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Terms in this set (44)
Absorbed in dermis, present all year long, leaves wrinkling.
Associated with skin cancer
Absorbed by skin epidermis = Sunburns
Associated with skin cancer
Refractive index of: Air
Refractive index of: Water
Refractive index of: Plastic CR39
Refractive index of: Crown glass
Refractive index of: Polycarbonate
Parallel then through F'
Toward F' then parallel
Through Optical Axis
What image is formed by Plus lens?
If Object distance is f, 2f, or more, Image is REAL and INVERTED, positive location (right)
If Object distance <f, Image is VIRTUAL and ERECT, negative location (left)
What image is formed by Minus lens?
ALWAYS forms VIRTUAL/ERECT image no matter where object is.
Negative location (left)
Image size: LESS than object
Image orientation: Erect
Location of exit pupil: Within tube
Tube length: Shorter
Weight: Generally lighter
Image orientation: Inverted
Location of exit pupil: Outside of tube
Tube length: Longer
Shape: May be bent*
Weight: Generally heavier
Limits the amount of light ENTERING the system
Can be the real aperture/ objective lens or image of one
Exit Pupil (Ramsden circle)
Limits the amount of light exiting the system
It is the image of objective lens as seen from the eyepiece side of the telescope
Myopes must ___ the length of a Galilean telescope. Magnification must be ____.
Myopes must ___ the length of a Keplerian telescope. Magnification must be ____.
Hyperopes must ___ the length of a Galilean telescope.
Magnification must be ____.
Hyperopes must ___ the length of a Keplerian telescope. Magnification must be ____.
Change in retinal image size due to lens refractive power
Change in retinal image size due to lens shape (front surface power, thickness, and index of refraction)
A lens that has no refractive power but does produce angular magnification
For the retinal image in corrected axial ametropia to be exactly the same size as in the emmetropic eye, the spectacle lenses should be positioned at the anterior focal point of the eye = 16.7 mm (1000/60 D)
How many degrees are in a Prism diopter?
A prism diopter is equivalent to 0.57 degrees, or conversely, 1 degree is equal to 1.75 prism diopters.
The intensity of the light transmitted when unpolarized light is incident on a polarizer is equal to which of the following percentages?
When unpolarized light reaches a polarizer, the intensity of transmitted light with respect the intensity of the incident light is 1/2 or 50%
-SV/MF: For powers up to +/- 6.50, tolerance is +/- 0.13D
-PALs: For powers up to +/- 8.00, tolerance is +/- 0.16D
-Powers higher than this have a tolerance of +/-2% for both SV/MF and PALs
-Tolerances are higher for PALs at all cylinder power ranges
-Tolerance ranges are the same for SV/MF and PALs
-Tolerance ranges are the same for SV/MF and PALs
-Unmounted prism, vertical prism imbalance, and horizontal prism imbalance tolerances are the same for SV/MF and PALs
-Vertical segment height (fitting point height) and vertical segment difference (fitting point difference) tolerances are the same for MF and PALs
-Horizontal segment location (fitting point location) tolerance is less for PALs than MF (1.0 mm vs. 2.5 mm)
Your -10.00 D patient wishes to have the thinnest edges possible for his glasses. Which of the following actions will result in a reduced edge thickness?
For a myope, the goal is to reduce the overall edge thickness. Decreased edge thickness can be achieved by decreasing the eyesize, increasing the refractive index of the lens, or by minimizing the center thickness.
UV Rays and Ophthalmic Materials
Polycarbonate blocks 100% of UV-B and most of UV-A light, does not require the addition of a UV protective coat.
CR-39 blocks 100% of UV-B light but only some of UV-A light.
Glass transmits both UV-A and UV-B light to a much greater degree than CR-39 or polycarbonate.
As the pupil size decreases, the amount of object defocus will decrease, as will the amount of light reaching the retina. Elderly patients often experience a phenomenon called senile miosis in which the size of the pupil is physiologically decreased, affording increased visual acuity secondary to a pinhole effect, along with an increased depth of field. However, the amount of light reaching the retina is diminished, which may cause a paradoxical and offsetting decrease in visual acuity.
A decreased pupil size increases the amount of diffraction experienced. A resulting retinal diffraction pattern caused by a point source appears as alternating light and dark bands of concentric circles in a bull's eye configuration. The measured distance between the center light circle (peak) to the center of the adjacent dark band (trough) is known as Airy's disk. The radius of Airy's disk increases with decreased pupil size. If viewed through a small pupil, two point sources of light will create two retinal diffraction patterns. The two objects must be sufficiently far away from each other to be resolved as two separate objects. The minimum required distance separating the two objects to allow for proper resolution is the radius of Airy's disk (or when the peak of one Airy's disk falls on the trough of the other Airy's disk). Because the radius of Airy's disk increases (as does the size of the diffraction pattern) with decreased pupil size, the distance between two objects must also increase in order for the perception of two point sources to occur.
In order to ensure proper measurements for a pair of progressive spectacles, which of the following should be completed during the fitting process?
