Chapter 26

The Refraction of Light: Lenses and Optical Instruments

Chapter 26 The Refraction of Light: Lenses and Optical Instruments26.1 The Index of Refraction

26.2 Snell’s law and the Refraction of Light

26.3 Total Internal reflection

26.6 Lenses

26.7 The formation of Images by lenses

26.8 The Thin-Lens equation and the Magnification equation

26.9 Lenses in combination

26.10 The Human eye

26.14 Lens aberrations

26.1 The Index of Refraction

• Light travels through a vacuum at a speed • Light travels through materials at a speed less than

its speed in a vacuum.• The change in speed as a ray of light goes from one

material to another causes the ray to deviate from its incident direction

• This is called 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:

𝑛=𝑆𝑝𝑒𝑒𝑑𝑜𝑓 h𝑙𝑖𝑔 𝑡 𝑖𝑛𝑣𝑎𝑐𝑢𝑢𝑚

𝑆𝑝𝑒𝑒𝑑𝑜𝑓 h𝑙𝑖𝑔 𝑡 𝑖𝑛 h𝑡 𝑒𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙=𝑐𝑣

26.2 Snell’s Law and the Refraction of Light

• When light strikes the interface, part of the light is reflected with the angle of incidence equalling the angle of reflection

• When a ray enters the second material and changes direction (is refracted) it behaves in one of the followinga) When light travels from a less

dense medium to a more dense one, the refracted light ray is bent toward the normal

b) When light travels from a more dense medium to a less dense one, the refracted light ray is bent away the normal

• The refraction that occurs at the interface between two materials obeys the Snell’s law of refraction

• The law states that1) the refracted ray, the incident ray, and the

normal to the interface all lie in the same plane

2) The angle of refraction is related to the angle of incidence according to

𝑛1 sin𝜃1=𝑛2 sin 𝜃2

Example 1 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.

Solution

𝑛1 sin𝜃1=𝑛2 sin 𝜃2

∴sin 𝜃2=𝑛1sin 𝜃1

𝑛2

sin 𝜃2=(1.00 ) (sin 46 )

1.33¿0.54

(a)

∴𝜃2=33 °

APPARENT DEPTH

• When an object lies under water, it appears to be closer to the surface than it actually is apparent depth

69.0

00.1

31sin33.1sinsin

1

221

n

n

441

313.30.2tan 12

Example 2 Finding a Sunken Chest

The searchlight on a yacht is being used to illuminate a sunkenchest. At what angle of incidence should the light be aimed?

Apparent depth, observer directly above object

𝑑′=𝑑 (𝑛2𝑛1 )

Conceptual Example 4 On the Inside Looking Out

A swimmer is under water and looking up at the surface. Someoneholds a coin in the air, directly above the swimmer’s eyes. To theswimmer, 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?

THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL (STUDY)

THE DERIVATION OF SNELL’S LAW

26.3 Total Internal Reflection

• If the incident ray is at the critical angle , the angle of refraction is 90°

• The critical is obtained from Snell’s law• When the incidence angle exceeds the critical

angle, all the incidence light ray is reflected back into the material from which it came

• This phenomenon is known as total internal reflection

sin 𝜃𝑐=𝑛2sin 90 °

𝑛1=𝑛2𝑛1

h𝑤 𝑒𝑟𝑒𝑛1>𝑛2

Example 5 Total Internal ReflectionA beam of light is propagating through diamond and strikes the diamond-air 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.

4.2442.2

00.1sinsin 1

1

21

n

nc(a)

(b) 3.3342.2

33.1sinsin 1

1

21

n

nc

Conceptual Example 6 The Sparkle of a DiamondThe diamond is famous for its sparkle because the light coming fromit 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?

