Interference of light occurs due to
A Reflection only
B Refraction only
C Superposition of coherent waves
D Diffraction only
Interference = superposition of two coherent waves with stable phase difference.
For interference, sources must be
A Intense
B Widely separated
C Phase-locked (coherent)
D Monochromatic only
In YDSE, fringe width β is given by
A λD/d
B d/(λD)
C λd/D
D Dd
β = λD/d.
Increasing slit separation (d) in YDSE causes fringe width β to
A Increase
B Decrease
C Become infinite
D Unchanged
In YDSE, central bright fringe is formed where path difference is
A λ/4
B λ/2
C λ
D 0
Condition for constructive interference is
A Δφ = (2n+1)π
B Δx = (2n+1)λ/2
C Δx = nλ
D Δx = λ/3
Condition for destructive interference is
A Δx = nλ
B Δx = (2n+1)λ/2
C Δx = λ
D Δx = 3λ
In YDSE, intensity at a point is proportional to
A a² + b²
B (a+b)²
C I = I₀ cos²(δ/2)
D I = I₀ sin²(δ/2)
Using white light in YDSE gives
A No fringes
B Coloured fringes
C Black fringes only
D Only central white fringe
In thin film interference, the phase change on reflection from denser medium is
A 0
B π
C π/2
D 2π
For thin film in reflected light, constructive interference occurs when
A 2μt cos r = mλ
B 2μt cos r = (2m+1)λ/2
C μt = λ
D t = 0
One reflection includes a π shift → modified condition.
For transmitted light, constructive interference in thin films is
A 2μt cos r = mλ
B 2μt cos r = (2m+1)λ/2
C t = λ
D μt = λ/4
Newton’s rings are formed due to
A Diffraction
B Refraction
C Interference between reflected rays from plano-convex lens and glass plate
D Polarization
In Newton’s rings, central spot is dark because
A Lens is absorbing
B Two rays have equal amplitude
C Path difference includes 180° phase shift on one reflection
D Light intensity is zero at center
Radius of m-th dark ring in Newton’s rings is
A rₘ² = mλR
B rₘ = mλR
C rₘ² = (2m+1)λR
D rₘ = √(mR)
Michelson interferometer fringe visibility is maximum when
A Intensities of two beams equal
B Intensities very unequal
C One path is blocked
D Mirrors misaligned
In Michelson interferometer, one fringe shift corresponds to mirror moved by
A λ
B λ/2
C 2λ
D λ/4
Optical path difference = 2Δd → one fringe shift for 2Δd = λ.
The basic requirement of Michelson interferometer is
A Two coherent beams formed by division of wavefront
B Lens to focus beams
C Two incoherent sources
D Always monochromatic light
Visibility (V) of fringes is defined as
A (Imax−Imin)/(Imax+Imin)
B Imax/Imin
C Imax+Imin
D ImaxImin
Coherence length is
A λ/2
B Speed of light × coherence time
C Infinite for white light
D Zero
Spatial coherence depends on
A Source size
B Wavelength only
C Temperature only
D Polarization
Temporal coherence depends on
A Source bandwidth
B Aperture size
C Path length only
D Lens diameter
Narrow bandwidth → long coherence time.
Phase difference in YDSE at angle θ is
A λ/D
B d sinθ
C 2π/λ
D d/λ
If one slit in YDSE is closed, fringe pattern becomes
A Brighter
B Darker
C Single-slit diffraction pattern
D Disappears completely
A soap film shows colours due to
A Scattering
B Thin film interference
C Diffraction
D Polarization
Path difference in reflected light from thin film of thickness t and refractive index μ is
A 2t
B 2μt
C 2μt cos r
D t/μ
Newton’s rings shape is
A Spiral
B Concentric circles
C Straight fringes
D Random
Michelson interferometer measures
A Wavelength
B Refractive index changes
C Very small distances
D All of the above
With increasing thickness of a thin film, interference fringes
A Vanish
B Get closer
C Become broader
D Become black
Effective optical path changes → fringe spacing decreases.
In YDSE, if slit separation doubles, number of fringes in same region
A Increases
B Decreases
C Stays same
D Zero
For maximum fringe contrast, amplitudes of interfering waves must be
A Equal
B Unequal
C Zero
D Constant
Fringe visibility is zero when
A Imax = Imin
B Imin = 0
C Imax ≫ Imin
D Intensities equal
If Imax = Imin → V = 0.
In Newton’s rings experiment, if plano-convex lens is replaced by a convex lens of larger radius, rings become
A Closer
B Wider
C Invisible
D Coloured
r² ∝ R → larger R → larger ring radii.
Michelson interferometer uses
A Transmission grating
B 50–50 beam splitter
C Diffraction lens
D Polarizer
In YDSE, if one slit is covered with thin transparent film, central fringe shifts because
A Intensity changes
B Coherence destroyed
C Additional optical path is introduced
D Fringe width changes
A localized fringe pattern appears in
A YDSE
B Newton’s rings
C Thin wedge interference
D Both B & C
Thin film interference occurs because of
A Reflection only
B Multiple reflections
C Scattering
D Diffraction
Newton’s rings diameter ratio for successive dark rings is
A Constant
B Square root proportional
C In arithmetic progression
D In geometric progression
rₘ ∝ √m.
Michelson fringes shift when
A One mirror moves
B Beam intensity varies
C Light is polarized
D Aperture changed
Optical path difference =
A μ × geometric path difference
B μ/t
C μ + t
D t – μ
Interference is absent if
A Path difference random
B Coherence lost
C Frequency differs
D All of the above
YDSE uses
A Division of amplitude
B Division of wavefront
C Division of frequency
D Division of coherence
Thin film of oil on water produces colours because
A Polarization
B Refraction
C Path difference varies with viewing angle
D Absorption
Michelson interferometer gives circular fringes when
A Mirrors perfectly perpendicular
B Path difference varies uniformly
C Mirrors slightly tilted
D No tilt present
Fringe width in interference depends on
A Wavelength
B Distance to screen
C Slit separation
D All of these
Interference energy distribution follows
A Energy creation
B Energy disappearance
C Redistribution without loss
D Destruction
In YDSE, if slit width increases greatly
A Fringe contrast improves
B Diffraction increases → washed-out fringes
C Fringes disappear completely
D Fringe width increases
Newton’s rings appear coloured with
A Monochromatic light
B White light
C Laser
D Yellow light only
Michelson interferometer fringes become very sharp when
A Bandwidth large
B Source highly monochromatic
C Mirrors rough
D Slits added
YDSE fringe shift due to path change Δx is
A Δx / λ fringes
B λ / Δx fringes
C Δx / d fringes
D d / Δx fringes
One fringe shift per λ change in path.