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26.1 The Index of Refraction

Light travels through a at a speed c = 00.3 ×10 8 m s Chapter 26 travels through materials at a speed less than its speed in a vacuum.

The Refraction of DEFINITION OF THE INDEX OF REFRACTION Light: and The index of refraction of a material is the ratio of the in a vacuum to the speed of light in the material: Optical Instruments Speed of light in vacuum c n = = Speed of light in the material v

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26.1 The Index of Refraction 26.2 Snell’s Law and the Refraction of Light

SNELL’S LAW

SNELL’S LAW OF REFRACTION

When light travels from a material with one index of refraction to a material with a different index of refraction, the angle of incidence is related to the angle of refraction by

n1 sin θ1 = n2 sin θ2

3 4 26.2 Snell’s Law and the Refraction of Light 26.2 Snell’s Law and the Refraction of Light

Example 1 Determining the Angle of Refraction n sin θ ( 00.1 )sin 46 o (a) sin θ = 1 1 = = 54.0 2 n 33.1 A light strikes an air/ surface at an 2 angle of 46 degrees with respect to the normal. Find the angle of refraction when θ = 33 o the direction of the ray is (a) from air to 2 water and (b) from water to air.

n sin θ ( 33.1 )sin 46 o (b) 1 1 sin θ2 = = = 96.0 n2 00.1

o θ2 = 74

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26.2 Snell’s Law and the Refraction of Light 26.2 Snell’s Law and the Refraction of Light

APPARENT DEPTH

−1 o θ2 = tan ( 0.2 3.3 )= 31

o Example 2 Finding a Sunken Chest n2 sin θ2 ( 33.1 )sin 31 sin θ1 = = = 69.0 n1 00.1 The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed? o θ1 = 44

7 8 26.2 Snell’s Law and the Refraction of Light 26.2 Snell’s Law and the Refraction of Light

Conceptual Example 4 On the Inside Looking Out

Apparent depth, A swimmer is under water and looking up at the surface. Someone observer directly holds a coin in the air, directly above the swimmer’s eyes. To the above object swimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin greater, less than, or the same as its actual height?

  n2 d′ = d   n1 

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26.2 Snell’s Law and the Refraction of Light 26.2 Snell’s Law and the Refraction of Light

THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL THE DERIVATIN OF SNELL’S LAW

11 12 26.3 Total Internal 26.3 Total Internal Reflection

Example 5 Total Internal Reflection When light passes from a medium of larger into one of smaller refractive index, the refracted ray bends away from the A beam of light is propagating through diamond and strikes the diamond-air normal. interface at an angle of incidence of 28 degrees. (a) Will part of the beam enter the air or will there be total internal reflection? (b) Repeat part (a) assuming that the diamond is surrounded by water.

n2 Critical angle sin θc = n1 > n2 n1

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26.3 Total Internal Reflection 26.3 Total Internal Reflection

Conceptual Example 6 The Sparkle of a Diamond

The diamond is famous for its sparkle because the light coming from it glitters as the diamond is moved about. Why does a diamond exhibit such brilliance? Why does it lose much of its brilliance when placed under water?

 n   00.1  −1 2  −1 o (a) θc = sin   = sin   = 24 4.  n1   42.2 

 n   33.1  (b) −1 2  −1 o θc = sin   = sin   = 33 3.  n1   42.2 

15 16 26.3 Total Internal Reflection 26.3 Total Internal Reflection

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26.4 Polarization and the Reflection and Refraction of Light 26.5 The of Light: and

The net effect of a is to change the direction of a light ray.

Light rays corresponding to different bend by different amounts.

n2 Brewster’s law tan θB = n1

19 20 26.5 The Dispersion of Light: Prisms and Rainbows 26.5 The Dispersion of Light: Prisms and Rainbows

Conceptual Example 7 The Refraction of Light Depends on Two Refractive Indices

It is possible for a prism to bend light upward, downward, or not at all. How can the situations depicted in the figure arise?

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26.5 The Dispersion of Light: Prisms and Rainbows 26.6 Lenses

Lenses refract light in such a way that an image of the light source is formed.

With a converging , paraxial rays that are parallel to the principal axis converge to the focal point.

23 24 26.6 Lenses 26.6 Lenses

With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point.

