer: C. 5.89 times ten to the power minus six metre

Inserting a plate of thickness t in one arm makes the light cross it twice, so the extra path introduced is twice the thickness times the quantity refractive index minus one.

Q4.General ScienceEasy

The brilliant colours seen in a soap bubble or in a thin film of oil spread on water are caused by

  1. A.Interference of light reflected from the two surfaces of the film
  2. B.Total internal reflection inside the film
  3. C.Dispersion of light as in a prism
  4. D.Scattering of light by the molecules of the film
Show answer
Correct answer: A. Interference of light reflected from the two surfaces of the film

The light reflected from the upper surface of the thin film and the light reflected from the lower surface travel slightly different distances and interfere.

Q5.General ScienceMedium

In Young's double slit experiment the width of a fringe will increase if

  1. A.The screen is moved closer to the slits
  2. B.The separation between the slits is reduced
  3. C.Blue light replaces red light
  4. D.The slits are made narrower without changing their separation
Show answer
Correct answer: B. The separation between the slits is reduced

The fringe width equals the wavelength multiplied by the distance of the screen and divided by the separation of the two slits.