physics 202 - university of virginiagalileo.phys.virginia.edu/~pqh/202_15n.pdf · 2010. 3. 25. ·...

Physics 202Professor P. Q. Hung

311B, Physics Building

Physics 202 – p. 1/35

Refraction

How does a light ray behave when it travels fromone transparent medium to another transparentmedium, e.g. from air to glass?

At the interface between the 2 media, part ofthe incident ray will reflect and part will betransmitted. All three rays are in the sameplane.

The transmitted ray is bent. The path of alight ray is reversible.

If the speed of light in medium 1 and medium2 is v1 and v2 respectively, it is found thatsin θ2/ sin θ1 = v2/v1

Physics 202 – p. 2/35

Refraction

How does a light ray behave when it travels fromone transparent medium to another transparentmedium, e.g. from air to glass?

At the interface between the 2 media, part ofthe incident ray will reflect and part will betransmitted. All three rays are in the sameplane.

The transmitted ray is bent. The path of alight ray is reversible.

If the speed of light in medium 1 and medium2 is v1 and v2 respectively, it is found thatsin θ2/ sin θ1 = v2/v1

Physics 202 – p. 2/35

Refraction

How does a light ray behave when it travels fromone transparent medium to another transparentmedium, e.g. from air to glass?

At the interface between the 2 media, part ofthe incident ray will reflect and part will betransmitted. All three rays are in the sameplane.

The transmitted ray is bent. The path of alight ray is reversible.

If the speed of light in medium 1 and medium2 is v1 and v2 respectively, it is found thatsin θ2/ sin θ1 = v2/v1

Physics 202 – p. 2/35

Refraction

Index of refraction:n = speed of light in vacuum

speed of light in medium= c

v

Since v = fλ and the frequency does notchange, it follows thatλ2

λ1

= n1

n2

The wavelength becomes shorter as onegoes from a medium with low index ofrefraction to one with high index of refraction.

Equation 2 becomes:n2 sin θ2 = n1 sin θ1

Snell’s law of refraction

Physics 202 – p. 3/35

Refraction

Index of refraction:n = speed of light in vacuum

speed of light in medium= c

v

Since v = fλ and the frequency does notchange, it follows thatλ2

λ1

= n1

n2

The wavelength becomes shorter as onegoes from a medium with low index ofrefraction to one with high index of refraction.

Equation 2 becomes:n2 sin θ2 = n1 sin θ1

Snell’s law of refraction

Physics 202 – p. 3/35

Refraction

Index of refraction:n = speed of light in vacuum

speed of light in medium= c

v

Since v = fλ and the frequency does notchange, it follows thatλ2

λ1

= n1

n2

The wavelength becomes shorter as onegoes from a medium with low index ofrefraction to one with high index of refraction.

Equation 2 becomes:n2 sin θ2 = n1 sin θ1

Snell’s law of refractionPhysics 202 – p. 3/35

Refraction

Physics 202 – p. 4/35

Refraction

Physics 202 – p. 5/35

Refraction

Physics 202 – p. 6/35

Refraction

Physics 202 – p. 7/35

Refraction

Physics 202 – p. 8/35

Refraction: Example 1

A light ray strikes an air/water surface at an angleof 460 with respect to the normal. The index ofrefraction for water is 1.33. Find the angle ofrefraction when the direction of the ray is (a) fromair to water and (b) from water to air.

From air to water: θ1 = 460, n1 = 1.sin θ2 = n1 sin θ1

n2

= 1.00 sin 460

1.33= 0.54 ⇒ θ2 = 330.

As the ray enters the medium with a higherindex of refraction, it bends toward thenormal.

Physics 202 – p. 9/35

Refraction: Example 1

θ1 = 460, n1 = 1.33

sin θ2 = n1 sin θ1

n2

= 1.33 sin 460

1.00= 0.96 ⇒ θ2 = 740.

As the ray enters the medium with a lowerindex of refraction, it bends away from thenormal.

Physics 202 – p. 10/35

Refraction: Example 2

A swimmer is treading water at the surface of a3.00-m-deep pool. She sees a coin on thebottom directly below. How deep does the coinappear to be?

For an observer directly above the object, theapparent depth is found to be (see drawing inclass)d

′

= d(n2

n1

)

n1: index of refraction of the medium wherethe object is located.n2: index of refraction associated with therefracted ray (where the observer is located).

Physics 202 – p. 11/35

Refraction: Example 2

d′

= d(n2

n1

) = (3.00m)(1.001.33

) = 2.26m

Physics 202 – p. 12/35

Total Internal reflection

The phenomenon of total internal reflection isfound in many applications such as fiber opticsfor example. The use of fiber optics in medicinehas an unquestionable importance.

