Chapter 8: Light and Optics

Question 1

electromagnetic spectrum

Answer 1

Radio waves on one end, (long wavelength, low frequency, low energy) and gamma rays on the other (short wavelength, high frequency, high energy) Between the two extremes we find, in order from lowest energy to highest energy, microwaves, infrared, visible light, ultraviolet, and X-rays.

Question 2

What is the range of Wavelengths corresponding to the visible spectrum of light?

Answer 2

400 nanometers to 700 nanometers.

Question 2

What are the common units for wavelengths?

Answer 2

Millimeters 10 to the -3 meters.

Micrometers 10 to the -6 meters.

Nanometers 10 to the -9 meters.

Angstrom, 10 to the –10 meters

Question 3

What is the speed of light in vacuum?

Answer 3

3 * 10^ 8 m/s

Question 4

Speed of light from frequency and wavelength.

Answer 4

C = f λ

C = speed of light in vacuum.

f = Frequency

Λ = Wavelength

Question 5

rectilinear propagation.

Answer 5

When light travels through a homogeneous medium, it travels in a straight line.

Question 6

Blackbody.

Answer 6

Refers to an I do absorber of all wavelengths of light which would appear completely back if it were at lower temperature than it’s surrounding.

Question 6

What is white?

Answer 6

Light that contains all the colors in equal intensity.

Question 7

Reflection.

Answer 7

Rebounding of incident light waves at the boundary of a medium. According to the law of reflection: θ1 = θ2

Question 8

Normal.

Answer 8

Line dropper, particularly to the boundary of a medium. All angles and objects are measured from the normal, not the surface of the medium.

Question 9

Plain mirrors.

Answer 9

They always create virtual images. They have parallel incident. Light rays remain parallel after reflection from a plain mirror. It causes neither convergence nor divergence of reflected light rays.

Question 10

Spherical mirrors.

Answer 10

It can be concave or convex. They have associated centre of curvature ( C ) and a radius of curvature ( r ). The center of curvature is a point on the optical axis located at distance equal to the radius of the curvature from the vertex of the mirror.

Question 11

Concave mirrors.

Answer 11

It’s like looking into a cave. Concave mirrors are converging mirrors.

Question 11

Focal length (f)

Answer 11

Distance between the focal point and the mirrored. Note that for all spherical mirrors f = r/2

Question 12

Convex mirrors.

Answer 12

They are diverging mirrors.

Question 13

Image distance.

Answer 13

If the image has a positive distance ( i > 0), it is a real image, which implies that the image is in front of the mirror. If the image has a negative distance ( i <0 ), it is virtual and thus located behind the mirror.

Question 14

Optics equation.

Answer 14

1/f = 1/o + 1/i = 2/r

Question 14

Magnification.

Answer 14

m = - i / o

A negative magnification signifies the inverted image, while a positive value signifies the upright image. If m < 1 The image is smaller than the object (reduced). If m >1 The image is larger than the object. (enlarged)

Question 15

What happens if an object is at the focal point of a converging mirror?

Answer 15

The reflected rays will be parallel and therefore the image will be at Infinity. The further away the object, the smaller the image will be.

Question 16

How to draw the rays for a mirror?

Answer 16

1) Ray parallel to the axis, reflects back through the focal point.

2) Ray through focal point, reflects back parallel to the axis.

3) Ray to center of mirror reflects back at the same angle relative to normal.

Draw the following race and find a point where any two intersect. This point of intersection marks the tip of the image. If the rays you draw do not appear to intersect, extend them to the other side of the mirror, creating a virtual image.

Question 17

Sign conventions for mirrors.

Answer 17

The focal length of converging mirrors, (and converging lenses) will always be positive. The focal length of diverging mirrors, (and diverging lens is) will always be negative.

Question 18

Refraction.

Answer 18

Bending of light as it passes from one medium to another and it changes speed.

Question 19

Snell’s Law

Answer 19

When light is in any medium besides a vacuum, its speed is less than c.

n = c / v

c = speed of light in vacuum.

v = speed of light in the medium.

n = Dimensionless quantity called the index of refraction of the medium.

Question 20

Snell’s Law As they passed from one medium to the other.

Answer 20

n1 sin θ1 = n2 sin θ2

When light enters the medium with a higher index of refraction (n2 > n1), It bends towards the normal (sin θ2<sin θ1). When light enters the medium with a lower index of refraction, it bends away from the normal.

Question 21

Critical angle

Answer 21

As the incident angle is increased, the reflected angle also increases and eventually a special incident angle called the critical angle (θc) Is reached for which the refracted angle (θ2) equals 90°. The reflected light ray passes along the interface between the 2 media.