Fiber optics

• Application of total internal reflection occurs is fiber optics

• Consists of cylindrical inner core that carries the light and the outer concentric shell, the cladding

• Little light is lost as a result of absorption by the core

• Light can travel many kilometres before its intensity diminishes appreciably

Endoscopy

• Optical fiber cables are used in medicine in endoscopy

• Used to peer inside the body

• Doctor is using bronchoscope to examine the lungs of a patient

Colonscope is used to examine the interior of the colon for diagnosing colon cancer in its early stage

• Optical fibres have made arthroscopic surgery possible

• Insert the instrument and the cable into a joint, with only a tiny incision and minimal damage to surrounding tissues

26.4 Polarization and the Reflection and Refraction of Light

Brewster’s law

• When light is incident on a non-metallic surface at the Brewster angle, the reflected light is completely polarized parallel to the surface

• The reflected and refracted rays are perpendicular to each other

tan𝜃𝐵=𝑛2𝑛1

26.5 The Dispersion of Light: Prisms and Rainbows (Study)

The net effect of a prism is to change the direction of a light ray.Light rays corresponding to different colors bend by different amounts.

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 situationsdepicted in the figure arise?

26.6 Lenses• Lenses refract light in such a way that an image of

the light source is formed.• With a converging (convex) lens, paraxial rays that

are parallel to the principal axis converge to the focal point.

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

http://www.wiley.com/college/cutnell/0470879521/image_gallery/ch26/pages/physics9e_fig_26_25.htm

26.7 The Formation of Images by Lenses

RAY DIAGRAMS for convex (converging) lens

• When the object is placed a distance further than twice the focal length from the lens, the image is real, inverted and smaller than the object.

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

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

RAY DIAGRAMS for diverging (concave) lens

• Regardless of the position of a real object, a diverging lens always forms an image that is upright, virtual and smaller than the object.

26.8 The Thin-Lens Equation and the Magnification Equation

𝑚=h𝑖h𝑜

=−𝑑𝑖

𝑑𝑜

1𝑓=1𝑑𝑜

+1𝑑𝑖

Summary of Sign Conventions for Lenses

NOTE THAT:• is positive for a converging (convex) lens and

negative for a diverging (concave) lens• is positive if the object is to the left of the lens and

negative if it is to the right of the lens • is positive for the image formed to the right of the

lens (real image) and negative if the image is formed to the left of the lens (virtual image)

• is positive for an upright image and negative for an inverted image

1𝑓=1𝑑𝑜

+1𝑑𝑖

𝑚=h𝑖h𝑜

=−𝑑𝑖

𝑑𝑜

Example 9 The Real Image Formed by a Camera Lens

A 1.70-m tall person is standing 2.50 m in front of a camera. The camera uses a converging lens whose focal length is 0.0500 m.

(a)Find the image distance and determine whether the image is real or virtual.

(b) Find the magnification and height of the image on the film.1𝑓=1𝑑𝑜

+1𝑑𝑖

1𝑑𝑖

=1𝑓−1𝑑0

1𝑑𝑖

=1

0.05𝑚−

12.5𝑚

∴𝑑𝑖=0.0510𝑚

𝑚=h𝑖h𝑜

=−𝑑𝑖

𝑑𝑜h 𝑖=

(−𝑑𝑖 ) (h𝑜 )𝑑𝑜

h 𝑖=(−0.0510𝑚 ) (1.70𝑚 )

2.50𝑚

The image is real since is +ve

m

26.9 Lenses in Combination

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

Example 11.The objective and eyepiece lenses (of a compound microscope) have focal lengths of 15.0mm and 25.5 mm respectively. A distance of 61.0 mm separates the lenses. The microscope is used to examine objects placed 24.1 mm from the objective lens. Determine the final image distance1𝑑𝑖2

=1𝑓 𝑒−1𝑑𝑜2

1𝑑𝑖1

=1𝑓 𝑜

−1𝑑𝑜1

1𝑑𝑖1

=1

15.0𝑚𝑚−

124.1𝑚𝑚

1𝑑𝑖1

=0.0252𝑚𝑚− 1

∴𝑑𝑖 1=39.7𝑚𝑚

𝑑𝑜2=61.0𝑚𝑚−𝑑𝑖 1=61.0𝑚𝑚−39.7𝑚𝑚=21.3𝑚𝑚

1𝑑𝑖2

=1𝑓 𝑒−1𝑑𝑜2

1𝑑𝑖2

=1

25.5𝑚𝑚−

121.3𝑚𝑚

1𝑑𝑖2

=−0.0077𝑚𝑚− 1

𝑑𝑖 2=−130𝑚𝑚

26.10 The Human Eye

• Far point:- is the location of the farthest object on which a fully relaxed eye can focus ( located nearly at infinity)

