Home Page Textbooks Nelson Science Perspectives 10 Page 590: Unit Review

Nelson Science Perspectives 10

1st Edition

Christy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe

ISBN: 9780176355289

Textbook solutions

All Solutions

Page 590: Unit Review

Exercise 1

Solution 1

Solution 2

Step 1

1 of 2

The angle of incidence is the angle between the “incident ray and the normal”.

Result

2 of 2

Option C.

Step 1

1 of 2

The angle of incidence is the angle between the incident ray and the normal.

Result

2 of 2

$textit{c.}$

Exercise 2

Solution 1

Solution 2

Step 1

1 of 2

Light reflecting of a plane mirror is an example of specular reflection since it has shiny surface.

Result

2 of 2

$textit{a.}$

Step 1

1 of 2

Light reflecting off a mirror is an example of “specular reflection” because of its shiny surface.

Result

2 of 2

Option A.

Exercise 3

Solution 1

Solution 2

Step 1

1 of 2

Index of refraction is the ratio of speed of light in a vacuum and speed of light in the medium.

Result

2 of 2

$textit{c.}$

Result

1 of 1

The index of refraction is the ratio of speed of light in a vacuum and speed of light in the medium, therefore option C is correct.

Exercise 4

Solution 1

Solution 2

Step 1

1 of 2

“The angle of refraction is the angle between the refracted ray and the normal” is statement that best describes angle of refraction.

Result

2 of 2

$textit{a.}$

Result

1 of 1

The angle of refraction is the angle between the refracted ray and the normal and therefore option a is correct.

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

Vacuum has index of refraction exactly $1$.

Result

2 of 2

$textit{c.}$

Result

1 of 1

The index of refraction is the ratio of speed of light in a vacuum and speed of light in the medium. For this to be equal to 1.0, both the numerator and the denominator must be equal and therefore option c is correct.

Exercise 6

Step 1

1 of 2

Refraction of light in a medium depends on its speed in the medium. The faster the speed of light in a medium, the lesser it refracts and vice versa. We know that the index of refraction is the ratio of speed of light in a vacuum and speed of light in the medium. This implies that the faster the speed of light in a medium, the smaller the value of the index of refraction and vice versa. Therefore material C with the least value of $n=1.58$ will refract light the least.

Result

2 of 2

Option C.

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

The object must be located between $F’$ and the lens.

Result

2 of 2

$textit{d.}$

Result

1 of 1

For a converging lens to form large, upright, virtual images of objects placed in front of them must have the objects within their focal length and therefore option d is correct.

Exercise 8

Solution 1

Solution 2

Step 1

1 of 2

A person that is far-sighted can clearly see distant object, but not nearby ones.

Result

2 of 2

$textit{a.}$

Result

1 of 1

Far-sighted, as the name suggests, is the ability to see distant objects clearly while near objects not clearly, therefore option a is correct.

Exercise 9

Solution 1

Solution 2

Step 1

1 of 2

“Light is a form of energy” is a true statement since it is always in motion, therefore, it is kinetic type of energy.

Result

2 of 2

Statement is true.

Step 1

1 of 2

Light is a form of energy is a true statement. The biggest example of this is the light from the sun which also warms up the planet earth and is essential for sustaining life on it.

Result

2 of 2

True.

Exercise 10

Solution 1

Solution 2

Step 1

1 of 2

Light is visible electromagnetic waves is a true statement. The electromagnetic spectrum is composed of a large variety of waves each having different wavelengths and frequency which are all invisible to the naked human eye except light.

Result

2 of 2

True.

Step 1

1 of 2

“Light is visible electromagnetic waves.” is true statement since wavelength at which light is can be detected by the human eye.

Result

2 of 2

The statement is true.

Exercise 11

Solution 1

Solution 2

Step 1

1 of 2

Light can only travel through a medium is a false statement. Light does not need a material medium for its propagation and can also travel through a vacuum.

Result

2 of 2

False.

Step 1

1 of 2

“Light can only travel through a medium.” is a false statement since light does not need a medium to travel through and can travel through the vacuum.

Result

2 of 2

The statement is false.

Exercise 12

Solution 1

Solution 2

Step 1

1 of 2

Statement “A light ray is either reflected or transmitted when it strikes a surface such as a piece of glass.”is false, since the ray of light which is falling on the transparent object partially reflects and partially refracts which means that both processes can happen.

Result

2 of 2

The statement is false.

The true one is: “A light ray is partially reflected and partially transmitted when it strikes a surface such as piece of glass.”

Step 1

1 of 2

A light ray is either reflected or transmitted when it strikes a surface such as a piece of glass is a false statement. Light falling on to an opaque object is entirely reflected, while light falling on a transparent object partially reflects while most of it refracts in it.

Result

2 of 2

False.

Exercise 13

Solution 1

Solution 2

Step 1

1 of 2

The centre of the lens is called the centre of curvature is a false statement because the centre of the lens is called the optical centre.

Result

2 of 2

False.

Step 1

1 of 2

Statement “The centre of the lens is called the centre of curvature.” is false since the centre of lens is the optical centre.

Result

2 of 2

Statement is false.

True statement will be:

The centre of the lens is called the optical centre.

Exercise 14

Solution 1

Solution 2

Step 1

1 of 2

A converging lens spreads parallel light rays apart on refraction is a false statement because as the name suggests, a converging lens “converges” parallel light rays onto a single point (called the focus of the lens) on refraction.

Result

2 of 2

False.

Step 1

1 of 2

Statement : “A converging lens spreads parallel light rays apart after reflection” is false, since converging lens converge parallel rays of light into a single point.

Result

2 of 2

Statement is false.

True statement will be:

“A converging lens converge parallel light rays after the reflection.

Exercise 15

Solution 1

Solution 2

Step 1

1 of 2

“A positive meniscus lens is a converging lens.” is a true statement.

Result

2 of 2

Statement is true.

Step 1

1 of 2

A positive meniscus lens is a converging lens is a true statement.

Result

2 of 2

True.

Exercise 16

Solution 1

Solution 2

Step 1

1 of 2

Statement: “A light ray that is parallel to the principal axis of a lens is not refracted.” is false, since rays which are parallel to the principal axis are refracted towards the focus in the case of converging lens and are dispersed outwards in case of diverging lens.

Result

2 of 2

Statement is false.

