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Physics 11Unit 8 – Geometric Optics

Part 1

1.Review of waves

• In the previous section, we have investigated the nature and behaviors of waves in general. We know that all waves possess the following characteristics:

• Reflection

• Refraction

• Diffraction

• Interference

• We have also spent a great deal of time looking at the behaviors of sounds, which is a longitudinal wave. In particular, we studied the working principles behind two types of musical instruments: the stringed instruments and wind instruments.

Unit 8 - Geometric Optics (Part 1) 2

• Recall there are two types of waves we encounter in nature depending on the way the wave is vibrating.

• (1) Longitudinal wave

• The particles vibrate in the same direction as it propagates.

• (2) Transverse wave

• The particles vibrate in the direction perpendicular to the direction of propagation.


• In this unit, we will be studying the second type of waves, trying to explore their nature and properties. Specifically, we will discuss lightbecause of its relevance to our daily lives.


2. Electromagnetic radiation• We have seen that all kinds of waves, either longitudinal or

transverse, carry energy (but not matter!) from the source to the receiver.

• A special type of wave, called electromagnetic (EM) wave, is made of alternating, perpendicular electric and magnetic fields. Since the electric field and magnetic field are at 90° to the direction the wave travels, it belongs to a transverse wave.


• There are many different types of EM waves depending on how they are generated, but they have the same speed when travelling in vacuum. This speed is called the speed of light, so named due to the fact that visible light is the most recognized EM wave. The experimental speed of light in vacuum is 3.0 × 108 m/s!

• Different sources produce the EM waves with different frequenciesand wavelengths, and the types of EM waves are identified by these factors. Note, however, that the values of frequency and wavelength of an EM wave are not random; they are determined by the general wave equation that we have seen in Unit 7:

• Since the speed of light is constant, higher the frequency, smaller the wavelength.

Unit 8 - Geometric Optics (Part 1)

𝑣 = 𝑓λ

6

• Technically, the type of an EM wave is classified by its associated wavelength.


(i) Radio waves and radars

• Wavelength: > 10 cm

• Used by TV, radios (AM and FM), cell phones, etc.

(ii) Microwaves

• Wavelength: 0.01 cm – 10 cm

• Used in microwave cooking and signal transmission over short distance.

(iii) Infrared radiation

• Wavelength: 0.0001 cm – 0.01 cm

• Used in remote controls, IR goggles and sensors.Unit 8 - Geometric Optics (Part 1) 8

(iv) Visible light

• Wavelength: 0.00004 cm – 0.00007 cm

• The form of EM waves that can be detected by human eyes.

(v) Ultraviolet radiation

• Wavelength: 10-7 cm – 10-5 cm

• Used to make vitamin D and sensors.

(vi) X-ray radiation

• Wavelength: 10-9 cm – 10-7 cm

• Used in medical diagnosis and imaging.


(vii) Gamma radiation

• Wavelength: < 10-9 cm

• Used in radiotherapy for cancer.

(viii) Cosmic rays

• Wavelength: extremely small

• Very energetic; the particles may carry energy as high as 1020 eV, which is about a hundred million times larger than what is produced by the Large Hadron Collider (LHC).


• Just like sound or water waves, light travels at different speeds in different media depending on their nature. The ratio of the speed of light travelling in vacuum to the speed of light travelling in a given material is called the index of refraction or refractive index. Mathematically,

• The higher the value of 𝑛, the more the light is slowed down in the material. For example, air has a refractive index of 1.0003 meaning that light travels at pretty much the same in air as in vacuum. On the other hand, water has a refractive index of 1.33, suggesting that light retains only about 75% of its speed when it enters into water from vacuum.


𝑛 =𝑐

𝑣

11

• The refractive indices for some substances are tabulated below.


Substance 𝒏 Substance 𝒏

Vacuum 1.0000 Crown glass 1.52

Air 1.0003 Sodium chloride 1.53

Water 1.33 Crystal glass 1.54

Ethyl alcohol 1.36 Ruby 1.54

Quartz 1.46 Flint glass 1.65

Glycerine 1.47 Zircon 1.92

Lucite 1.51 Diamond 2.42

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• The values in the table are sometimes called absolute refractive indices as they refer to the light travelling from a vacuum to substances.

• When light travels between two substances with different refractive indices, the ratio of the speeds is called the relative refractive index. This value is also equal to the ratio of the absolute refractive indices of these substances.

