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KMK > Optics > Flashcards

Optics Flashcards

1
Q

Aperture stop

A

controls the amt of light passing through the optic system when viewing an object
- first lens in the telescope (objective lens)

Ex. pupils

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2
Q

Entrance pupil

A

image of aperture stop formed by the lens in FRONT of it
- if no lenses, aperture stop is the entrance pupil

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3
Q

Exit pupil

A

image of aperture stop formed by lenses BEHIND it

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4
Q

Where’s the exit pupil on a Galilean and keplerian telescope

A
  • Galilean inside the telescope
  • Keplerian outside the telescope
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5
Q

Field of view

A
  • increased with minus lens
  • decreased with plus lens
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6
Q

Depth of focus

A

interval surrounding the retina in which the eye sees an object as in focus

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7
Q

Depth of field

A

surrounds fixation plan in which object can reside and be in focus. Increases with decreased pupil size and decreased focal lengths (myope)

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8
Q

Pantoscopic tilt

A

rotating lens around 180 degree
meridian

induces minus cyl with axis 180 for a - lens and axis 090 for a + lens

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9
Q

Faceform tilt

A

rotating lens around 90 degree meridian induces minus cyl with axis 090 for - lens and axis 180 for + lens

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10
Q

Circle of least confusion

A

point of best focus for a lens
- located dioptrically halfway between the 2 line images that are formed in the principle meridians

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11
Q

Interval of sturm

A

the distance btw 2 foci of the principle meridian

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12
Q

Telescope Know this chart

Galilean/Keplerian
type of lens
exit pupil
FOV
Mag

A

Galilean
Lens: object (+) ocular(-)
exit pupil: inside
FOV: small
Mag: up to 4x

Keplerian
Lens: objective (+) ocular (+)
exit pupil: outside
FOV: Large
Mag: Up to 10x

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13
Q

What causes spherical aberrations?

A

spherical lens (both plus and minus)
- light located further away from the center bends more

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14
Q

How would you minimize spherical aberration

A
  • prescribe an aspheric lens for higher powered Rx
  • decr aperture/pupil
  • AVOID biconvex lenses
  • plus powered lens = steeper centrally, flatter peripherally
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15
Q

What is the power of the planoconvex thin lens whose curved surface has a radius of 3 cm?

A

+16.67

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16
Q

Nodal point

A

light passes undeviated
- located at the center of curvature

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17
Q

Consider a very simple eye model that consists of only a single spherical refracting interface (cornea) and a screen (retina). Assume that the radius of curvature of the SSRI is 6mm, and the axial length of the eye is 23mm. If an object, 35cm tall is placed 1m in front of the eye, how large is the retinal image?

A

5.9mm

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18
Q

A pt looks at an object embedded in a glass 30cm below the flat surface. What is the apparent depth of an object?

A

20cm

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19
Q

Snells law

A

relationship btw angle of refraction and angle of incidence

20
Q

Total internal reflection

A

light > critical angle will be reflected internally
- only possible if n2 < n1
- ex. gonio

21
Q

What is the power of a thin lens (n2 = 1.5) in air whose front radius of curvature is 3 cm (with a convex surface) and whose back radius of curvature is 2cm (with a concave surface)

A

-8.33D

22
Q

A real object is 35cm from a +3.00 thin lens placed in contact with a +1.00 thin lens. Find the conjugate image and focal lengths

A

f’ = +25cm and f= -25cm

image is inverted, magnified, real

23
Q

When a stand magnifier is used to view reading material, light leaving the magnifier has a vergence of -3.00D. That is, an img of the material is located 33cm on the opposite side of the magnifier from the pt (33 cm to the left of the magnifier). What add is required at the cornea for this pt to use the magnifier, assuming the pt is 40cm from the magnifier?

A

+1.36
- use down vergence equation

24
Q

A pts ametropia is perfectly corrected by cls that are +10.00D. What power would be required at a vertex distance of 12mm?