True statement will be:

“A light ray that is parallel to the principal axis of lens can be refracted.”

Step 1

1 of 2

A light ray that is parallel to the principal axis of a lens is not refracted is a false statement. This is because light rays parallel to the principal axis are refracted towards the focus of the lens in case of converging lens and disperse outwards in case of a diverging lens. A light ray that is passing through the principal axis or that is lying on the principal axis is not refracted.

Result

2 of 2

False.

Exercise 17

Step 1

1 of 2

Statement “Einstein predicted that gravitational forces can change path of light.” is true.

Result

2 of 2

Statement is true.

Exercise 18

Solution 1

Solution 2

Step 1

1 of 2

Statement: “The function of film in a camera is most like the function of iris in the human eye” is false since image is being formed on the retina.

Result

2 of 2

Statement is false.

True statement would be:

“The function of film in a camera is most like the function of the retina in the human eye”

Step 1

1 of 2

The function of film in a camera is most like the function of iris in the human eye is a false statement because image is formed on the retina of the eye.

Result

2 of 2

False.

Exercise 19

Solution 1

Solution 2

Step 1

1 of 2

An object located at the secondary principal focus F’, of a converging lens has an image that is exactly the same size as the object is a false statement. If the object is placed at the principal focus of a converging lens than its image is infinitely large in and located at infinity. The size of the image and object is identical when the object is located at 2F of the converging lens.

Result

2 of 2

False.

Step 1

1 of 2

Statement: “An object located at the secondary principal focus F’, of a converging lens has an image that is exactly the same size as the object.” is false since, in the case of converging lens, image will be exact same size as the object when the object is located at $2F$, otherwise it will be located in infinity and infinitely large.

Result

2 of 2

Statement is false.

True statement would be:

“An object located at the secondary principal focus 2F, of a converging lens has an image that is exactly the same size as the object.”

Exercise 20

Solution 1

Solution 2

Step 1

1 of 2

A light source that produces its own light is called a $textit{luminous source}$.

Result

2 of 2

Missing word is: luminous source.

Result

1 of 1

A light source that produces its own light is called a luminous object/source.

Exercise 21

Result

1 of 1

All electromagnetic waves travel at the speed of light equal to $3times10^8$ meters per second.

Exercise 22

Solution 1

Solution 2

Step 1

1 of 2

$textit{Gamma rays}$ are the type of electromagnetic waves with the greatest energy.

Result

2 of 2

Missing words are: gamma rays.

Result

1 of 1

Gamma rays are the type of electromagnetic waves with the greatest energy.

Exercise 23

Solution 1

Solution 2

Step 1

1 of 2

A diverging convex mirror can only produce images of a $textit{smaller}$ size and an $textit{upright}$ attitude.

Result

2 of 2

Missing words are: smaller and upright.

Result

1 of 1

A diverging convex mirror can only produce images of a smaller size and an upright attitude.

Exercise 24

Solution 1

Solution 2

Step 1

1 of 2

A magnifying glass is an example of a $textit{converging}$ lens.

Result

2 of 2

Missing word is: converging.

Result

1 of 1

A magnifying glass is an example of a converging lens.

Exercise 25

Result

1 of 1

At the critical angle, the angle of refraction is $90text{textdegree}$.

Exercise 26

Solution 1

Solution 2

Step 1

1 of 2

A lens that is thinnest at the middle and thickest at the edge is called a $textit{diverging}$ lens.

Result

2 of 2

Missing word is: diverging.

Result

1 of 1

A lens that is thinnest at the middle and thickest at the edge is called a diverging lens.

Exercise 27

Solution 1

Solution 2

Result

1 of 1

Light rays that are parallel to the principal axis of a converging lens meet at the principal focus.

Step 1

1 of 2

Light rays that are parallel to the principal axis of a converging lens meet at the $textit{principal}$ focus.

Result

2 of 2

Missing word is: focus.

Exercise 28

Solution 1

Solution 2

Result

1 of 1

Total internal reflection will occur when the angle of incidence is greater than the critical angle.

Step 1

1 of 2

Total internal reflection will occur when the angle of incidence is greater than the $textit{critical}$ angle.

Result

2 of 2

Missing word is critical.

Exercise 29

Solution 1

Solution 2

Result

1 of 1

The term for near-sightedness is myopia.

Step 1

1 of 2

The term for near-sightedness is $textit{myopia}$.

Result

2 of 2

Missing word is: myopia.

Exercise 30

Solution 1

Solution 2

Step 1

1 of 2

$textit{a.}$, Ultraviolet light – “black” lights

$textit{b.}$, Infrared light – DVD player remote controls

$textit{c.}$, Visible light – human vision

$textit{d.}$, Radio waves – radar

$textit{e.}$, X-rays – dental imaging

Result

2 of 2

$textit{a.}$, “black” lights

$textit{b.}$, DVD player remote controls

$textit{c.}$, human vision

$textit{d.}$, radar

$textit{e.}$, dental imaging

Step 1

1 of 6

a. Ultraviolet light- “black” lights

Step 2

2 of 6

b. Infra-red light – DVD player remote control

Step 3

3 of 6

c. Visible light – human vision

Step 4

4 of 6

d. Radio waves – radar

Step 5

5 of 6

e. X-rays – dental imaging

Result

6 of 6

Click to see answers.

Exercise 31

Solution 1

Solution 2

Step 1

1 of 2

$textit{a.}$, Clean sheet of glass is transparent material.

$textit{b.}$, Clear, cold winter sky is transparent.

$textit{c.}$, Frosted glass is translucent material.

$textit{d.}$, Brick is opaque material.

$textit{e.}$, Tree is opaque material.

$textit{f.}$, Hazy, polluted air is translucent.

$textit{g.}$, Glass of water is transparent.

Result

2 of 2

$textit{a.}$, transparent

$textit{b.}$, transparent

$textit{c.}$, translucent

$textit{d.}$, opaque

$textit{e.}$, opaque

$textit{f.}$, translucent

$textit{g.}$, transparent

Step 1

1 of 8

a. A clean sheet of glass is transparent.

Step 2

2 of 8

b. A clear cold winter sky is transparent.

Step 3

3 of 8

c. Frosted glass is translucent.

Step 4

4 of 8

d. A brick is opaque.

Step 5

5 of 8

e. A tree is opaque.

Step 6

6 of 8

f. Hazy polluted air is translucent.

