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Foundation Physics > Optics > Flashcards

Optics Flashcards

1
Q

What is optics?

A

Interaction of light with matter

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2
Q

What is macroscopic?

A

visible to the naked eye

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3
Q

What are the three examples of the wave nature of light?

A
  • Diffraction
  • Interference
  • Polarisation
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4
Q

Light has a wave nature, and when is this shown?

A

Diffraction

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5
Q

What is Diffraction?

A

Light waves deviate from straight path and “spread out” as they pass by an obstacle or through an opening.

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6
Q

During diffraction how does the opening affect the dispersal of the waves?

A

Smaller the opening – the more the waves spread

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7
Q

How does the length of the wavelength affect the amount of diffraction?

A

The longer the wavelength of the wave, the larger the amount of diffraction

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8
Q

What is interference?

A

in which two waves come together in such a way that they completely cancel each other out to form a resultant wave of greater, lower, or the same amplitude

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9
Q

If you have two or more light, sound, or electromagnetic waves of the same frequency combine to reinforce or cancel each other, what would the amplitude be of the resulting wave?

A

The amplitude of the resulting wave being equal to the sum of the amplitudes of the combining waves

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10
Q

What is destructive interference?

A

A type of interference that occurs at any location along the medium where the two interfering waves have a displacement in the opposite direction. In destructive interference, the crest of one wave meets the trough of another, and the result is a lower total amplitude

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11
Q

How does the Interference pattern (bright & dark regions) work?

A

Each slit acts as a source of the waves, where the waves interfere constructively or destruvively at the screen. A pattern of light and dark fringes are observed on the screen

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12
Q

What colour are the

a) constructive interference
b) destructive interference

A

a) bright

b) dark

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13
Q

What is constructive interference?

A

When two waves meet in such a way that their crests line up together, then it’s called constructive interference. The resulting wave has a higher amplitude.

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14
Q

Name two benefits of diffraction and why?

A

Radio wave diffraction : Mountainous area - Longer wavelength waves diffracted around and between mountains As radio waves undergo diffraction it means that a signal from a transmitter may be received even though it may be “shaded” by a large object between them
X-ray diffraction : used to investigate internal structure of important biological molecules - example, proteins and DNA. In addition, used for compound identification,

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15
Q

What does diffraction depend on?

A

The relationship between the wavelength and the size of the obstacle.

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16
Q

What is polarisation?

A

The process of transforming unpolarized light into polarized light

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17
Q

Is general light polarised or unpolarised?

A

Unpolarised

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18
Q

What is unpolarised light?

A

When the waves produced usually have their oscillation moving at all different angles - all the light waves have different planes of oscillation

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19
Q

What is polarised light?

A

When light passes through a filter and the waves of light are limited to one plane of movement and the rest is absorbed, leaving polarized light to only have one plane of oscillation.

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20
Q

How can light be polarised?

A

Light can become polarised by :
•Reflection
•Refraction
•Scattering

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21
Q

How is light polarised by reflection?

A

It is done by reflecting off the unpolarized light in a nonmetallic reflecting surface. Light bounces off surface – unpolarized and once bounced it becomes partly polarized

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22
Q

The amount of polarisation done by relfection depends on what?

A

It will depend on the angle of incidence of the light and the composition of the material used for the reflecting surface.

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23
Q

If you reflect polarised light which is horizontal will a reflected polarised light appear?

A

Yes

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24
Q

If you reflect polarised light which is vertical, will a reflected polarised light appear?

A

No

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25
Q

Is light reflected off a mirror polarized?

A

No, light reflecting off a mirror or other metallic surfaces are not polarized.

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26
Q

Give an application of polarised light

A

Applications : 3D movies
2 cameras, a short distance apart, photograph original scene therefore, 2 slightly different images projected on screen
Each image linearly polarised in mutually perpendicular directions
3D glasses have perpendicular polarisation axis, so each eye sees a different image associated with different viewing angle from each camera
Brain perceives the compound image as having depth or 3 dimensions.

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27
Q

What is geometrical optics?

A

subset of optics concerning interaction of light with macroscopic material

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28
Q

What is a medium?

A

Materials through which light can travel through

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29
Q

When will light travel in a straight line?

