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CBSE NOTES CLASS 12 PHYSICS

CHAPTER 10 WAVE OPTICS

Newton’s corpuscular theory

Wave theory - nature of electromagnetic waves

Wave front

Huygens principle

Refraction of a plane wave from rarer to denser medium

Refraction of a plane wave from denser to rarer medium

Reflection of a plane wave by a plane surface

Behaviour of a plane wave front with different surfaces

The Doppler effect

Superposition principle

Coherent sources of light

Interference of light

Young’s double slit experiment

Fringe width in double slit experiment

Diffraction of light

Single slit experiment

Double slit vs single slit patterns

Interference vs diffraction due to single slit

Constraints for diffraction due to single slit

Viewing the diffraction pattern

Energy is conserved during interference and diffraction

When can we consider the light beam to be parallel beam in single slit experiment?

Resolving power of an objective lens

Fresnel distance

Polarisation

Polarisation by transmission

Polarisation by scattering

Polarisation by reflection - Brewster’s law

Law of Malus

Polaroid and uses of polaroids

CBSE NOTES CLASS 12 PHYSICS

CHAPTER 10 WAVE OPTICS

Young’s double slit experiment

Thomas Young made two pinholes S1 and S2 (very close to each other) on an opaque screen. These were illuminated by another pinhole that was in turn, lit by a bright source. Light waves spread out from S and fall on both S1 and S2. S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2.

???

Thus, the two sources S1 and S2 will be locked in phase; i.e., they will be coherent like the two vibrating needles.

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Spherical waves originating from S1 and S2 will produce interference fringes on the screen GG’.

For an arbitrary point P on the line GG’ to have a maximum,

S2P – S1P = nλ; n = 0, 1, 2 ...

From the geometry of the figure,

 S2P2  S1P2=D2 + x+d22  D2+x-d22=2xd

which gives

S2P-S2PS2P+S2P=2xd

S2P-S2P=2xdS2P+S2P

If x << D and d << D, then,

S2P+S2P ≈ 2D

Hence,

S2P-S2P=xdD

For constructive interference or bright fringes,

xn=nλDd             [n = 0, 1, 2.  ]

For destructive interference or dark fringes,

xn=n+12λDd             [n = 0, 1, 2.  ]

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