A

+8.93D

25
Q

A crown glass spec lens (n=1.5) has a front surface power of +15.00D and a back surface power of -5.00D. the central thickness of the lens is 2mm. What is the back vertex of this lens?

A

+10.30D

26
Q

When does the nodal point coincide with the principal points?

A

only if the nodal points are the same on BOTH sides then the principal planes are coincident

27
Q

Field stop

A

limits the FOV or limits the size of the object that can be imaged by the system

28
Q

In most telescopes used for distance viewing, the objective lens is the aperture stop. What is the exit pupil? What is the entrance pupil? Where is the exit pupil located for a Galilean telescope? A Keplerian Telescope?

A

The objective lens is the aperture stop. There are no lenses on the object side of the objective lens, so it is also the entrance pupil. The exit pupil is the image of the objective lens through the ocular lens. For a Keplerian telescope (with a converging ocular lens). the exit pupil is located outside of the telescope (closer to the pts eye), while the Galilean telescope (with diverging ocular lens), the exit pupil is located inside of the telescope

29
Q

Entrance port

A

image of the field stop formed by the lens in front of it

30
Q

Exit port

A

Image of the field stop formed by the lens behind it

31
Q

short focal length

A

larger depth of field

32
Q

increase aperture size

A

decr depth of field and decr depth of focus

33
Q

How will the following lens affect the field of view?

minus
plus

A

minus = incr FOV
plus = decr FOV

34
Q

Consider a lens whose power is given by +5.00+2.00x180. Find the circle of least confusion. What is the interval of sturm?

A

IOS = 6cm
COLC = 17cm (or use spherical equivalent to find COLC)

35
Q

A lens prescription is given by +2.00-3.00x180. Draw the power cross. Which meridian has the most plus power? If you place this lens over a fogged emmetropic eye and show the pt a clock dial, which line will appear sharpest?

A
  • the horizontal meridian will have the most plus power.
  • 3-9 o’ clock will appear the sharpest
36
Q

What is the (approximate) power in the vertical meridian of the following lens: +5.00-2.00x135

A

+4.00

37
Q

Consider a real object placed 40cm in front of a concave mirror with a radius of curvature of 18cm. Find the conjugate img and magnification.

A

L’ = +8.61
M = -0.29

38
Q

We are using +25.00 standard lens. To achieve appropriate focus, we must move target 2.0mm forward

A

Fv = +1.25

39
Q

When using a lensometer, the sphere power is found to be -3.00D and the axis is 125. The power wheel is then rotated in the minus direction until reaching a power of -4.00D, at which point the cyl lines becomes clear. What is the power of the lens?

A

-3.00-1.00x125

40
Q

A mire is 8cm from the cornea and 5.5 cm tall. The reflected img is measured to be 2.3 mm. What is the radius of curvature for cornea? What would the read out of the know keratometer be?

A

r = -7.0mm
K = 48.21

41
Q

You use a lens clock calibrated with n=1.5. You want to measure the power of a lens made from material n=1.7. The power read out of the lens clock is +6.00D. What is the actual power of the lens?

A

+8.40D

42
Q

A cls has a BC measured to be +42.00 DK. What is the radius of curvature of this BC in mm?

A

8.04mm

43
Q

If seg appers too high on the pt how would you adjust it?

A
  • incr pantoscopic tilt
  • Decr vertex distance
  • spread pads
  • move pads up
  • stretch the bridge
44
Q

IF seg appears too low to the pt

A
  • Narrow the pads
  • move mads down
  • increase vertex distance
  • Reduce pantoscopic tilt
  • shrink bridge
45
Q

The front vertex power through the distance portion of a lens is measured to be +3.45-2.00x090. The front vertex power through the nea portion of the lens is measured to be 5.45-2.00x090.

A

The add power is simply the difference in sphere powers, which in this example is 5.45-3.45 or 2.00D

KMK (18 decks)

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