A

Light rays will travel in a straight line if they remain in the same medium

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30
Q

When does the speed/path of light change?

A

As light goes into a different medium

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31
Q

At the boundary between two media, the light ray can change direction by what?

A

Reflection or refraction

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32
Q

What are the three laws of reflection?

A
  • Angle of incidence(θi) = angle of reflection(θr)
  • Angles measured with reference to the normal to the surface
  • Incident and reflected rays and normal all lie in the same plane
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33
Q

What is the difference between reflection on a smooth and a rough surface?

A

On a smooth surface: reflection at a definite angle : Specular reflection
On rough surface : No unique angle of reflection for all rays. Light reflected in all directions

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34
Q

Majority of objects (clothing, plants, people) are visible to us, why?

A

because they reflect light in a diffuse manner manner (scattered at many angles)

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35
Q

At the surface of a transparent media, glass, water, will

a) reflection occur?
b) refraction occur?
c) both

A

both reflection and refraction occur

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36
Q

What is refraction?

A

Deflection from a straight path in passing obliquely from one medium (such as air) into another (such as glass)

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37
Q

The way the light bends after passing through a medium depends on what?

A

Depends on the speed of light in both media

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38
Q

What is the refractive index?

A

The amount by which a medium reduces the speed of light (n) -> n = c/v (v = speed of light) (c = Speed of light in a vacuum: c = 3x108 ms-)

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39
Q

Calculate the speed of light

in diamond, when its refractive index is 2.42 and c =3x108 ms-

A

𝒗 =𝒄/𝒏
=𝟑 × 𝟏𝟏𝟖 𝒎𝒔−𝟏 / 𝟐. 𝟒𝟒
= 𝟏. 𝟐𝟐 × 𝟏𝟏𝟖 m/s

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40
Q

How long does it take light to travel 394cm in glass

of refractive index 1.52 ?

A 
Calculate the speed of light in glass
𝒗 =𝒄/𝒏
𝒗 =𝟑 × 𝟏𝟎𝟖𝒎𝒔−𝟏 / 𝟏. 𝟓𝟓
= 𝟏. 𝟗𝟗 × 𝟏𝟎𝟖 𝒎/𝒔
Calculate the travel time based on 𝑣 =𝑑/𝑡
𝒕 =𝒅/𝒗
𝒕 =𝟑.𝟗𝟗𝟗 /𝟏.𝟗𝟗 × 𝟏𝟎𝟖 𝒎
= 2 × 𝟏𝟎^−𝟖
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41
Q

What is monochromatic light?

A

Light of one colour or frequency or wavelength

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42
Q

When the refraction index of the second medium is greater than the first, is the angle

a) small
b) big

A

a) the angle is small (slower)

43
Q

When the refraction index is smaller, the angle is

a) small
b) big

A

b ) greater (faster)

44
Q

When the refractive index is zero aka perpendicular to surface, is the angle

a) small
b) big
c) none of the above

A

Transmitted ray is not deflected and it goes straight through as it is independent of the materials on either side of the interface. The refractive index of two then, is also zero.

45
Q

What does Snell’s Law/Law of Refraction state?

A

Incident and refracted rays and the normal are all in the same plane
n1sin(a1) = n2sin2(a2)

46
Q

A laser beam is directed upwards from below the surface of a lake at an angle of 35º to the vertical. Determine the angle at which the light emerges into the air. n1(air) =1.0003 and n2 (water) =1.33

A 
n1sin(a1) = n2sin2(a2)
sin(a1) = n2sin(a2)/n1
sin(a1) = 1.33 x sin(35) / 1.0003 = .76 (sin-1)
a1 = 49.7
47
Q

If an object is placed in a denser medium, when viewed from rarer medium what is the apparent depth?

A

appears to be at a lesser depth than its real depth due to refraction of light.

48
Q

What is the equation for real and apparent depth

A 
d' = (n1/n2)d 
d' = apparent depth
d = real depth
49
Q

Give an example of refraction

A

Setting sun appears flattened (top to bottom) because light from lower part of the sun undergoes greater refraction upon passing through denser air (higher refractive index) in lower part of the Earth’s atmosphere.

50
Q

What is the critical angle?