• For example, for a light that moves from substance 1 to substance 2, the relative refractive index is:


1𝑛2 =𝑣1𝑣2

=Τ1 𝑣2Τ1 𝑣1

=Τ𝑐 𝑣2Τ𝑐 𝑣1

=𝑛2𝑛1

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• Example: Calculate the speed of light in Lucite.

• From the table on p. 12, the refractive index of Lucite is 1.51. Therefore,

which yields


𝑛 = 1.51 =𝑐

𝑣=3.0 × 108

𝑣

𝑣 =𝑐

𝑛=3.0 × 108

1.51= 1.99 × 108 m/s

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• Example: What is the relative refractive index for light traveling from water into glass and from glass to water?

• Note that 𝑛𝑔 = 1.52 for glass and 𝑛𝑤 = 1.33 for water.

• Hence, for water to glass:

• Yet for glass to water:


𝑤𝑛𝑔 =𝑛𝑔

𝑛𝑤=1.52

1.33= 1.14

𝑔𝑛𝑤 =𝑛𝑤𝑛𝑔

=1.33

1.52= 0.88

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3. Optical effects(a) Camera obscura

• Also known as pinhole image, this natural phenomenon refers to the formation of an inverted image on a screen inside a light-proof box opposite to a pinhole, or aperture, through which light from a scene outside enters into the box.

• An illustration of a pinhole camera:


• Pinhole camera is a camera that does not have lens. The size of the image it produces depends on the diameter of the aperture and the length of the camera (called the focal length).

• By similar triangles, it can be seen that

• This ratio is called the magnification, 𝑀, of the pinhole camera.Unit 8 - Geometric Optics (Part 1)

ℎ𝑖ℎ0

=𝑑𝑖𝑑0

17

• When an image is inverted, a negative sign has to be added to the magnification. This gives rise to the magnification equation:

• Example: Calculate the size of the image of the tree that is 8.0 m high and 80 m from a pinhole camera that is 20 cm long.

• Since ℎ0 = 8.0, 𝑑0 = 80, 𝑑𝑖 = 0.20, we have


ℎ𝑖 = ℎ0𝑑𝑖𝑑0

= 8.00.20

80= 0.02 m

𝑀 =ℎ𝑖ℎ0

= −𝑑𝑖𝑑0

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• In physics, there are 3 characteristics we have to mention when describing an image: attitude, size and type.

• Real image is an image which can be formed on a screen. On the other hand, the image that cannot be formed on a screen is called virtual image.

• All images formed by pinhole cameras are real, inverted and smaller.


Characteristics Description

Attitude Upright or inverted

Size Smaller, same or larger

Type Real or virtual

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• Some photos taken by pinhole cameras:


(b) Reflection

• When light strikes a surface, it makes an angle 𝜃𝑖, called angle of incidence, to the normal of the surface. The light is bounced back by the surface at the angle 𝜃𝑟, called angle of reflection, to the normalwhich is equal to the angle of incidence.

• We can then formulate the law of reflection mathematically:


𝜃𝑖 = 𝜃𝑟

• Note that the incident ray, the reflected ray, and the normal should lie in the same plane.

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• Depending on the nature of the reflecting surface, the incident light rays may undergo either specular reflection or diffuse reflection.

(1) Specular reflection

• Occurs on flat surfaces.

• All incident rays are reflected at the same angle.

(2) Diffuse reflection

• Occurs on rough surfaces or due to scattering of the surface materials.

• Incident rays are reflected at many different angles.


• The effects of specular and diffuse reflections can be clearly shown in the following pictures:


Image formed on calm water Image formed on wavy water23

• The idea of reflection can be used to explain the formation of images by mirrors.

(i) Plane mirror

• Rays from each point of the cake go in all directions; only those that form the green bundle enter into the eye and can be seen.

• Our brain interprets these rays as coming from the image points behind the mirror.

• The image is as far behind the mirror as the object is in front.



• Due to the congruency, not only do the object and image are equally apart from the mirror, but also they are of the same size (or height).

• Note also that the lights that the eye receives do not truly pass through the image behind the mirror. Such misinterpretation arises from the fact that our brains always perceive lights as having traveled a straight-line path. Because of this, the images formed by a plane mirror are virtual images.

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• Example: A woman 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror and how high must its lower edge be above the floor if she is to be able to see her whole body? Assume her eyes are 10 cm below the top of her head.


(ii) Spherical mirrors

• A spherical, curved mirror can be thought of as a section of hollow sphere. Depending on either the inner or outer side of the mirror is polished to reflect light, it is called a concave (inner) or convex (outer) mirror.

• A concave mirror makes parallel light rays converge on each other, thus being called converging mirror.