Step 7

7 of 8

g. A glass of water is transparent.

Result

8 of 8

Click to see answers.

Exercise 32

Solution 1

Solution 2

Step 1

1 of 2

Converging lens has a thick middle part and two thin endings, on the other hand diverging lens has thin middle part and two thick endings.

Result

2 of 2

Converging lens has a thick middle part and two thin endings, on the other hand diverging lens has thin middle part and two thick endings.

Result

1 of 1

A converging lens is thickest in its middle and thinnest on its edges, while a diverging lens is thinnest in its middle and thickest on its edges.

Exercise 33

Step 1

1 of 7

a. When light strikes a surface, a perpendicular line is drawn from the point of incidence and the surface. This perpendicular line is called the normal. Angle of incidence is the angle between the incident ray and this normal.

Step 2

2 of 7

b. Angle of refraction is the angle between the refracted ray and the normal.

Step 3

3 of 7

c. Focus of a converging lens is the point lying on the principal axis of a converging lens, where all the light rays in a parallel beam of light meet on refraction.

Focus of a diverging lens is the point lying on the principal axis of a diverging lens, from where all the light rays in a parallel beam of light appear to diverge from on refraction.

Step 4

4 of 7

d. Magnification is the ratio the image size and the object size or image distance from the lens and object distance from the lens. It is the measure of how big(or small) the image is with respect to the object.

Step 5

5 of 7

e. A mirage is a naturally occurring optical illusion in which on a hot day, the water vapor in air reflect the sky above it on plain surface such as roads so that it appears like wet water surface to a distant observer.

Step 6

6 of 7

f. A virtual image is an image formed in mirrors and lenses in which the light rays do not intersect (as in the case of real images) but seem to be diverging from, where the eye imagines the image to be.

Result

7 of 7

Click to see answers.

Exercise 34

Solution 1

Solution 2

Step 1

1 of 2

$textit{a.}$, Stone is an opaque object.

$textit{b.}$, Clean glass slab is a transparent object.

$textit{c.}$, Frosted glass window is translucent object

Result

2 of 2

$textit{a.}$, stone – opaque

$textit{b.}$, clean glass – transparent

$textit{c.}$, frosted glass window – translucent

Step 1

1 of 4

a. A stone is an opaque object.

Step 2

2 of 4

b. A clean glass slab is a transparent object.

Step 3

3 of 4

c. A frosted glass window is a translucent object.

Result

4 of 4

Click to see answers.

Exercise 35

Step 1

1 of 2

$textit{a.}$, MUM, will be the three-letter word that appears exactly same when viewed in a plane mirror.

$textit{b.}$, DAD, will be the word that appears different in the plane mirror.

Result

2 of 2

$textit{a.}$, “MUM”

$textit{b.}$, “DAD”

Exercise 36

Solution 1

Solution 2

Step 1

1 of 4

a. Light is produced by electric discharge when electricity passes through a gas. The electric current energizes the gas particles and causes them to glow and emit light.

Step 2

2 of 4

b. Light is produced by bioluminescence by chemical processes taking place inside living beings.

Step 3

3 of 4

c. Light is produced by chemiluminescence by chemical reactions. Light is produced as a by-product of such reactions and usually such reactions do not dissipate heat and therefore this light is of immense application in areas where an naked spark can cause fire.

Result

4 of 4

Click to see answers.

Step 1

1 of 2

$textit{a.}$, Electric discharge produces light when electricity passes through the gas and in this process electric current gives energy to the surrounding particles of gas which causes them to glow, therefore, they emit light.

$textit{b.}$, Bioluminescence produces light by chemical processes which take place in living organisms.

$textit{c.}$, Chemiluminescence produce light by chemical reactions and light is produced as a product of these reactions after which there is no dissipation of heat, therefore, this light is safe to use in areas where there is a high risk of fire etc.

Result

2 of 2

$textit{a.}$, Light is produced when electricity passes through the gas.

$textit{b.}$, Light is produced by chemical processes inside living organisms.

$textit{c.}$, Light is produced as a product of chemical reactions after which the heat is not dissipated.

Exercise 37

Solution 1

Solution 2

Step 1

1 of 1

The word OPTICS will appear as shown in a plane mirror: Exercise scan

Step 1

1 of 2

Result

2 of 2

Word “OPTICS” will appear like this in the plane mirror.

Exercise 38

Step 1

1 of 1

All electromagnetic waves propagate with the same speed of $3times10^8$ meters per second. They do not require a material medium to propagate and can even travel through a vacuum.

Exercise 39

Solution 1

Solution 2

Step 1

1 of 2

$textit{a.}$, In the process of incandescence light is produced when there is a heating up of metal through the electric current.

Metal, which is heated up, will emit light.

In the process of fluorescence, materials absorbs ultraviolet light and emit white light after.

$textit{b.}$, In the process of tribolumiscence when it comes to rubbing, scratching or crushing crystals, life is being emitted.

In the process of phosphorence light is emitted after the ultraviolet light is absorbed.

Result

2 of 2

$textit{a.}$, In the process of incadescence light is produced by metal which is heated up.

In the process of fluorescence light is emitted by materials which are absorbing UV light.

$textit{b.}$, In the process of triboluminescence, light is emitted when crystals are rubbed, scratched or crushed.

In the process of phosphorescence light is emitted after the UV light is absorbed.

Step 1

1 of 2

a. Light in incandescence is produced by heating up a metal through electric current. The hot metal emits light. In fluorescence however, light is emitted by florescent materials which emit white light after absorbing ultraviolet light.

Step 2

2 of 2

b. Light is produced by triboluminescence when certain crystals are rubbed, scratched or crushed where by visible light is emitted by phosphorence by absorbing ultraviolet light from the electromagnetic spectrum.

Exercise 40

Solution 1

Solution 2

Step 1

1 of 5

a. A highly waxed floor is polished to be glossy and therefore a beam of light falling on it will reflect regularly causing specular reflection.

Step 2

2 of 5

b. A crumbled sheet of aluminum foil is no longer shiny because of the multiple folds on it and therefore light falling on it will disperse causing diffuse reflection.

Step 3

3 of 5

c. A dense carpet is dull in appearance and therefore will cause diffuse reflection.

Step 4

4 of 5

d. A flat sheet of aluminum foil with the shiny side up is shiny and cause specular reflection.

Result

5 of 5

Click to see answers.

