Chapter 9: Optics – Interference, Diffraction & Polarization (Set-2)

Interference of light occurs due to

A Reflection only
B Refraction only
C Superposition of coherent waves
D Diffraction only

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

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

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

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
D λ/4

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

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

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

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

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

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