θ c = sin^-1 (n2/n1)

Question 22

Lenses

Answer 22

Lenses refract light while mirrors reflected. Convex lenses are converging and concave lenses are diverging.

Question 22

Total internal reflection.

Answer 22

A phenomenon in which all the light incident on a boundary is reflected back into the original material, resulting with any angle of incidence greater than the critical angle.

Question 23

Lensmaker’s equation.

Answer 23

1 / f = (n – 1) (1/r1 - 1/r2)

Where n is the index of refraction of the lenses material, R1 is the radius of curvature of the first lens surface and R2 is the radius of curvature of the second lens surface.

Question 24

How to draw the ray diagrams for lenses?

Answer 24

1) Ray parallel to axis, refracts through focal point of front face of the lens.

2) Ray through or Towards focal point before reaching lengths, refracts parallel to axis

3) Ray to center of lengths, continues straight through with no refraction.

To find where the image is throughout the following race and find a point where any 2 intersect. This point of intersection marks the tip of the image. If they’re ways you draw do not appear to intersect, extend them to the same side of the lens from which the light came, creating a virtual image.

Question 25

Farsighted and nearsighted.

Answer 25

Converging lenses, (reading glasses) are needed by people who are far sighted. Diverging lenses, (standard glasses) are needed by people who are nearsighted.

Question 25

Convention sign e for lenses.

Answer 25

Upright = real = +m

Inverted = virtual = -m

Question 26

Concave lens.

Answer 26

They are diverging.

Question 26

Convex lens.

Answer 26

They are converging.

Question 27

Relationship between mirrors and lenses.

Answer 27

It is important to realize that concave mirrors and convex lenses are both converging and thus have similar properties. Convex mirrors and concave lenses are both diverging and also have similar properties.

Question 28

Converging (convex) lenses

Answer 28

The radius and focal point length are always positive.

Question 29

Divergent (concave) lenses.

Answer 29

The radius and focal point length are always negative.

Question 30

If the image is in the opposite side of lines from the light source, are they real or virtual?

Answer 30

They are real and the image sign is positive.

Question 31

If the images are on the same side of lenses as light source, is it real or virtual?

Answer 31

It is virtual and the image sign is negative.

Question 32

Are concave mirrors converging or diverging?

Answer 32

It is converging.

Question 33

Are convex mirrors converging or diverging?

Answer 33

It is diverging.

Question 34

Are convex lens is converging or diverging?

Answer 34

It is converging.

Question 34

Are concave lenses converging or diverging?

Answer 34

It is diverging.

Question 35

Nearsighted (Myopia)

Answer 35

people can see near objects clear. They need diverging lenses.

Question 35

Power (P)

Answer 35

P = 1 / f

The unit is diopters.

Question 36

Farsighted (Hyperopia)

Answer 36

people can see distance objects and they need converging lenses.

Question 37

Multiple lens system.

Answer 37

These systems behave as a single lens with equivalent focal length:

1/f = 1/f1 + 1/f2 + 1/f3 + … + 1/fn

P = P1 + P2 + P3 + … Pn

M = m1 x m2 x m3 x … x mn

Question 38

Positions of dark fringes in slit-lens set up.

Answer 38

a sin θ = n λ

a = width of the slit

θ = Angle between the line draw from the center of the lens to the dark fringe and the axis of the lens.

n = Enter your indicating the number of the fringe.

Λ = Wavelength of the incident wave.

Question 38

Positions of dark fringes in double slit setup.

Answer 38

d sin θ = (n + ½) λ

d = Distance between the two slits.

Λ = Wavelength of the incident wave.

θ = Angle between the line drawn from the midpoint between the two slits to the dark fringe of the normal.

n = Integer indicating the number of fringe.

Question 39

Diffraction gratings.

Answer 39

Consist of multiple slates arranged in patterns. It creates colorful patterns similar to prism, as the different wavelengths interfere in characteristic patterns, for example CD’s or DVD’s.

Question 40

Plane-polarized light.

Answer 40

All of the light rays have electric fields with parallel orientation. It is created by passing unpolarized light through a polarizer.

Question 41

order of electromagnetic spectrum

Answer 41

from increased wavelenght to small wavelenght:

ratio waves -> microwave -> IR -> visible light -> UV -> x- ray -> gamma rays

Question 42

relationship between wavelengh, frequency and energy

Answer 42

higher frequency = higher energy = small wavelenght

lower frequency = lower energy = big wavelenght