Step 1

1 of 2

$textit{a.}$, A highly waxed floor is exhibiting specular reflection since this is the reflection from smooth surfaces and reflection is regular.

$textit{b.}$, A crumpled sheet of aluminium foil is exhibiting diffuse reflection since it is the rough surface and there will be dispersion or light rays.

$textit{c.}$, A dense carpet will exhibit diffuse reflection since it has a dull appearance.

$textit{d.}$, A flat sheet of aluminium foil, shiny side up, will exhibit specular reflection since the surface is “smooth”, therefore, there will be regular reflection.

Result

2 of 2

$textit{a.}$, specular reflection

$textit{b.}$, diffuse reflection

$textit{c.}$, diffuse reflection

$textit{d}$, specular reflection

Exercise 41

Solution 1

Solution 2

Step 1

1 of 2

Laser light would not be a useful tool for illuminating a dark room since laser light is monochromatic in nature and does not scatter, it travels in straight lines instead.

Result

2 of 2

Because it travels in straight lines, and do not scatter.

Result

1 of 1

Laser light is monochromatic in nature and does not scatter instead travels in straight lines and therefore will not be useful to illuminate a dark room.

Exercise 42

Solution 1

Solution 2

Step 1

1 of 4

Diverging lenses and diverging mirrors can never form a real image since they both have a focus from which the rays of light diverge from, they never actually intersect.

Step 2

2 of 4

1., Ray diagram of diverging lens forming a virtual image: Exercise scan

Step 3

3 of 4

2., Ray diagram of diverging mirror forming a virtual image: Exercise scan

Result

4 of 4

Diverging lenses and diverging mirrors can never form a real image since they both have a focus from which the rays of light diverge from, they never actually intersect.

Step 1

1 of 1

Diverging mirrors and lenses both have a focus from which the light rays seem to diverge from and therefore these light rays never intersect so they can never form a real image and the resulting images are always virtual in nature from where the light rays appear to be coming from.

Exercise 43

Step 1

1 of 7

Description

Angle of incidence

Angle of reflection

Step 2

2 of 7

The angle between the incident ray and the normal is $38text{textdegree}$

$$
38text{textdegree}
$$

Step 3

3 of 7

The angle between the incident ray and the normal is $12text{textdegree}$

$$
12text{textdegree}
$$

Step 4

4 of 7

The angle between the reflected ray and the flat mirror surface is $43text{textdegree}$

$$
90text{textdegree}-43text{textdegree}=47text{textdegree}
$$

$$
47text{textdegree}
$$

Step 5

5 of 7

The angle between the reflected ray and the normal is $23text{textdegree}$

$$
23text{textdegree}
$$

Step 6

6 of 7

The angle between the incident ray and the normal is $0text{textdegree}$

$$
0text{textdegree}
$$

Result

7 of 7

Click to see table.

Exercise 44

Step 1

1 of 8

$textit{A}$, Ray diagrams:

$textit{(a)}$

Exercise scan

Step 2

2 of 8

$textit{(b)}$

Exercise scan

Step 3

3 of 8

$textit{(c)}$

Exercise scan

Step 4

4 of 8

$textit{B}$, Image description:

Plane mirror always forms virtual, upright, same shape and size as the object, image where left and right sides are inverted.

Step 5

5 of 8

$textit{(a)}$: Image is virtual, placed behind the mirror, formed by the extensions of three reflected rays. It is same shape and size as “T”, upright and, we may not notice this since this letter is symmetrical, but it is also laterally inverted.

Step 6

6 of 8

$textit{(b)}$: Image is virtual, placed behind the mirror, formed by the extensions of three reflected rays. It is same shape and size as “P”, upright and, laterally inverted.

Step 7

7 of 8

$textit{(c)}$: Image is virtual, placed behind the mirror, formed by the extensions of three reflected rays. It is same shape and size as the actual object (arrow in this case), upright and, laterally inverted.

Result

8 of 8

See the description.

Exercise 45

Step 1

1 of 4

$textit{A}$, Ray diagrams:

$textit{(a)}$

Exercise scan

Step 2

2 of 4

$textit{(b)}$ and $textit{c}$

Exercise scan

Step 3

3 of 4

$textit{B}$, Image description:

$textit{(a)}$, Image is formed by concave mirror and since the object is placed between focal point and centre, it will be real, enlarged and upside-down image.

Step 4

4 of 4

$textit{(b)}$ and $textit{(c)}$ are the same, therefore: Image is formed by the convex mirror and is an virtual image since it is formed by the extensions of the reflected rays, it is upright and smaller in size than the actual object.

Exercise 46

Step 1

1 of 2

Draw a normal perpendicular to the two medium boundary. Measure the angles between the rays and the normal. It will be observed that the angle between the normal and the ray in medium A is greater than the angle between the normal and the ray in medium B. This implies that the ray bent towards the normal when it entered medium B. Therefore:

a. Material B has a greater index of refraction.

b. Light will travel slower in material B.

Result

2 of 2

Click to see answers.

Exercise 47

Step 1

1 of 6

Ray diagram: Exercise scan

Step 2

2 of 6

The image is real, inverted, larger than the object and on the other side of the lens as the object.

Step 3

3 of 6

Ray diagram: Exercise scan

Step 4

4 of 6

The image is real, inverted, smaller than the object and on the other side of the lens as the object.

Step 5

5 of 6

Ray diagram: Exercise scan

Step 6

6 of 6

The image is virtual, upright, smaller than the object and on the same side of the lens as the object.

Exercise 48

Step 1

1 of 1

Some detergents have fluorescent material added to them. These on washed clothes absorb ultraviolet light from the electromagnetic spectrum and emit them as visible light. This emission make the clothes “appear” cleaner. The word appear is used to emphasis that the vibrant colors of these clothes is because of the emission of this visible light and not because the clothes are actually cleaner. In this scenario, this detergent will be useful because the concern here is not the cleanliness of the shirt but its white appearance that can be further enhanced by using a detergent with added fluorescent.

Exercise 49

Step 1

1 of 3

a.
The ultraviolet radiations of the electromagnetic spectrum are the primary cause of skin cancer in the world.

Step 2

2 of 3

b.
Using generous quantities of sunblock creams and lotions can help create a barrier between the exposed skin and electromagnetic spectrum, protecting the skin from the possible adverse effects of electromagnetic radiations.

