fT
f
/12
Acceleration:
tydt
yda sin2max2
2
rcaqEradiative 2
041
jsin4
12
2max
0
trc
qyEradiative
Sinusoidal E/M field
Sinusoidal Electromagnetic Radiation
Why there is no light going through a cardboard?
Electric fields are not blocked by matterElectrons and nucleus in cardboard reradiate lightBehind the cardboard reradiated E/M field cancels original field
Cardboard
1. Radiative pressure β too small to be observed in most cases2. E/M fields can affect charged particles: nucleus and electrons
Both fields (E and M) are always present β they βfeedβ each other
But usually only electric field is considered (B=E/c)
Effect of E/M Radiation on Matter
Effect of Radiation on a Neutral Atom
Main effect: brief electric kick sideways
Neutral atom: polarizes
Electron is much lighter than nucleus:can model atom as outer electron connected to the rest of the atom by a spring:
F=eEResonance
Radiation and Neutral Atom: Resonance
tEE y sin0
tFeEF yy sin0
Amplitude of oscillation will depend on how close we are to the natural free-oscillation frequency of the ball-spring system
Resonance
E/M radiation waves with frequency ~106 Hz has big effect on mobile electrons in the metal of radio antenna: can tune radio to a single frequency
E/M radiation with frequency ~ 1015 Hz has big effect on organic molecules: retina in your eye responds to visible light but not radio waves
Very high frequency (X-rays) has little effect on atoms and can pass through matter (your body): X-ray imaging
Importance of Resonance
In transparent media, the superposition can result in change of wavelength and speed of wavefront
Index of refraction of medium,
Depends upon wavelengthand properties of medium
Refraction: Bending of Light
Rays perpendicular to wavefront bend at surface
A ray bends as it goes from one transparent media to anotherRefraction: Snellβs Law
sin (π1 )=π£1π /ππ1π1
π2
π2
π£1π
π£2π
πsin (π1 )π£1
=sin (π2 )π£2
sin (π2)=π£2π /π
sin (π1 )π/π1
=sin (π2 )π /π2
A ray travels from air to water
Example of Snellβs Law
π1
π2
ππππ=45 Β°ππ€ππ‘ππ β ?33 Β°
Reflection and transmission
Total Internal Reflection
πππππ π
πππππ π β 1.5
=.75
ππππ
πππππ π For small
W?
ππππ β si nβ1 [πππππ π sin (πππππ π ) ]
=.96
=1.15
ππππ β 49 Β°
ππππ ππππ πβ² π‘ ππ₯ππ π‘β¦πππ‘ππππ πππ π πππ
ππππ β 75 Β°
Prisms and Lens
Convergent lens Divergent lens
Lens is flat in center and prism angle steadily increases as y increases
Prisms and Lens
Thin Lenses How does the deflection angle depend on the height, ?
2 πΏ2y
πΏ=π¦π
π
For converging lenses parallel rays cross the axis at the focal distance from the lens
πΏy
When changes by factor of 2 change prism angle changes by factor of 2
πΏβπ
π2+π3=π
π1+π4=πΏ+πFor small angles, using Snellβs law
and
ππ2+ππ3=πΏ+ππ(πΒΏΒΏ2+π3)=πΏ+π ΒΏ
ππ=πΏ+ππΏ=π(πβ 1)
So the deviation angle is independent of the
; is the incident angle (air to glass)
; is the refracted angle (air to glass)
; is the refracted angle (glass to air) ; is the incident angle (glass to air)
π1
π
π
π2
π3
πΏ
π
Deviation doesnβt depend on incident angle
π4
Add to the 2nd perpendicular
πΏ=π¦πy
π π π ππΌ π½
πΌβ π¦π π
π½ β π¦π π
πΌ+π½=πΏ
π¦π π
+π¦π π
=πΏ=π¦π
1π π
+1π π
=1π Thin lens formula
Images
β’ Images are formed where rays intersectβReal image: rays of light actually intersect
βVirtual image: rays of light appear to intersect
Lensesβ’ A lens consists of a piece of glass or plastic,
ground so that each of its two refracting surfaces is a segment of either a sphere or a plane
β’ Converging lensesβ’ Thickest in the middle
β’ Diverging lensesβ’ Thickest at the edges
Focal Length of a Converging Lens
β’ The parallel rays pass through the lens and converge at the focal point
β’ Focal length is positive.
Focal Length of a Diverging Lens
β’ The parallel rays diverge after passing through the diverging lens
β’ The focal point is where the rays appear to have originated (focal length is negative)
Converging Lens,
β’ The image is real and inverted
π hobject
π β²
h β²
imageπ
Converging Lens,
β’ The image is virtual and uprightπ
hobject
π β²h β²
image
π
β’ Magnifying glass
Magnification
π
Diverging Lens
β’ The image is virtual and upright
π hobject
PhotolithographyA photomask is imaged onto the surface of a semiconductor substrate in the production of an integrated circuit. The mask is 0.25 m in front of a lens (0.25m), and the focal length of the lens is 0.05m. What should be the distance of the semiconductor surface behind the lens, ?
Choice (m)A 0.05
B 0.0625
C 0.01
D 0.125
E 0.25
1π π
+1π π
=1π
Plane or Flat Mirror
π =βπ β² h=h β²Magnification
π hobject
π β²h β²
image
Spherical Mirrors
β’ A spherical mirror has the shape of a segment of a sphere
β’ A concave spherical mirror has the silvered surface of the mirror on the inner, or concave, side of the curve
β’ A convex spherical mirror has the silvered surface of the mirror on the outer, or convex, side of the curve