The Refraction of Light: Lenses and Optical Instruments Chapter 26 The Refraction of Light: Lenses and Optical Instruments

Chapter 26 The Refraction of Light: Lenses and Optical Instruments 26.1 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 𝑐=3× 10 8 𝑚/𝑠 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 𝑛 of a material is the ratio of the speed of light in a vacuum to the speed of light in the material: 𝑛= 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = 𝑐 𝑣

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 following When light travels from a less dense medium to a more dense one, the refracted light ray is bent toward the normal 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 that the refracted ray, the incident ray, and the normal to the interface all lie in the same plane The angle of refraction 𝜃 2 is related to the angle of incidence 𝜃 1 according to 𝑛 1 sin 𝜃 1 = 𝑛 2 sin 𝜃 2

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 from air to water and from water to air. Solution (a) 𝑛 1 sin 𝜃 1 = 𝑛 2 sin 𝜃 2 ∴ sin 𝜃 2 = 𝑛 1 sin 𝜃 1 𝑛 2 sin 𝜃 2 = 1.00 sin 46 1.33 =0.54 ∴𝜃 2 =33°

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

Example 2 Finding a Sunken Chest The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed?

Apparent depth, observer directly above object 𝑑 ′ =𝑑 𝑛 2 𝑛 1

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, 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 𝜃 𝑐 , 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 𝜃 𝑐 = 𝑛 2 sin 90° 𝑛 1 = 𝑛 2 𝑛 1 𝑤ℎ𝑒𝑟𝑒 𝑛 1 > 𝑛 2

Example 5 Total Internal Reflection A beam of light is propagating through diamond and strikes the diamond-air interface at an angle of incidence of 28 degrees. Will part of the beam enter the air or will there be total internal reflection? Repeat part (a) assuming that the diamond is surrounded by water.

(a) (b)

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?

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 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 tan 𝜃 𝐵 = 𝑛 2 𝑛 1 Brewster’s law

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 situations depicted 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.

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 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 𝑚= ℎ 𝑖 ℎ 𝑜 =− 𝑑 𝑖 𝑑 𝑜

Summary of Sign Conventions for Lenses 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 𝑚= ℎ 𝑖 ℎ 𝑜 =− 𝑑 𝑖 𝑑 𝑜 NOTE THAT: 𝑓 is positive for a converging (convex) lens and negative for a diverging (concave) lens 𝑑 𝑜 is positive if the object is to the left of the lens and negative if it is to the right of the lens 𝑑 𝑖 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) 𝑚 is positive for an upright image and negative for an inverted image

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. 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. 1 𝑑 𝑖 = 1 𝑓 − 1 𝑑 0 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 1 𝑑 𝑖 = 1 0.05𝑚 − 1 2.5𝑚 ∴ 𝑑 𝑖 =0.0510𝑚 The image is real since 𝑑 𝑖 is +ve 𝑚= ℎ 𝑖 ℎ 𝑜 =− 𝑑 𝑖 𝑑 𝑜 ℎ 𝑖 = − 𝑑 𝑖 ℎ 𝑜 𝑑 𝑜 ℎ 𝑖 = −0.0510𝑚 1.70𝑚 2.50𝑚 ∴ ℎ 𝑖 =−0.0347 m

26.9 Lenses in Combination The image produced by one lens serves as the object for the next lens.

Determine the final image distance 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 distance 1 𝑑 𝑖2 = 1 𝑓 𝑒 − 1 𝑑 𝑜2 1 𝑑 𝑖1 = 1 𝑓 𝑜 − 1 𝑑 𝑜1 1 𝑑 𝑖1 = 1 15.0𝑚𝑚 − 1 24.1𝑚𝑚 1 𝑑 𝑖1 =0.0252 𝑚𝑚 −1 ∴ 𝑑 𝑖1 =39.7 𝑚𝑚

𝑑 𝑜2 =61.0𝑚𝑚− 𝑑 𝑖1 =61.0𝑚𝑚−39.7𝑚𝑚=21.3 𝑚𝑚 1 𝑑 𝑖2 = 1 𝑓 𝑒 − 1 𝑑 𝑜2 1 𝑑 𝑖2 = 1 25.5 𝑚𝑚 − 1 21.3 𝑚𝑚 1 𝑑 𝑖2 =−0.0077 𝑚𝑚 −1 𝑑 𝑖2 =−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

This defect is corrected by the use of a diverging (concave) lens The lens creates an image of the distance object at the far point of the nearsighted 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. 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 𝑑 𝑜 =∞ 𝑑 𝑖 =521𝑐𝑚 1 𝑓 = 1 ∞ + 1 −519𝑐𝑚 𝑓=−519𝑐𝑚

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. 𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑑𝑖𝑜𝑝𝑡𝑒𝑟𝑠 = 1 𝑓 (𝑖𝑛 𝑚𝑒𝑡𝑒𝑟𝑠) A converging lens has a positive refractive power, and a diverging lens has a negative refractive power

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? Using the lens equation 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 Can read the magazine when the image formed by the lens is at the near point of the eye ∴ 1 𝑑 𝑜 = 1 𝑓 − 1 𝑑 𝑖 𝑑 𝑖 =−142 𝑐𝑚 1 𝑑 𝑜 = 1 35.1 𝑐𝑚 − 1 −142 𝑐𝑚 ∴𝑑 𝑜 =28.1 𝑐𝑚

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 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 Chromatic aberration is greatly reduced by the use of compound lens The lens combination is known as an achromatic lens

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 By what distance has his near point changed? What focal-length lenses does he require at age 45 to read a book at 25cm ∴ 𝑑 𝑖 =−40.6𝑐𝑚 At age 40 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 1 𝑑 𝑖 = 1 𝑓 − 1 𝑑 𝑜 1 𝑑 𝑖 = 1 65𝑐𝑚 − 1 25𝑐𝑚 At age 45 1 𝑑 𝑖 = 1 𝑓 − 1 𝑑 𝑜 1 𝑑 𝑖 = 1 65𝑐𝑚 − 1 29𝑐𝑚 ∴ 𝑑 𝑖 =−52.4𝑐𝑚

The man’s near point has shifted by: 52.4𝑐𝑚−40.6𝑐𝑚=11.8𝑐𝑚 ∴ 𝑑 𝑖 =−40.6𝑐𝑚 ∴ 𝑑 𝑖 =−52.4𝑐𝑚 The man’s near point has shifted by: 52.4𝑐𝑚−40.6𝑐𝑚=11.8𝑐𝑚 b) 1 𝑓 = 1 𝑑 𝑜 + 1 𝑑 𝑖 1 𝑓 = 1 25 + 1 −52.4𝑐𝑚 𝑑 𝑜 =25𝑐𝑚 𝑓=47.8𝑐𝑚

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