Result

3 of 3

Click to see answers.

Exercise 50

Step 1

1 of 1

Images formed in a plane mirror are virtual, upright and same size as the object and they are also an equidistant behind the mirror as the object in front of it. Therefore, the image of the cat will be an equal distance behind the mirror as the cat in front of it. The cat will never be able to touch its image and the closest the two can get is equal to the thickness of the mirror.

Exercise 51

Step 1

1 of 5

Data:

$$
d_o=41
$$

$$
f=27
$$

Step 2

2 of 5

Concave (converging) mirror equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{27(41)}{41-27}=dfrac{1107}{14}
$$

Evaluate:

$$
d_i=79.071
$$

The image formed is $79.071$ centimeters away from the mirror.

Step 3

3 of 5

Data:

$$
d_o=20
$$

$$
f=27
$$

Step 4

4 of 5

Concave (converging) mirror equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{27(20)}{20-27}=dfrac{540}{-7}
$$

Evaluate:

$$
d_i=-77.143
$$

The image formed is $77.143$ centimeters away from the mirror. The negative sign implies that it is virtual in nature.

Result

5 of 5

Click to see solutions.

Exercise 52

Step 1

1 of 3

Data:

$$
c=3times10^8
$$

$$
n=1.46
$$

Step 2

2 of 3

Formula of index of refraction:

$$
n=dfrac{c}{v}
$$

Rearrange to solve for $v$:

$$
v=dfrac{c}{n}
$$

Substitute values:

$$
v=dfrac{3times10^8}{1.46}
$$

Simplify and evaluate:

$$
v=2.055times10^8
$$

Result

3 of 3

Speed of light is $2.055times10^8$ meters per second in fused quartz.

Exercise 53

Step 1

1 of 3

Data:

$$
c=3times10^8
$$

$$
v=2.04times10^8
$$

Step 2

2 of 3

Formula of index of refraction:

$$
n=dfrac{c}{v}
$$

Substitute values:

$$
n=dfrac{3times10^8}{2.04times10^8}
$$

Simplify and evaluate:

$$
n=dfrac{3}{2.04}=1.4706
$$

Result

3 of 3

Index of refraction of turpentine is 1.4706.

Exercise 54

Step 1

1 of 3

Data:

$$
theta_i=45text{textdegree}
$$

$$
theta_r=30text{textdegree}
$$

Step 2

2 of 3

Formula of index of refraction using Snell’s law:

$$
n=dfrac{sin theta_i}{sin theta_r}
$$

Substitute values:

$$
n=dfrac{sin 45text{textdegree}}{sin 30text{textdegree}}
$$

Evaluate:

$$
n=1.414
$$

This implies that the speed of light in the medium is $dfrac{3times10^8}{1.414}=2.121times10^8$ meters per second.

Result

3 of 3

Speed of light in the medium is $2.121times10^8$ meters per second.

Exercise 55

Step 1

1 of 3

Data:

$$
d_o=23
$$

$$
f=-30
$$

This is a diverging lens and therefore it has a virtual focus and hence the negative value of f.

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{-30(23)}{23-(-30)}=dfrac{-690}{53}
$$

Evaluate:

$$
d_i=-13.02
$$

The image formed is $13.02$ centimeters and virtual in nature indicated by the negative sign of $d_i$.

Result

3 of 3

The image formed is $13.02$ centimeters from the lens and virtual in nature indicated by the negative sign of $d_i$.

Exercise 56

Step 1

1 of 3

Data:

$$
d_o=45
$$

$$
f=34
$$

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{34(45)}{45-34}=dfrac{1530}{11}
$$

Evaluate:

$$
d_i=139.09
$$

The image formed is $139.09$ centimeters from the lens and is real in nature and inverted in attitude.

Result

3 of 3

The image formed is $139.09$ centimeters from the lens and is real in nature and inverted in attitude.

Exercise 57

Step 1

1 of 3

Data:

$$
d_o=34
$$

$$
d_i=21
$$

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{f}=dfrac{1}{d_0}+dfrac{1}{d_i}
$$

Substitute values:

$$
dfrac{1}{f}=dfrac{1}{34}+dfrac{1}{21}
$$

Simplify using least common denominator:

$$
dfrac{1}{f}=dfrac{21+34}{34(21)}=dfrac{55}{714}
$$

Solve for $f$:

$$
f=dfrac{714}{55}
$$

Evaluate:

$$
f=12.982
$$

The focal length of the lens is 12.982 centimeters.

Result

3 of 3

The focal length of the lens is 12.982 centimeters.

Exercise 58

Step 1

1 of 4

Data:

$$
d_o=17
$$

$$
f=24
$$

Step 2

2 of 4

Lens equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{24(17)}{17-24}=dfrac{408}{-7}
$$

Evaluate:

$$
d_i=-58.286
$$

The image formed is $58.286$ centimeters from the lens on the same side of the lens as the object. It is virtual in nature and upright in attitude.

Step 3

3 of 4

b.
Equation of magnification in terms of distance:

$$
M=dfrac{d_i}{d_o}
$$

Substitute values:

$$
M=dfrac{58.286}{17}
$$

Evaluate:

$$
M=3.429
$$

The magnification of the lens is $3.429$

Result

4 of 4

a. $d_i=58.286$ centimeters

b. $M=3.429$

Exercise 59

Step 1

1 of 3

Data:

$$
d_o=13
$$

$$
d_i=-5
$$

The image is on the same side of the lens as the object, implies that this is a virtual image and hence the negative sign for $d_i$.

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{f}=dfrac{1}{d_0}+dfrac{1}{d_i}
$$

Substitute values:

$$
dfrac{1}{f}=dfrac{1}{13}+dfrac{1}{-5}
$$

Simplify using least common denominator:

$$
dfrac{1}{f}=dfrac{-5+13}{13(-5)}=dfrac{8}{-65}
$$

Solve for $f$:

$$
f=-dfrac{65}{8}
$$

Evaluate:

$$
f=-8.125
$$

The focal length of the lens is $8.125$ centimeters. The negative sign of $f$ implies that the focus is itself virtual in nature; a characteristic property of diverging lenses.

Result

3 of 3

The focal length of the lens is 8.125 centimeters.

Exercise 60

Step 1

1 of 3

Data:

$$
d_i=-12
$$

$$
f=-27
$$

This is a diverging lens and therefore it has a virtual focus and hence the negative value of f.

