1. Lenses and the Refraction of Light

2. 26.1 The Index of Refraction
Light travels through a vacuum at at c
Light travels through materials such as water or glass at a speed less than c.
Light will bend, or refract, if it enters a medium with a different speed. It may reflect as well.
The amount of bending is quantified by the index of refraction.
The index of refraction of a material is the ratio of the speed of light in a vacuum to the speed of light in the material. As a ratio, it is unitless:

3. 26.2 Snell’s Law and the Refraction of Light
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
where n1 is the index of refraction in the incident ray medium, and n2 is that index of refraction in the refracting ray medium.
There is also a reflected ray at an angle of θ1.
A mirror surface does not refract, it only reflects.

4. Determining the Angle of Refraction
A light ray strikes an air/water surface at an
angle of 46 degrees with respect to the
normal. Find the angle of refraction when
the direction of the ray is (a) from air to
water and (b) from water to air.
From air to water From water to air
n1 = 1 n1 = 1.33
θ1 = 46° θ1 = 46°
n2 = n2 = 1
Find θ Find θ2

5. 26.2 Snell’s Law and the Refraction of Light
When a ray is transmitted into a material with a higher index of refraction it bends to make a smaller angle with the normal.
(b)
When a ray is transmitted into a material with a lower index of refraction it bends to make a larger angle with the normal.

6. 26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH – when an object is in a medium with a different index of refraction that that of the observer, the observer will see a virtual image of the object which is at a different location than the object
Apparent depth,
observer directly
above object
Where n1 is for the medium of the incident ray coming from the object and n2 is for the medium of the refracted ray

7. Workbook exercises

8. 26.2 Snell’s Law and the Refraction of Light
Conceptual Example 4 On the Inside Looking Out
A swimmer is under water and looking up at the surface. Someone
holds a coin in the air, directly above the swimmer’s eyes. To the
swimmer, the coin appears to be at a certain height above the
water. Is the apparent height of the coin:
Greater than its actual height
Less than its actual height
Equal to its actual height

9. 26.2 Snell’s Law and the Refraction of Light
THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL
Incident ray and emergent ray (doubly refracted) are equal.

10. We can’t compare n1 to n3 without knowing n2.
A light ray travels from medium 1 to medium 3 as shown. For these media,
n3 = n1.
n3 > n1.
n3 < n1.
We can’t compare n1 to n3 without knowing n2.
Answer: B

11. 26.3 Total Internal Reflection
When light passes from a medium of larger refractive index into one
of smaller refractive index, the refracted ray bends away from the
normal.
At some point, the incident angle will be such that the reflected angle is 90° from the normal, i.e. no part of the ray will refract out of the 2nd medium.
That incident angle is the incident angle, θc.
Recall sin 90° = 1
Critical angle

12. 26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an index of refraction of The cladding has an index of refraction of What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber?

13. 26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an index of refraction of The cladding has an index of refraction of What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber?
Known Find
ncore = θ1
nclad = 1.48
n1 sinθ1 = n2 sin θ2 Law of refraction
θ2 = 90 – θc Geometry – see illustration
sin θc = nclad /ncore Total internal reflection

14. 26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an index of refraction of The cladding has an index of refraction of What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber?
Known Find
ncore = θ1
nclad = 1.48
nair θ1 = ncore θ2 Law of refraction
θ2 = 90 – θc Geometry – see illustration
sin θc = nclad /ncore Total internal reflection
θ1 =37.44 degrees.

15. 26.5 The Dispersion of Light: Prisms and Rainbows
The net effect of a prism is to change the direction of a light ray.
Light rays corresponding to different colors bends by different amounts.

16. 26.5 The Dispersion of Light: Prisms and Rainbows
A narrow beam of sunlight is incident on a slab of crown glass at 45 degrees. Find the angle between the violet ray and the red ray in the glass.

17. 26.5 The Dispersion of Light: Prisms and Rainbows
A narrow beam of sunlight is incident on a slab of crown glass at 45 degrees. Find the angle between the violet ray and the red ray in the glass.
Answer: degrees

18. 26.5 The Dispersion of Light: Prisms and Rainbows

19. 26.6 Lenses
Mirrors reflect light according to the Law of Reflection and create a real or virtual image of the light source.
Lenses refract light according to Snell’s Law of Refraction and create a real or virtual image of the light source.
Converging and diverging lenses are analogous to concave and convex mirrors.

20. 26.6 Lenses
With a converging lens, paraxial rays that are parallel to the principal axis converge to the focal point of the lens on the opposite side from the object.

21. 26.7 The Formation of Images by Lenses
Ray diagrams for converging lenses:
Ray 1 is parallel when it leaves the object and is refracted so that it passes through the focal point on the other side of the lens.
Ray 2 passes through the focal point on the same side as the object and is refracted so that it is parallel on the other side of the lens.
Ray 3 passes though the principal axis in the middle of the lens and continues on in a straight line.

22. 26.7 The Formation of Images by Lenses
There are 3 different types of images formed by a converging lens, depending on the relationship between the object and the focal point of the lens.

23. Object more than twice the focal length from the lens
In this example, when the object is placed further than twice the focal length from the lens, the real image is inverted and smaller than the object.
When using a digital camera, the film is replaced by a charge coupling device (CCD), a semi-conductor diode that acts much like a solar cell and changes light into electrical current.