A

Angle of incident for which refracted ray emerges tangent to the surface is called the critical angle
Sin(ac) = n2/n1

51
Q

When does total internal reflection occur?

A

(a dense medium to a less dense medium) If the angle of incidence is greater than the critical angle

52
Q

Determine the critical angle for water and diamond with respect to air

A

ac = sin -1 (n1/n2)
Water : sin-1 (1.0003 / 1.33) = 49 degrees
ac = sin-1 (1.0003/ 2.42) = 24.4

53
Q

In diamond, light enters from any direction, why is this so?

A

Diamond has large refractive index and consequently small critical angle

54
Q

a) What happens to light ray at the glass-air interface in prism, if the angle of incidence is 45 degrees.
Refractive index of glass =1.52
Refractive index of air =1.0003
b) What happens the beam if the prism is immersed in water? Refractive index of water =1.33

A

a) sin-1 = (1.0003/1.52) = 41 degrees
Total internal reflection at glass air interface occurs due to incident angle is 41 degrees
b) sin-1 (1.33/1.52) = 61 degrees > 45
Total internal reflection at glass-water interface does not occur

55
Q

What is rainbow formation caused by?

A

A combination of refraction and reflection. Incoming white light (broad spectrum of wavelength) is separated into its component colours

56
Q

Give an application for Total internal reflection

A

Fibre optic cables used for telecommunications and for diagnostic tools in medicine

57
Q

What is Total Internal Reflection?

A

When light meets the interface between the medium in which it is traveling and a medium of smaller refractive index at an angle of incidence greater than the critical angle, all light being reflected back to the first medium.

58
Q

Light in air is incident on a glass block at an angle of 35 degrees. The sides of the glass block are parallel. At what angle does the light emerge into the air from the lower surface of the glass block?
(take the angles on the sides to be a3 + a2)

A 
glass block has parallel sides, therefore
𝜶𝟑 = 𝜶2
Let n1= refractive index of air
& n2 = refractive index of glass
Using Snell’s Law on both interfaces
Refraction
𝒏𝟏𝐬in(𝟑𝟓)𝒐 = 𝒏𝟐𝐬in(𝜶𝟐)
𝒏𝟐𝐬in(𝜶𝟑) = 𝒏𝟏𝐬in(𝜶𝟒)
𝒏𝟐sin(𝜶𝟐) = 𝒏𝟐𝐬𝐬𝐬(𝜶𝟑)
𝒏𝟏𝐬in𝟑𝟓𝒐 = 𝒏𝟏𝐬in(𝜶𝟒)
𝜶𝟒 = 𝟑𝟓degrees
59
Q

What laws can be used to explain how lenses and mirrors operate

A

laws of refraction and reflection

60
Q

When might we see sunlight concentrated by magnifying glass burning a hole in paper?

A

Sunlight focused by a converging magnifying glass can burn paper. Light rays from the sun are nearly parallel and cross at the focal point of the lens. The more powerful the lens, the closer to the lens the rays will cross.

61
Q

What is the Refractive power (optical power) of the lens

A

is the degree to which a lens, mirror, or other optical system converges or diverges light, measured in diopters (dpt) =1/f ( f is in metres)

62
Q

A farsighted person requires an eyeglass of strength

  1. 5 diopters. What is the focal length of the eyeglass lens?
    (a) 4 cm, (b) 40 cm or (c) 400cm
A

D =2.5m-1
f = 1/D =1/2.5m = 0.4 m
= 40cm.

63
Q

If an object is at a distance greater than the focal length from the lens, what is the image formed?

A

Real (may be projected and displayed on a screen) inverted image formed

64
Q

What is the focal point?

A

The point at which the light rays cross is called the focal point F of the lens

65
Q

What is the focal length?

A

The distance from the center of the lens to its focal point is called focal length f.

66
Q

Suppose you take a magnifying glass out on a sunny day and you find that it concentrates sunlight to a small spot 8.00 cm away from the lens. What are the focal length and power of the lens?

A

The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus,
f = 8.00 cm.
To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. This gives
P = 1 / f
= 1/ 0.0800 m
=12.5 D

67
Q

When an object is at a distance greater than the focal length from the lens, how will the rays enter?