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{d_o}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_o$:

$$
dfrac{1}{d_o}=dfrac{1}{f}-dfrac{1}{d_i}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_o}=dfrac{d_i-f}{fd_i}
$$

Rearrange:

$$
d_o=dfrac{fd_i}{d_i-f}
$$

Substitute values:

$$
d_o=dfrac{-27(-12)}{-12-(-27)}=dfrac{324}{15}
$$

Evaluate:

$$
d_o=21.6
$$

The fruit bowl is $21.6$ centimeters away from the lens.

Result

3 of 3

The fruit bowl is $21.6$ centimeters away from the lens.

Exercise 61

Step 1

1 of 3

Data:

$$
h_o=19
$$

$$
h_i=58
$$

Step 2

2 of 3

Equation of magnification in terms of height:

$$
M=dfrac{h_i}{h_o}
$$

Substitute values:

$$
M=dfrac{58}{19}
$$

Evaluate:

$$
M=3.053
$$

The magnification of the lens is $3.053$

Result

3 of 3

The magnification of the lens is 3.053

Exercise 62

Solution 1

Solution 2

Step 1

1 of 1

Fluorescent lights are the most efficient light source. They produce almost no heat and consume very little electrical energy. They do however, required proper recycling as they contain mercury that needs to be disposed off properly. In current time, LED lights from light emitting diodes is quickly becoming a popular choice because of its low energy consumption and long life.

Step 1

1 of 2

Fluorescent lights are most efficient when it comes to light sources.

They produce almost no heat and not consume much electrical energy, but what they need is a proper recycling since they contain mercury which has to be disposed properly because of its toxic properties.

LED lights are becoming popular as well since they consume low energy and have a long life.

Result

2 of 2

Fluorescent lights and LED lights are most useful in saving energy.

Exercise 63

Solution 1

Solution 2

Step 1

1 of 1

Convex mirrors form smaller, virtual images of objects in front of them. This implies that they have a wide field of view and therefore are installed on sharp turns on roads to visually inform drivers on the turn about the oncoming traffic.

Step 1

1 of 2

Images formed by convex mirrors are smaller in size than the actual object and virtual in nature.

As a result, they give wide field of view which makes them good for installing on sharp turns on roads which visually inform drivers on the turn about incoming vehicles.

Result

2 of 2

Their purpose is to give wide field of view and warn drivers on the turn about the incoming traffic.

Exercise 64

Solution 1

Solution 2

Step 1

1 of 3

$textit{a.}$, Sun is considered to be the original source of energy used by the living organisms since the heat and sunlight are essential for their survival.

Plants use sunlight to produce their own food by photosynthesis, which as a result gives us oxygen which is important for living, also plants are becoming food for many species.

If there was not heat of the sun, temperatures will be extremely low and there will be almost impossible for life to develop .

Step 2

2 of 3

$textit{b.}$, Public notice:

There is nothing better than spending your free time at the pool, bath in the sunlight, enjoying its warmth, or is it?
Behind that delightful sunbath, enemy is hiding and this enemy are ultraviolet rays which are very harmful while in a long contact with our skin. They can tear our cells which can cause several mutations which can later lead to cancer. Protect yourself from the enemy! Don’t spend too much time tanning, use sunscreen creams and lotions which can provide a protective barrier between the skin and harmful radiations!

Result

3 of 3

$textit{a.}$, Because of the sunlight and heat which are essential for life.

$textit{b.}$, People should not spend too much time tanning because of the harmful ultraviolet rays which can cause mutations and therefore cancer. In order to protect themselves people should use sunscreen creams and lotions which can cause permanent barrier between skin and harmful radiation.

Step 1

1 of 3

a.
Sun IS the source of life on this planet earth. The heat from the sun keeps the planet warm enough to sustain life. Its light is essential for photosynthesis in plants. Plants manufacture their own food using the process of photosynthesis. This sustains plant life on earth. These plant also become the source of food for all other living organisms on earth and therefore in a sense, it can be said that the sun is the source of life on planet earth without its warmth and light, life on planet earth will cease to exist.

Step 2

2 of 3

b.
Ultraviolet rays from the sun can be harmful to the human skin and can even lead to skin cancer. Exposure to this can be reduced by staying out of the sun (indoors) during its peak hours and using sunscreen creams and lotions to provide a protective barrier between the skin and these radiations.

Result

3 of 3

Click to see answers.

Exercise 65

Solution 1

Solution 2

Step 1

1 of 3

Diagram which shows how the change in direction is related to the change in speed: Exercise scan

Step 2

2 of 3

There is a reduction in speed when ray of light enters a new medium with different optical density, therefore it is being bent.

Ray which forms angle of incidence with the principal axis is the incident ray which is entering a new medium and being refracted at the angle of refraction.

Result

3 of 3

There is a reduction in speed when ray of light enters a new medium with different optical density, therefore it is being bent.

Step 1

1 of 1

Light bends on entering a medium where it travels with a slower speed because of its reduction in speed. Blue ray shows how the ray would have passed through the other medium if the speed had not changed, while the red ray shows the actual path of the refracted light ray. Exercise scan

Exercise 66

Solution 1

Solution 2

Step 1

1 of 6

Human Eye

Camera

Step 2

2 of 6

Light inlet

Human eye lens

Camera lens

Step 3

3 of 6

Amount of light entering the eye

Controlled by the dilation and reduction of the iris

Controlled by the aperture settings

Step 4

4 of 6

Type of image formed

Real image

Step 5

5 of 6

Location of image

Retina of the eye

Charge coupled device or photographic film

Result

6 of 6

Click to see table.

Step 1

1 of 2

begin{tabular}{|l|l|l|}
hline
& human eye & camera \ hline
light inlet & lens in the eye & camera lens \ hline
which amount will enter the eye/camera & begin{tabular}[c]{@{}l@{}}controlled by dilation\ and reduction of the irisend{tabular} & begin{tabular}[c]{@{}l@{}}controlled by the aperture \ settingsend{tabular} \ hline
type of the image which is formed & real image & real image \ hline
location of the image & retina of the eye & begin{tabular}[c]{@{}l@{}}photographic film or\ charged coupled deviceend{tabular} \ hline
end{tabular}

Result

2 of 2

Light inlet is reduced by lens in the eye and by the camera lens in the case of camera.