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26.7 The Formation of Images by Lenses 26.7 The Formation of Images by Lenses

RAY DIAGRAMS IMAGE FORMATION BY A CONVERGING LENS

In this example, when the object is placed further than twice the from the lens, the real image is inverted and smaller than the object.

27 28 26.7 The Formation of Images by Lenses 26.7 The Formation of Images by Lenses

When the object is placed between F and 2F, the real image is When the object is placed between F and the lens, the virtual image is inverted and larger than the object. upright and larger than the object.

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26.7 The Formation of Images by Lenses 26.8 The Thin-Lens Equation and the Magnification Equation

IMAGE FORMATION BY A DIVERGING LENS

1 1 1 hi di + = m = = − d d f ho do A diverging lens always forms an upright, virtual, diminished image. o i

31 32 26.8 The Thin-Lens Equation and the Magnification Equation 26.8 The Thin-Lens Equation and the Magnification Equation

Summary of Sign Conventions for Lenses Example 9 The Real Image Formed by a Lens f is + for a converging lens. A 1.70-m tall person is standing 2.50 m in front of a camera. The f is − for a diverging lens. camera uses a converging lens whose focal length is 0.0500 m. (a)Find the image distance and determine whether the image is

do is + if the object is to the left of the lens. real or virtual. (b) Find the magnification and height of the image on the film. d is − if the object is to the right of the lens. o 1 1 1 1 1 (a) = − = − =19 6. m−1 di f do .0 0500 m 50.2 m di is + for an image formed to the right of the lens (real image).

di = .0 0510 m real image di is − for an image formed to the left of the lens (virtual image). d .0 0510 m m is + for an upright image. (b) m = − i = − = − .0 0204 do 50.2 m m is − for an inverted image.

hi = mh o = (− .0 0204 )( 50.2 m)= − .0 0347 m 33 34

26.9 Lenses in Combination 26.10 The ANATOMY

The image produced by one lens serves as the object for the next lens.

35 36 26.10 The Human Eye 26.10 The Human Eye NEARSIGNTEDNESS

The lens only contributes about 20-25% of the refraction, but its function is important. The lens creates an image of the distance object at the far point of the nearsighted eye. 37 38

26.10 The Human Eye 26.10 The Human Eye

Example 12 Eyeglasses for the Nearsighted Person

A nearsighted person has a far point located only 521 cm from the eye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the so the person can see distant objects.

1 1 1 1 1 = + = − f do di ∞ 519 cm

f = −519 cm

39 40 26.10 The Human Eye 26.10 The Human Eye FARSIGNTEDNESS THE REFRACTIVE POWER OF A LENS – THE DIOPTER

Optometrists who prescribe correctional lenses and the opticians who make the lenses do not specify the focal length. Instead they use the concept of refractive power.

1 Refractive power (in diopters) = f ()in meters

The lens creates an image of the close object at the near point of the farsighted eye. 41 42

26.11 Angular Magnification and the Magnifying 26.11 Angular Magnification and the Magnifying Glass

h θ ()in radians = Angular size ≈ o do

The size of the image on the retina determines how large an object appears to be.

43 44 26.11 Angular Magnification and the Magnifying Glass 26.11 Angular Magnification and the Magnifying Glass

Example 14 A Penny and the Moon Angular magnification Compare the angular size of a penny held at arms length with that of the moon. θ′ M = θ

h 9.1 cm Penny θ ≈ o = = .0 027 rad do 71 cm

Angular magnification of a magnifying glass 6 ho 5.3 ×10 m Moon θ ≈ = = .0 0090 rad  1 1  8 ≈  −  do 3.9 ×10 m M  N  f di 

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26.12 The Compound 26.13 The

To increase the angular magnification beyond that possible with a magnifying glass, an additional converging lens can be included to “premagnify” the object.

Angular magnification of a compound microscope

f Angular magnification of M ≈ − o (L − fe )N M ≈ − an astronomical telescope fe fo fe

47 48 26.14 Lens Aberrations 26.14 Lens Aberrations

In a converging lens, prevents light rays parallel to the principal axis from converging at a single point.

Spherical aberration can be reduced Chromatic aberration arises when different colors are focused at by using a variable-aperture diaphragm. different points along the principal axis.

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