A

Ray 1 : entering lens parallel to optic axis will exit and pass through the focal point
Ray 2 : passing through the focal point will exit the lens and travel parallel to optic axis
Ray 3 : in between 1 + 2 and go straight through centre of lens and will undergo small deviation (displacement (not shown)(thin lens)

68
Q

If an object is at a distance less than the focal

length from the lens, what is the image formed?

A
  • virtual
  • upright
  • magnified.
69
Q

What is a diverging lens and give an example

A

a diverging lens is where light diverges away from the lenses central optic axis and generates a virtual image and has a negative f. (concave)

70
Q

What does the Law of Refractionsay about the paths of light rays

A

They are exactly reversible -This means that the direction of the arrows could be reversed for all of the rays.For example, if a point light source is placed at the focal point of a convex lens, parallel light rays emerge from the other side

71
Q

Why can’t a virtual image always produced by a concave lens can’t be viewed on screen?

A

In optics, a virtual image is an image formed when the outgoing rays from a point on an object always diverge. The image appears to be located at the point of apparent divergence. Because the rays never really converge, a virtual image cannot be projected onto a screen.

72
Q

When will a converging lens produce an upright image?

A

if the object is located in front of the focal point.

73
Q

We know that a concave lens always forms a virtual and diminished image, then we use this lens in myopic eye, then we must be able to see the image is diminished, but why is this not so?

A

Yes, concave lens cannot form an image of a real object, but you have to consider that there is a second lens involved, the eye lens. In the case of myopic eye, where far objects cannot focus, the eye lens can still form a real image of a virtual object formed by the concave lens because the far object (actually virtual object) appears to be closer though its smaller.

74
Q

What is a real image?

A

The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.

75
Q

What is a virtual image?

A

an optical image formed from the apparent divergence of light rays from a point, as opposed to an image formed from their actual divergence.

76
Q

What is the thin lens equation?

A

1/s + 1/s’ = 1/f (1/p + 1/q = 1/f)
s / p = object distance
s’ / q= image distance
f = focal length

77
Q

What does it mean if the object distance p is positive if

A

The object is in front of lens

78
Q

What does it mean if the object distance p is negative

A

The object is behind lens

79
Q

What does it mean if the focal length f is positive?

A

convex lens

80
Q

What does it mean if the focal length f is negative?

A

concave lens

81
Q

What does it mean if the image distance q is positive?

A

image is formed behind the lens (real)

82
Q

What does it mean if the image distance q is negative?

A

image is formed in front of the lens (virtual)

83
Q

What is the object distance?

A

distance in cm between the object and centre of the lens

84
Q

What is the image distance?

A

distance in cm between the centre of the lens and screen

85
Q

What is magnification?

A

Optical magnification is the ratio between the apparent size of an object (or its size in an image) and its true size, and thus it is a dimensionless number.
𝑴 = − 𝒉𝑰/𝒉0 (height image/height object)
𝑴 = − 𝒒/p (-s’/s)

86
Q

If M is Negative, what is the image formed?

A

inverted image

87
Q

If M is positive, what is the image formed?

A

upright image

88
Q

What is a converging lens?

A

where the incoming incident parallel light rays leave the lens bend towards the central optic axis and eventually converge to form a real inverted image. always has a positive f

89
Q

An object 0.5 cm in height is placed 8 cm from a convex lens of focal length 10 cm. Determine the position, magnification, orientation and
height of the image.

A 
a) 1/s + 1/s' = 1/f
1/s' = 1/f-1/s
1/s' = 1/10 - 1/8
-1/40
= -40cm
b) -s'/s 
-40/8 = 5 -> image upright
c) 𝒉𝑰 = 𝑴 × 𝒉𝑶 = 𝟓 × 𝟎. 𝟓𝟓𝟓 = 𝟐. 𝟓cm
90
Q 
A concave spherical mirror has a radius of curvature of 30 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If an answer does not exist, enter DNE. If an answer is infinity, enter INFINITY.)
p = 15 cm (s)
a) image distance    	
b)image orientation    	 
c) magnification
A 
Concave = r -> +30cm
a) 1/q = 2/30 - 1/15 
= 0 -> INFINITY
b) image at infinity
c) Diminished
91
Q 
A concave spherical mirror has a radius of curvature of 30 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If an answer does not exist, enter DNE. If an answer is infinity, enter INFINITY.)
p = 30 (s)
a) image distance    	
b)image orientation    	 
c) magnification
A 
a) 1/q = 2/30 - 1/30
q = 30cm
b) real and inverted (M = negative and image distance is positive)
c) − s'/s
-30/30 = -1
92
Q 
A concave spherical mirror has a radius of curvature of 30 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If an answer does not exist, enter DNE. If an answer is infinity, enter INFINITY.)
p = 45 (s)
a) image distance    	
b)image orientation    	 
c) magnification
A