Which amount will enter the eye or camera is controlled by dilation and reduction of the iris in the case of human eye and by aperture settings in the case of camera.

Real image is formed in both of the cases.

Image is placed in the retina of the eye and on photographic film or charged coupled device in the case of camera.

Exercise 67

Solution 1

Solution 2

Step 1

1 of 2

Word “radiation” mostly has a negative meaning since it is being associated with harmful part of electromagnetic spectre such as UV rays, gamma rays etc.

Radiation is actually defined as a procedure of heat and light transfer through a vacuum.

Therefore, since radiation represents whole spectre, it does not have to be specifically dangerous, there are part of spectre which are not harmful at all.

Result

2 of 2

Since the radiation is defined as a procedure of heat and light transfer through a vacuum, it does not represent only dangerous part of electromagnetic spectre, it represents whole spectre, and there is radiation which is not harmful.

Step 1

1 of 1

Radiation is usually associated with ultraviolet rays, gamma, alpha, beta rays and nuclear radiations all of which are harmful. Radiation however is defined as the procedure of heat (and light) transfer through a vacuum. This is the way by which heat and light from the sun reach the planet earth. Radiation itself is therefore not harmful although its types maybe.

Exercise 68

Step 1

1 of 5

a. Her vision problem is known as far-sightedness and hyperopia.

Step 2

2 of 5

b. Ray diagram: Exercise scan

Step 3

3 of 5

In hyperopia, the light rays meet behind the retina to form and image and therefore the image formed on the retina is blurred as shown in figure A.

Step 4

4 of 5

c. The optometrist is likely to recommend a convex lens or its modified version; the positive meniscus lens for the correction of this vision problem. Figure B shows how the convex lens converges, the light rays towards the eye lens so that a properly focused image is formed on the retina of the eye instead of behind it.

Result

5 of 5

Click to see answers.

Exercise 69

Solution 1

Solution 2

Step 1

1 of 5

a. Converging lenses usually make real images of objects placed in front of them. These real images are inverted in attitude and on the other side of the lens as the objects. Converging lenses only form virtual and upright images when the object is placed within their focal length.

Step 2

2 of 5

Converging mirrors also known as concave mirrors usually form real images of objects placed in front of them. These real images are inverted in attitude and located in front of the mirror on the same side as the object. Converging mirrors only form virtual images when the object is placed with in their focal length.

Step 3

3 of 5

b. Diverging lenses always make virtual images of objects placed in front of them. These virtual images are upright in attitude, smaller in size than the object and on the same side of the lens as the objects.

Step 4

4 of 5

Diverging mirrors also known as convex mirrors always form virtual images of objects placed in front of them. These virtual images are upright in attitude, smaller in size than the object and located behind the mirror.

Result

5 of 5

Click to see answers.

Step 1

1 of 3

$textit{a.}$, Converging lenses usually make real images of the objects which are placed in front of them.

These images are inverted in altitude and located on the other side of the lens than where the object is.

They will only form virtual and upright image when the object is placed within the focal length.

Converging mirrors usually form real images which are inverted in altitude and located in front of the mirror on the same side as the object.

They only form virtual images when the object is placed within the focal length.

Step 2

2 of 3

$textit{b.}$, Diverging lenses always make virtual images which are upright in altitude, smaller in size than the actual object and on the same side of the lens as the object.

Diverging mirrors always form virtual images which are upright in altitude, smaller in size than the actual object and located behind the mirror.

Result

3 of 3

$textit{a.}$, Converging lenses: real, inverted in altitude, located on the other side of the lens image or virtual, upright image when the object is placed within the focal length.

Converging mirrors: real, inverted in altitude, located on the other side of the lens image or virtual, upright image when the object is placed within the focal length.

$textit{b.}$, Diverging lenses: virtual, upright in altitude, smaller in size, on the same side of the lens image.

Diverging mirrors: virtual, upright in altitude, smaller in size and located behind the mirror image.

Exercise 70

Step 1

1 of 7

If all forms of electromagnetic waves with the exception of visible light disappeared it would make great impact on our life, we will be only able to see and energy from the sun will be partially transfered.

Step 2

2 of 7

If the radio waves did not exist, there will be no communication, most of devices we are using nowdays will not be able to work, there will be no radar navigation etc.

Step 3

3 of 7

If the microwaves did not exist, we will not be able to heat up or cook our food in microwave ovens etc.

Step 4

4 of 7

If infrared light did not exist, we would not be able to use devices for the night vision, television would not work etc.

Step 5

5 of 7

If there was not ultraviolet light tanning, phototherapy, killing bacteria, creating of fluorescent effects, etc. will be impossible.

Step 6

6 of 7

If X-rays did not exist, we would be unable to make X-ray photograph of our bones, teeth etc.

Step 7

7 of 7

If gamma rays did not exist we would not be able to kill cells which are cancerous, also nuclear reactions will be impossible which would really impact the world we live in etc.

Exercise 71

Step 1

1 of 7

Our life would be completely different if we could see all types of the electromagnetic spectrum.

Step 2

2 of 7

$textit{Radio waves}$

If we were able to see radio waves, any device which involves technology will shine so bright, also we would be able to see Sun and all the planets which change magnetic field clearly.

Step 3

3 of 7

$textit{Microwaves}$

If we were able to see microwaves, when we look at the sky we will see light basically shooting at us uniformly, which would not be so pleasant.

Step 4

4 of 7

$textit{Ultraviolet light}$

If we were able to see ultraviolet light, everything around us will seem like covered in some haze with a blueish tone.

Step 5

5 of 7

$textit{Infrared light}$

If we were able to see infrared light, we will be unable to see colours, we would only see difference in temperature of the objects we are observing; ones with greater temperature would appear brighter.

Step 6

6 of 7

$textit{X-rays}$

If we were able to see X-rays, we would basically have an X-ray vision which we have seen only in Sci-Fi movies.

Since X-rays have pretty high energy, we will not be able to sleep, since when we close our eyes, they could penetrate through our skin easily, we will need some metal protection to turn of the “X-ray vision.”

Step 7

7 of 7

$textit{Gamma rays}$

If we were able to see gamma rays, night sky will look completely different, stars would appear as circles with pulsating energy.

Exercise 72

Step 1

1 of 2

1., Application of light in surgery

Until I have learnt what the actually laser was and its properties, it has been fascinating to me how the light can be used to make cuts in skin, or flesh, but now that I know that beam of laser light travels in the straight line, each ray with the same speed and wavelength, and therefore there is a very high energy, I can understand this type of application.