• A convex mirror, on the other hand, diverges the parallel light rays; it is therefore called a diverging mirror.


(1) Concave mirror

• Every concave mirror consists of five major components as shown in the following diagram:

• Center of curvature (C): The center of the sphere from which the mirror is produced.

• Vertex (P): The center of the mirror.

• Principal axis (PA): The line that connects the center and the vertex.

• Focus (F): The midpoint between the center and the vertex.

• Focal length (f): The distance between the focus and the vertex.


• There are three rules that govern the construction of the ray diagram for concave mirrors.

(1) A ray that is parallel to the principal axis is reflected through the focus.

(2) A ray passing through the focus is reflected parallel to the principal axis.

(3) A ray passing through the center is reflected back along the samepath.


• Example: The formation of the image of a candle by a concave mirror.

• Note: Usually two rays are sufficient to determine where the image will be formed. The image appears at the position where the two rays converge. To verify, the third ray can be drawn passing through the center of curvature C.


• Example: Draw the ray diagram for the following and describe the image formed.

• The image is:


(2) Convex mirrors

• Convex mirrors are also composed of the essential parts that concave mirrors have: C, F, PA, P, and f.

• The rules of constructing the ray diagram for convex mirrors are exactly the same as those used for concave mirrors.


• Please note that while different types of images (large/small, real/virtual) can be formed by concave mirrors, convex mirrors always form upright, smaller and virtual images!

• Example: Find the image formed by a convex mirror.


(3) Mirror equation

• There is a formula, called the mirror equation, that allows for a quick prediction of where an image will be formed by a curved mirror.

• Simple derivation:

Since ∆𝑂′𝐴𝑂~∆𝐼′𝐴𝐼:

Similarly, since ∆𝑂′𝐹𝑂~∆𝐵𝐹𝐴:


ℎ0ℎ𝑖

=𝑑0𝑑𝑖

ℎ𝑜ℎ𝑖

=𝑑0 − 𝑓

𝑓

Equating these two gives

which yields after rearrangement:

• This equation can be applied to either concave or convex mirrors.

• The magnification of a mirror can be defined as before:


𝑑0𝑑𝑖

=𝑑0 − 𝑓

𝑓

1

𝑓=

1

𝑑0+1

𝑑𝑖

𝑀 =ℎ𝑖ℎ0

= −𝑑𝑖𝑑0

• There are sign conventions when these equations are used:

• (1) The image height ℎ𝑖 is positive if it is upright, or negative if it is inverted relative to the object.

• (2) The image distance 𝑑𝑖 and the object distance 𝑑0 are positive if both are on the reflecting side of the mirror, or negative if they are behind the reflecting side. That means, the image is real if 𝑑𝑖 is positive while it is virtual if 𝑑𝑖 is negative.

• (3) The focal length 𝑓 is positive if it is on the reflecting side of the mirror. This is always the case for concave mirrors.


• Example: A concave mirror has a radius of curvature of 12.0 cm. A 1.2 cm tall object is placed a distance of 8.2 cm away from the mirror.

(a) Locate the image.

(b) Calculate the height of the image.

(c) Describe the image.

• It is better to first sketch the ray diagram to get some insights.


• (a) Since 𝑓 = Τ12.02 = 6.0 cm, and 𝑑0 = 8.2 cm, the location of the

image is

• (b) Using the magnification formula:

• (c) The image is inverted, larger, and real. This is exactly what is predicted by the ray diagram.


1

6.0=

1

8.2+1

𝑑𝑖→ 𝑑𝑖 = 22.0 cm

ℎ𝑖ℎ0

= −𝑑𝑖𝑑0

→ ℎ𝑖 = − 1.222.0

8.2= −3.2 cm

• Example: A 5.3 cm tall object is placed 6.4 cm away from a spherical mirror. A virtual image is formed 4.2 cm from the mirror.

(a) What is the focal length of the mirror?

(b) What kind of mirror is it?

(c) What is the height of the image?

• (a) Since a virtual image is formed by the mirror, 𝑑𝑖 = −4.2 cm. By the mirror equation, the focal length is:


1

𝑓=

1

𝑑0+1

𝑑𝑖=

1

6.4+

1

−4.2→ 𝑓 = −12.2 cm

• (b) Since 𝑓 = −12.2 cm, the focal point is located behind the reflecting side, suggesting that the mirror is convex.

• (c) By the magnification formula:

The image is upright, smaller, and virtual.


ℎ𝑖ℎ0

= −𝑑𝑖𝑑0

→ ℎ𝑖 = − 5.3−4.2

6.4= 3.5 cm

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