• Near point:- the point nearest the eye at which an object can be placed and still produce a sharp image at the retina (located 25 cm from the eye)

• Accommodation:- the process in which the lens changes its focal length to focus on objects at different distances

NEARSIGHTEDNESS (Myopia)

• The person can focus on near objects but not on distant object

• The far point is no longer at infinity but some distance closer to the eye

• The focal length of the eye is shorter than it should be and the distant object forms an image in front of the retina

• The lens creates an image of the distance object at the far point of the nearsighted eye.

This defect is corrected by the use of a diverging (concave) lens

http://www.wiley.com/college/cutnell/0470879521/image_gallery/ch26/pages/physics9e_fig_26_36.htm

Example 12 Eyeglasses for the Nearsighted PersonA nearsighted person has a far point located only 521 cm from theeye. 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 glassesso the person can see distant objects.

1𝑓=1𝑑𝑜

+1𝑑𝑖

1𝑓=1∞

+1

−519𝑐𝑚

𝑓 =−519𝑐𝑚

𝑑𝑜=∞ 𝑑𝑖=521𝑐𝑚

FARSIGHTEDNESS

THE REFRACTIVE POWER OF A LENS – THE DIOPTER

• The extent to which rays of light are refracted depends on its focal length

• 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.

𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒𝑝𝑜𝑤𝑒𝑟 (𝑖𝑛𝑑𝑖𝑜𝑝𝑡𝑒𝑟𝑠 )= 1𝑓 (𝑖𝑛𝑚𝑒𝑡𝑒𝑟𝑠)

• A converging lens has a positive refractive power, and a diverging lens has a negative refractive power

Using the lens equation

1𝑓=1𝑑𝑜

+1𝑑𝑖

• Can read the magazine when the image formed by the lens is at the near point of the eye

∴ 1𝑑𝑜

=1𝑓−1𝑑𝑖

1𝑑𝑜

=1

35.1𝑐𝑚−

1−142𝑐𝑚

∴𝑑𝑜=28.1𝑐𝑚

𝑑𝑖=−142𝑐𝑚

Exercise 26-77 (Ex 26-76 in the 8th Ed)Your friend has a near point of 142 cm, and she wears contact lenses that have a focal length of 35.1 cm.How close can she hold a magazine and still read it clearly?

26.14 Lens Aberrations

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

• Lens aberrations limit the formation of perfectly focused or sharp images by the optical lens

• Spherical aberration can be reduced by using a variable-aperture diaphragm.

• Chromatic aberration is greatly reduced by the use of compound lens

• The lens combination is known as an achromatic lens

Chromatic aberrations

• The index of refraction of material from which the material is made varies with lens

• Different colours are focused at different points along the principal axis

1𝑓=1𝑑𝑜

+1𝑑𝑖

1𝑑𝑖

=1𝑓−1𝑑𝑜

1𝑑𝑖

=1

65𝑐𝑚−

125 𝑐𝑚

∴𝑑𝑖=−40.6𝑐𝑚At age 40

At age 45

1𝑑𝑖

=1𝑓−1𝑑𝑜

1𝑑𝑖

=1

65𝑐𝑚−

129𝑐𝑚

∴𝑑𝑖=−52.4𝑐𝑚

Exercise 26-121 (Ex 26-119 in the 8th Ed)At age 40, a certain man requires contact lenses, with f= 65cm, to read a book held 25cm from his eyes. At age 45, while wearing these contacts he must now hold a book 29cm from his eyes(a)By what distance has his near point changed?(b)What focal-length lenses does he require at

age 45 to read a book at 25cm

∴𝑑𝑖=−40.6𝑐𝑚 ∴𝑑𝑖=−52.4𝑐𝑚

The man’s near point has shifted by:52.4𝑐𝑚−40.6𝑐𝑚=11.8𝑐𝑚

b)

1𝑓=1𝑑𝑜

+1𝑑𝑖

1𝑓=125

+1

−52.4𝑐𝑚

𝑑𝑜=25𝑐𝑚𝑓 =47.8𝑐𝑚

THE END