What is it?

When light is shined at the interface of twomedia at some critical angle, the refracted rayis 900 away from the normal. For angles largerthan the critical angle, the incident light ray isreflected back ⇒ Total internal reflectionn1 sin θc = n2 sin 900 = n2 ⇒

sin θc = n2

n1

Physics 202 – p. 13/35

Total Internal reflection

The phenomenon of total internal reflection isfound in many applications such as fiber opticsfor example. The use of fiber optics in medicinehas an unquestionable importance.

What is it?

When light is shined at the interface of twomedia at some critical angle, the refracted rayis 900 away from the normal. For angles largerthan the critical angle, the incident light ray isreflected back ⇒ Total internal reflectionn1 sin θc = n2 sin 900 = n2 ⇒

sin θc = n2

n1Physics 202 – p. 13/35

Total Internal Reflection

Physics 202 – p. 14/35

Total Internal Reflection

Physics 202 – p. 15/35

Total Internal reflection: Example

What does a fish see in a still pond when it looksupward toward the water surface at an angle of400, 48.80, and 600?

The critical angle for water-air boundary isfound fromsin θc = n2

n1

= 1

1.33= 0.752 ⇒ θc = 48.80

Physics 202 – p. 16/35

Total Internal reflection: Example

The path of light ray is reversible.So at 400, the fish can see light from a bugflying above the surface.At 48.80, the fish can see light coming fromthe shore (i.e. at 900).At 600, the fish can see light coming from thebottom of the pond.

Physics 202 – p. 17/35

Brewster’s angle

tan θB = n2

n1

and the relected and refracted raysare perpendicular to each other.

Physics 202 – p. 18/35

Lenses

Physics 202 – p. 19/35

Convex lenses

Physics 202 – p. 20/35

Concave lenses

Physics 202 – p. 21/35

Types of lenses

Converging lenses: double convex orplano-convex or convex meniscus.

Ray 1 coming from the left parallel to axis,gets refracted by the lens toward the focalpoint on the axis on the right-hand side of thelens.

Ray 2 coming from the left toward the focalpoint on the left, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 22/35

Types of lenses

Converging lenses: double convex orplano-convex or convex meniscus.

Ray 1 coming from the left parallel to axis,gets refracted by the lens toward the focalpoint on the axis on the right-hand side of thelens.

Ray 2 coming from the left toward the focalpoint on the left, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 22/35

Types of lenses

Converging lenses: double convex orplano-convex or convex meniscus.

Ray 1 coming from the left parallel to axis,gets refracted by the lens toward the focalpoint on the axis on the right-hand side of thelens.

Ray 2 coming from the left toward the focalpoint on the left, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 22/35

Types of lenses

Diverging lenses: double concave orplano-concave or concave meniscus

Ray 1 coming from the left parallel to axis,gets refracted by the lens away from the axisand appears to come from the focal point onthe left-hand side of the lens.

Ray 2 coming from the left toward the focalpoint on the right, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 23/35

Types of lenses

Diverging lenses: double concave orplano-concave or concave meniscus

Ray 1 coming from the left parallel to axis,gets refracted by the lens away from the axisand appears to come from the focal point onthe left-hand side of the lens.

Ray 2 coming from the left toward the focalpoint on the right, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 23/35

Types of lenses

Diverging lenses: double concave orplano-concave or concave meniscus

Ray 1 coming from the left parallel to axis,gets refracted by the lens away from the axisand appears to come from the focal point onthe left-hand side of the lens.

Ray 2 coming from the left toward the focalpoint on the right, gets refracted and emergesparallel to the axis on the right of the lens.

Ray 3 coming from the left toward the centerof the lens just goes through withoutappreciable bending.

Physics 202 – p. 23/35

Types of lenses

The intersection of any two rays gives the sizeand location of the image.What kind of images are produced?

For a diverging lens, it is always virtual,upright, and reduced in size.

For a converging lens, it depends on thelocation of the object.

Physics 202 – p. 24/35

Types of lenses

The intersection of any two rays gives the sizeand location of the image.What kind of images are produced?

For a diverging lens, it is always virtual,upright, and reduced in size.

For a converging lens, it depends on thelocation of the object.

Physics 202 – p. 24/35

Convex lenses

Physics 202 – p. 25/35

Concave lenses

Physics 202 – p. 26/35

Concave lenses: Image formation

Physics 202 – p. 27/35

Convex lenses: Image formation

Physics 202 – p. 28/35

Thin Lens equation

Let p and q (d0 and di in the book) be thedistance of the object and image from the centerof the lens respectively.

