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
- A.5.89 m
- B.5.89 times ten to the power minus three metre
- C.5.89 times ten to the power minus six metre
- D.5.89 times ten to the power minus four metre
Correct answer
Explanation
The correct answer is 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. Setting that equal to the number of fringes shifted times the wavelength gives a thickness of ten times 5890 angstrom divided by twice 0.5, that is 5890 nanometres, which is 5.89 times ten to the power minus six metre, roughly six micrometres. Option A is wrong by a factor of a million; a plate almost six metres thick is absurd for an interferometer. Option B is wrong because a thickness of about six millimetres would shift many thousands of fringes, not ten. Option D is wrong by a factor of a hundred and comes from dropping two powers of ten while converting the angstrom to the metre.
Read the full article: Interference, Diffraction and Polarisation of Light Notes
Practice Questions
View allBrewster's law is (where the symbol mu stands for the refractive index and i_p for the polarising angle)
- A.i_p = 90 degrees
- B.mu = tan i_p
- C.i_p minus r = tan mu
- D.i_p = tan mu
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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.
Read the full explanation →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.
- A.3448 m
- B.68.96 m
- C.34.48 m
- D.6896 m
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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.
Read the full explanation →The phenomenon that establishes the transverse nature of light waves is
- A.Interference
- B.Diffraction
- C.Polarisation
- D.Dispersion
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Only a transverse wave, in which the vibrations are at right angles to the direction of travel, can have those vibrations restricted to a single plane by a Polaroid.
Read the full explanation →The brilliant colours seen in a soap bubble or in a thin film of oil spread on water are caused by
- A.Interference of light reflected from the two surfaces of the film
- B.Total internal reflection inside the film
- C.Dispersion of light as in a prism
- D.Scattering of light by the molecules of the film
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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.
Read the full explanation →In Young's double slit experiment the width of a fringe will increase if
- A.The screen is moved closer to the slits
- B.The separation between the slits is reduced
- C.Blue light replaces red light
- D.The slits are made narrower without changing their separation
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The fringe width equals the wavelength multiplied by the distance of the screen and divided by the separation of the two slits.
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