a) 1/q = 2/30 - 1/45 = 2/45 ( /2 )
22. 5cm
b) real ( q is positive) and inverted (mag is negative)
c) -s’/s = -.5

93
Q

When you are combining lenses, what is the effective focal length (feff) of combination of a number of thin lenses close together?

A

1/f(eff) = 1/f1 + 1/f2 + ..

94
Q

When you are combining lenses, what is the effective refractive power (or strength) (Seff) of combination of
thin lenses close together?

A

S(eff) = S1 + S2…

95
Q

Determine the combined strength of a thin convex lens and a thin concave lens
placed close together if their respective focal lengths are 10cm and -20cm.

A

Strength S, in diopters (D) = 1/f ( f is in metres)
1/f(eff) = s(eff)
s(eff) = 1/.1m - 1/.2m = 5 diopters

96
Q

An object is placed 45 cm from a lens of focal length -25 cm. Determine the position, magnification, and orientation of the image.

A 
a) 1/p + 1/q = 1/f
1/q = 1/f - 1/p
1/q = 1/-25 - 1/45 = -14/225 (225/-14 = -16.1 cm
b) -q/p = .36 = image is upright
c) virtual + upright
97
Q

What is the radius of curvature

A

R = 2f (2/f)

98
Q

An object is positioned 5 cm in front of a concave mirror of focal length 10 cm. Determine the location of its image and its characteristics.

A

a) p = 5 cm
f = 10 cm
1/q = 1/f - 1/p
q = -10cm
b) magnification : - (- q/p) =- -10/5 ) = 2
virtual + located behind mirror + upright

99
Q

What are the two lenses which make up a simple compound microscope

A

Two convex lenses (image formed by objective lens is

inside focal length of eyepiece lens)

100
Q

In a simple refracting telescope, what is the image formed?

A

Virtual image at infinity, Magnified and inverted

101
Q

Effective focal length of the objective in the Hubble telescope is 57.8 m. What focal length eyepiece is required to give a magnification of -> 8.0 x 10^3 ?

A

M = F0/Fe
fe = f0/M
57.8 / 8.0 x 10^3 = 7.23 x 10^-3

102
Q

How is the camera used as a optical instrument?

A

Lens translated to change image distance to adjust for different object distances.
Real, inverted image formed on CCD array

103
Q

Microscope has the following characteristics
• Objective lens focal length = 0.7 cm, eyepiece lens focal length = 4.5 cm.
• Separation between lenses = 21 cm, Object size = 0.01 cm.
• Object to objective lens distance = 0.73 cm.
• Calculate total magnification produced and size of the final image.

A

NO. 1
1/q = 1/f - 1/p
1/q = 1/.7 - 1/.73 = 30/551 -> 551/30 = 18.37
18.37/.73 = -25.16
Magnification negative because image is inverted
𝒉𝑰 = 𝑴 × 𝒉𝑶 = −𝟐𝟐. 𝟑 × 𝟎. 𝟎𝟎𝟎𝟎 = −𝟎. 𝟐33cm
NO.1
Image produced by objective is situated at a distance (21-17) cm = 4 cm from the eyepiece.
1/4.5 - 1/4 = -36cm
q negative: image is on same side of eyepiece as object
Magnification Me = -q/p = -(-36/4) = 9
Size of the final image = 𝒉𝑰’ = 𝑴e × 𝒉1= (9)( -0.233) = -2.1cm
Total magnification MT = M0 x Me = (-23.3) x 9 = - 210

Foundation Physics (8 decks)

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