Step 2

2 of 2

2., Application of light in communication.

I have been surprised when I found out that light is being used for communications in the places where there is a high risk that radio signals interfere with waves of machines which are surrounding the communication device.

Exercise 73

Step 1

1 of 3

$textit{a.}$, Mirrors

Exercise scan

Step 2

2 of 3

$textit{b.}$, Lenses

Exercise scan

Result

3 of 3

See the image.

Exercise 74

Step 1

1 of 4

Reflective telescopes are telescopes which use reflection from curved mirrors in order to get image of far-away objects in space.

Step 2

2 of 4

Gregorian telescope.

James Gregory has only made design for reflective telescope which later, Newton has constructed.

Step 3

3 of 4

Newtonian telescope.

This telescope, which was made by Sir Isaac Newton uses concave mirror as the primary one and flat, diagonal mirror as the secondary one.

Step 4

4 of 4

After the Newton, Cassegrain has made some changes to the reflective telescope.

This telescope uses concave mirror as a primary mirror and convex mirror as the secondary one.

Exercise 75

Step 1

1 of 2

textit{(a)}, Medical uses of laser:\

begin{tabular}{|l|l|l|l|l|l|}
hline
multicolumn{6}{|l|}{Laser} \ hline
multicolumn{2}{|l|}{begin{tabular}[c]{@{}l@{}}why are they\ used in medicineend{tabular}} & begin{tabular}[c]{@{}l@{}}ablating tissue\ by vaporisation of \ the tissue surfaceend{tabular} & begin{tabular}[c]{@{}l@{}}internal surgery \ without open woundend{tabular} & begin{tabular}[c]{@{}l@{}}producing a \ precise wound \ while cutting deeplyend{tabular} & begin{tabular}[c]{@{}l@{}}reducing blood \ lossend{tabular} \ hline
multicolumn{2}{|l|}{uses in medicine} & general surgeries & burning tissue & dental procedures & cosmetic procedures \ hline
end{tabular}

Step 2

2 of 2

$textit{(b)}$, Applications of lasers for cosmetic purposes:

Lasers are being used for treating scars, wrinkles, damages on the skin which are caused by sun etc.

Also it is used for tattoos removal, lifting etc.

This burning procedure with laser during which layers of skin are being vaporised, removes wrinkles, scars, stains etc., but cosmetician which is performing this type of surgery has to be very careful since this surgeries can cause permanent damage on the skin.

Exercise 76

Step 1

1 of 2

Galileo Galiei is considered to be the first one who invented telescope.

The telescope that he has been using was a refracting telescope which had convex lens at the objective and concave eyepiece at the end of the long tube.

Step 2

2 of 2

By using the telescope he has discovered sunspots, four largest moons of Jupiter.

Also he has been observing Venus and Moon, and he discovered phases of Venus and valleys and hills on the Moon.

Exercise 77

Step 1

1 of 2

In this solution, I’m going to explain what is MOST’s first telescope and how it is constructed, and what are his research goals.

Step 2

2 of 2

The MOST telescope was created for the study of $textbf{asteroseismology}$(the study of star vibrations) by the Canadian team of the scientist and was launched in space in 2003.
The MOST team was created in 1996., with the main goal to search the age of the universe and to find and search for extrasolar planets.
The telescope was also designed to be light-weighted with weighing only 54 kilograms and was constructed with a highly precise telescope that measures 15 centimeters in diameter.
Their purpose was to research stars while orbiting around the earth and did make success upon it became operational. It discovered that Procyon didn’t have any pulsations at all, and with that founding contradicted opinion older than 20 years, and made the astronomers rethink their model for stars.

Exercise 78

Step 1

1 of 7

$textit{(a)}$,
Compound microscope is microscope which is using combination of lenses to enlarge image of the sample which is being observed.

Step 2

2 of 7

In the 16th century, Dutch maker of lens have designed some devices which were magnifying objects.

Step 3

3 of 7

First scientist who has perfected microscope was Galileo Galilei and scientists who are considered as first makers of compound microscope were Zaccharias Janssen and Hans Lipperhey bye experimenting with different types and sizes of lenses.

Step 4

4 of 7

Later in that same century, Antony van Leewnhoek has improved microscope by grinding and polishing lenses which lead him to the conclusion that specific shape of lens makes enlarged image.

He was the first to notice bacteria, blood cells etc.

Step 5

5 of 7

In 18th century, achromatic lens has been developed by Chester Moore Hall, which have improved microscope even more since before that refraction of light caused trouble.

Step 6

6 of 7

Later in 18th and in 19th century microscopes have been even more improved and August Kohler has created uniform illumination in microscope and, therefore, observed subjects could be photographed.

Ernst Leitz found a way with which the magnifications of multiple lenses could be changed.

Ernst Abbe and Zeiss have developed UV microscope.

Step 7

7 of 7

$textit{(b)}$,
Difference between compound and electron microscope is that in the case of compound one source of illumination is visible light and in the case of electron one is current, also in electron microscope are used electromagnetic lenses and in the case of compound one glass lenses are used.

Exercise 79

Step 1

1 of 3

$textit{(a)}$

Optics has made positive contributions to society.

Mirrors are not just for decoration or checking self-appearance, now they play main role in the construction of many useful devices, such as telescopes, for safety etc.

Lenses are used for eyeglasses without which we will not be able to fix our sight, also for microscopes and telescopes etc.

Optics has helped in developing transfer of information by signals which are travelling through the optic fibres, illumination has been improved by the invention of different kind of bulbs, lasers helped people in many different fields of work etc.

Step 2

2 of 3

$textit{(b)}$

Usage of lasers in medical purposes is interesting application of light.

Laser is beam of light which travels in straight line, rays have same speed and wavelength, therefore their energy is being focused, which gives it a high energy.

That is why lasers can easily and precisely cut through the skin or flesh during surgeries.

Also, laser light is used in cosmetic purposes such as removing scars or tattoos, even wrinkles.

This application is beneficial to the society since our life may depend on those surgeries and laser can help during them.

Step 3

3 of 3

Transferring information through the optic fibres by reflection of light while travelling through it is really useful for the society, since the signals are transferred much faster with the invention of them.

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Page 590: Unit Review

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