•Measure distance PDs monocularly (not binocularly)
°Recommended method is to use a pupillometer (most accurate)
•Fit and fully adjust the desired frame on the patient's face
°Pantoscopic tilt, frame height, vertex distance, face-form wrap, and nose-pad alignment
°If the frame is not pre-adjusted, fitting height measurements may be incorrect
•Ensure that the clear demo lenses are in place in the desired frame
°If lenses are not present, or are darkly tinted, place clear (transparent) tape across the eyewire of the empty frame
•Dispenser should be positioned with his or her eyes at the level of the patient's eyes
°The patient should look at the bridge of the fitter's nose
°Draw a horizontal line on the lens or tape through the center of the pupil for each eye (not the lower eyelid margin as for bifocals)
•Place the frame on the specific manufacturer's centration chart (there is no standard progressive lens cut-out chart)
°Line up the bridge so that it is centered on the diagonally converging central alignment pattern
°Line up the marked horizontal fitting lines so that they are aligned with the horizontal axis of the chart
°Mark the previously measured PD for each eye as a vertical line that crosses the horizontal fitting line
•The fitting height of the lenses should then be measured as the vertical distance between the fitting cross and the level of the inside bevel of the lower eyewire of the frame
•Move the frame over to the lens picture portion of the centration chart so that the cross on the lens matches with the cross on the chart
°Verify that the add power circle fits within the boundaries of the frame
°Verify that one of the large overall circles completely encloses the frame's lens shape
Your patient states that he requires safety glasses for work. According to the American National Standards Institute (ANSI), what is the minimum center thickness allowable for high-impact prescription safety lenses made from polycarbonate?
Seg OC for FT 22, 25, 28
Seg OC for FT 35
Seg OC for FT 45
Seg OC for Executive
Seg OC for Round
Seg OC is centered (divide round by 2)
Which of the following alterations of a plus powered lens will INCREASE the spectacle magnification experienced by a patient?
Spectacle magnification is a difference in the size of the retinal image caused by the use of corrective lenses. The higher the prescription, the greater the magnification (or minification) experienced by a patient. There are several ways to decrease spectacle magnification. For a plus powered lens, increasing the index of refraction, decreasing the base curve, vertex distance or center thickness will reduce magnification. For a minus powered lens, spectacle minification can be lessened by decreasing the vertex distance or the index of refraction, and increasing the center thickness or base curve. Although this sounds backwards, remember that the goal for minus lenses is to reduce minification, or to increase magnification; and increasing the center thickness of the lens or decreasing the index will achieve this goal.
Determine the reverse slab-off amount for the following Rx. Assume the reading level is 10 mm below the distance optical center.
Reverse slab-off is often prescribed to correct a vertical prism imbalance due to anisometropia. Reverse slab-off adds base down prism to the most plus lens or the least minus lens. To solve this problem, calculate the vertical power difference between the lenses. Unless specified, assume the reading level is 10 mm below the distance optical centers. Using Prentice's rule, calculate the reverse slab-off amount. The total vertical power of the right lens is calculated using the F=sphere + cylinder (sine2 theta) equation. For the right lens, the the calculation is -4.00 + -1.00(sine2 45)= -4.50 D in the 90 degree meridian.
The left lens has -1.00 D in the vertical meridian. The difference between the two in the vertical meridian is 3.50 D.
Prism = 3.5 x 1 or 3.5 reverse slab-off.
If a thick lens with a positive front and back surface increases in thickness, what effect will this have on the net equivalent power of the optical system?
If both of the surface powers of a thick less are either negative or positive, increasing the separation between the surfaces will cause the net equivalent power of the optical system to become more negative (or less positive).
The effective power equation is below:
De = D1 + D2 - (t/n) x D1D2
A polycarbonate lens measures 6 mm thick at its nasal edge and 8 mm thick at its temporal edge. The power of the lens is -3.50 D and it measures 45 mm horizontally. Given the lens parameters, calculate the amount of horizontal prismatic effect that is present within the middle of the lens.
Using the formula P=100g(n-1)/d, where P= the power of the prism, g= the difference in thickness between the apex and the base, n= the index of the lens, and d= the distance between the apex and the base, one is able to calculate the power of the prism that is located in the middle of the lens. Input the appropriate values into the equation and solve for P. P= 100 (2) (1.59-1)/45, P=2.62. Be sure to memorize the index of refraction for polycarbonate (1.59), and CR-39 (1.50) because these values may not always be given. Also, be aware of distractors such as extra information that is not needed but is provided in the question, as in this case where the power of the lens was not required to calculate the correct answer.
Frame Adjustment Lens sitting too close or far
If the right lens is in -> move the right temple in (or left temple out)
If the left temple is in -> move the left temple in (or right temple out)
If the right lens is out -> move the right temple out (or left temple in)
If the left lens is out -> move the left temple out (or right temple in)
Frame Adjustment Lens crooked
If the right lens is up -> bend the right temple up (or left temple down)
If the left lens is up -> bend the left temple up (or right temple down)
If the right lens is down -> bend the right temple down (or left temple up)
If the left lens is down -> bend the left temple down (or right temple up)
How to add yoked based down prism to PALs
Multiply add power by 2/3
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