Thin Lens equation1

p+ 1

q= 1

f

Magnification

M = h′

h= −

q

p

Physics 202 – p. 29/35

Thin Lens equation

Let p and q (d0 and di in the book) be thedistance of the object and image from the centerof the lens respectively.

Thin Lens equation1

p+ 1

q= 1

f

Magnification

M = h′

h= −

q

p

Physics 202 – p. 29/35

Thin Lens Equation

Physics 202 – p. 30/35

Thin Lens equation

Sign conventions:

f > 0: converging lens; f < 0: diverging lens

p > 0: left of lens (real object); p < 0: right oflens (virtual object for combination of lenses)

q > 0: right of lens (real image); q < 0: left oflens (virtual image)

M > 0: upright; M < 0: inverted.

Physics 202 – p. 31/35

Thin Lens equation

Sign conventions:

f > 0: converging lens; f < 0: diverging lens

p > 0: left of lens (real object); p < 0: right oflens (virtual object for combination of lenses)

q > 0: right of lens (real image); q < 0: left oflens (virtual image)

M > 0: upright; M < 0: inverted.

Physics 202 – p. 31/35

Thin Lens equation

Sign conventions:

f > 0: converging lens; f < 0: diverging lens

p > 0: left of lens (real object); p < 0: right oflens (virtual object for combination of lenses)

q > 0: right of lens (real image); q < 0: left oflens (virtual image)

M > 0: upright; M < 0: inverted.

Physics 202 – p. 31/35

Thin Lens equation

Sign conventions:

f > 0: converging lens; f < 0: diverging lens

p > 0: left of lens (real object); p < 0: right oflens (virtual object for combination of lenses)

q > 0: right of lens (real image); q < 0: left oflens (virtual image)

M > 0: upright; M < 0: inverted.

Physics 202 – p. 31/35

Thin Lens equation: Example

A 1.70-m-tall person is standing 2.50 m in frontof a camera which uses a converging lens withfocal length 0.0500 m. Find the image distance(between the lens and the film), themagnification, and the height of the image.

1

q= 1

f−

1

p= 1

0.0500m−

1

2.50m= 19.6m−1

⇒

q = 0.0510m ⇒ Real image.

M = h′

h= −

qp

= −0.0510m2.50m

= −0.0204 Invertedand reduced.

h′

= Mh = (−0.0204)1.70m = −0.0347m.

Physics 202 – p. 32/35

Thin Lens equation: Example

A 1.70-m-tall person is standing 2.50 m in frontof a camera which uses a converging lens withfocal length 0.0500 m. Find the image distance(between the lens and the film), themagnification, and the height of the image.

1

q= 1

f−

1

p= 1

0.0500m−

1

2.50m= 19.6m−1

⇒

q = 0.0510m ⇒ Real image.

M = h′

h= −

qp

= −0.0510m2.50m

= −0.0204 Invertedand reduced.

h′

= Mh = (−0.0204)1.70m = −0.0347m.

Physics 202 – p. 32/35

Thin Lens equation: Example

A 1.70-m-tall person is standing 2.50 m in frontof a camera which uses a converging lens withfocal length 0.0500 m. Find the image distance(between the lens and the film), themagnification, and the height of the image.

1

q= 1

f−

1

p= 1

0.0500m−

1

2.50m= 19.6m−1

⇒

q = 0.0510m ⇒ Real image.

M = h′

h= −

qp

= −0.0510m2.50m

= −0.0204 Invertedand reduced.

h′

= Mh = (−0.0204)1.70m = −0.0347m.

Physics 202 – p. 32/35

Dispersion

The index of refraction for a given materialdepends also on the frequency of lightincident upon it. Higher frequency ⇒ Higherindex of refraction. This is referred to asdispersion.

Violet light with a higher frequency than redlight will be refracted more than red lightwhen it strikes a droplet of water for example.For that situation, look at Figure 29-37,38 tosee why a rainbow appears as it is.

Physics 202 – p. 33/35

Dispersion

The index of refraction for a given materialdepends also on the frequency of lightincident upon it. Higher frequency ⇒ Higherindex of refraction. This is referred to asdispersion.

Violet light with a higher frequency than redlight will be refracted more than red lightwhen it strikes a droplet of water for example.For that situation, look at Figure 29-37,38 tosee why a rainbow appears as it is.

Physics 202 – p. 33/35

Rainbow

Physics 202 – p. 34/35

Rainbow

Physics 202 – p. 35/35