24. Object further away than focal length
When the object is placed between F and 2F, the real image is
inverted and larger than the object.
To see the real image from a projector on the screen, you have to be between the lens and the screen and look at the reflection from the screen

25. 26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A CONVERGING LENS (MAGNIFYING GLASS) 2
When the object is placed between F and the lens, the virtual image is upright and larger than the object.
Now you look through the lens, but you don’t see the object itself, but rather it’s virtual image.
Note the lens is not high enough for #2 , but I drew it in.

26. 26.6 Lenses
With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point on the same side as the object.

27. 26.7 The Formation of Images by Lenses
Ray diagrams for diverging lenses:
Ray 1 is parallel when it leaves the object and is refracted so that it’s continuation passes through the focal point on the same side of the lens as the object.
Ray 2 is refracted so that it is parallel and its continuation passes through the focal point on the opposite side of the lens.
Ray 3 passes though the principal axis in the middle of the lens and continues on in a straight line.

28. 26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A DIVERGING LENS 2
A diverging lens always forms an upright, virtual, reduced image.
Ray 2 leaves the object and travels toward the focal point on the other side of the lens.

29. The image will be as it was, but much dimmer.
A lens produces a sharply-focused, inverted image on a screen. What will you see on the screen if the lens is removed?
The image will be as it was, but much dimmer.
The image will be right-side-up and sharp.
The image will be right-side-up and blurry.
The image will be inverted and blurry.
There will be no image at all.
STT23.4

30. 26.8 The Thin-Lens Equation and the Magnification Equation

31. 26.8 The Thin-Lens Equation and the Magnification Equation
Summary of Sign Conventions for Lenses
di is + for a real image formed on the opposite side of the lens as the object.
di is - for a virtual image formed on the same side of the lens as the object.

32. A converging lens
A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the nature of the image?
real, inverted and reduced
real, inverted and enlarged
virtual, upright and enlarged
virtual, upright and reduced.

33. A converging lens
A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the a) position and b)magnification of the image?
Solve for di and m

34. Converging lens From the thin-lens equation, we obtain:
The negative answer shows it is on the same side of the lens as the object. The magnification equation gives the magnification to be:
The positive number shows it is upright.

35. A converging lens
A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the a) position and b)magnification of the image?
Solve for di and m

36. The cornea and the lens act to bend the incoming light waves.
26.10 The Human Eye
Anatomy of the Eye
The cornea and the lens act to bend the incoming light waves.
The lens only contributes about 20-25% of the refraction, but its function
is important, as it provides the fine-tuning mechanism.
The retina acts as the film, or CCD and receives the image, which it then sends, via the optic nerve, to the brain, which interprets the image.

37. When the ciliary muscles relax, the lens becomes less curved.
26.10 The Human Eye
OPTICS
When the ciliary muscles relax, the lens becomes less curved.
The radius of curvature, and therefore the focal length increases.
The eye can focus on more distant objects.
Of course there is a limit to how far away an object can be that can be focused at the retina, even for people with good vision.

38. When the ciliary muscles become tensed, the lens becomes more convex.
26.10 The Human Eye
OPTICS
When the ciliary muscles become tensed, the lens becomes more convex.
The radius of curvature, and therefore the focal length, decreases.
This allows the eye to focus on closer objects.
Of course there is a limit to this close focus.

39. 26.10 The Human Eye
NEARSIGHTEDNESS
For some people, the lens cannot become “relaxed” enough to focus far-away objects on the retina. Instead, the eye focuses the image in front of the retina.
For the purposes of this discussion, “a distant” object is considered to be infinitely far away (d0 = ∞).

40. 26.10 The Human Eye
NEARSIGHTEDNESS
A divergent lens in front of the eye creates a virtual image in front of the object, close enough for the nearsighted eye to see.
The virtual image acts as a virtual object, from which the eye makes a real image on the retina.

41. The lens creates an image of the close object behind the retina.
26.10 The Human Eye
FARSIGHTEDNESS
The lenses of some people cannot tense up enough to focus on a close object.
The lens creates an image of the close object behind the retina.
This happens to all people as they age !!
Now the objective is to create a virtual image further away than the object.

42. 26.10 The Human Eye
FARSIGHTEDNESS
The lens creates a virtual image of the close object at the near point
of the farsighted eye.

43. Example 12 Eyeglasses for the Nearsighted Person
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 glasses
so the person can see “distant” objects.

44. 26.10 The Human Eye
d0 = ∞
di = (521-2)= 519 cm
Find f

45. THE REFRACTIVE POWER OF A LENS – THE DIOPTER
26.10 The Human Eye
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.

46. 26.11 Angular Magnification and the Magnifying Glass
The size of the image on the retina determines how large
an object appears to be.

47. 26.11 Angular Magnification and the Magnifying Glass

48. 26.11 Angular Magnification and the Magnifying Glass
Example 14 A Penny and the Moon
Compare the angular size of a penny held at arms length with that of
the moon.
Penny
Moon

49. 26.11 Angular Magnification and the Magnifying Glass
of a magnifying glass

50. 26.12 The Compound Microscope
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

51. Angular magnification of an astronomical telescope
26.13 The Telescope
Angular magnification of
an astronomical telescope