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COT Exam review- Clinical Optics| Questions with 100% Solutions| A Rated $13.99   Add to cart

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COT Exam review- Clinical Optics| Questions with 100% Solutions| A Rated

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The ray of light that enters a transparent medium is termed: a) Incident ray b) Divergent ray c) Emergent ray d) Parallel ray - a) The incident ray is the ray that first strikes and enters a medium. A divergent ray has been refracted outward. The emergent ray is the ray as it exits the medium....

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  • September 7, 2024
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COT Exam review- Clinical Optics|
Questions with 100% Solutions| A
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The ray of light that enters a transparent medium is termed:

a) Incident ray

b) Divergent ray

c) Emergent ray

d) Parallel ray - ✔✔a) The incident ray is the ray that first strikes and enters a medium. A
divergent ray has been refracted outward. The emergent ray is the ray as it exits the medium. A
parallel ray is straight.



If light passes through a lens and the rays are spread apart on exiting, this is known as:

a) a) Index of refraction (IR)

b) Convergence

c) Zero vergence

d) Divergence - ✔✔d) If the light is spread apart by the lens, the rays are said to be divergent.



If light passes through a lens and the rays are bent toward each other on exiting, this is known as:

a) IR

b) Convergence

c) Zero vergence

d) Divergence - ✔✔b) Convergence occurs when light rays are brought together (bent inward) by
the lens.



A comparison of the speed of light in air to the speed of light through a substance is:

a) IR (Snell's Law)

b) Angle of refraction

c) Internal reflection

,d) Optical interference - ✔✔a) The IR of a substance is found by dividing the speed of light in air
by the speed of light through the substance. This is known as Snell's Law.



The denser the substance, the more slowly light passes through it, and:

a) The lower the IR

b) The higher the IR

c) The more transparent it is

d) The more suitable it is for use as an ophthalmic lens - ✔✔b) A substance with the high IR is
more dense, and slows down the light passing through it.



The IR of crown glass is:

a) 0

b) 1.00

c) 1.33

d) 1.50 - ✔✔d) Depending on which reference is consulted, the IR of crown glass is 1.50.
Light traveling through a vacuum is zero, through air is 1.0, and through water is 1.33.



Light travelling through a prism will be bent toward the prism's:

a) Apex

b) Base

c) Center

d) Smallest angle - ✔✔b) Light traveling through a prism is bent toward the prism's base.



The image of an object viewed through a prism:

a) Is real and shifted toward the base

b) Is virtual and shifted toward the base

c) Is real and shifted toward the apex

d) Is virtual and shifted toward the apex - ✔✔d) Because light is bent toward the base, the image
appears to be shifted toward the apex. A real image can be projected on a screen, while a virtual image
cannot.

,A 1.00 diopter prism bends light:

a) 1 centimeter cm at a distance of 1 cm from the prism

b) 1 m at a distance of 1 cm from the prism

c) 1 cm at a distance of 1 m from the prism

d) 1 m at a distance of 1 m from the prism - ✔✔c) A 1.00 diopter prism bends light 1 cm at a
distance of 1m from the prism.



A 2.00 diopter prism displaces an object 1 cm at a distance of:

a) 0.2 m

b) 0.5 m

c) 24 cm

d) 5.0 cm - ✔✔b) The formula for figuring prism displacement is P+ C/D, where P is the prism power,
C is the displacement of the object in cm, and D is the distance from the prism in meters. To plug in this
problem: 2 = 1/D. Do algebra and multiply both sides by D/1, yielding 2D=1. Now divide both sides by 2:
D = 1/2. 1/2 equals 0.5 m.



This displacement of an object 4 cm at a distance of 1 m would require prism of:

a) 0.5 diopters

b) 1.0 diopters

c) 2.0 diopters

d) 5.0 diopters - ✔✔d) Plug in again: P = 5/1. Thus P=5.0 diopters



At a distance of 2 m a 12 diopter prism would displace an object:

a) 6 cm

b) 2.4 m

c) 24 cm

d) 6 m - ✔✔c) Plug in to get: 12 = C/2. Multiply each side by 2/1, and 24= C. The answer is 24 cm.
(Be sure to pay attention to units of measurement! Another correct answer would have been 0.24m.)

, A spherical lens refracts light:

a) Not at all

b) In one direction only

c) Equally in every direction

d) At 90 degrees from its axis - ✔✔c) A spherical lens refracts light equally in very direction. This
is evidenced by the fact that you can shine a light through a plus lens and focus it to a point.



A 1.00 diopter spherical lens focuses light at:

a) 1/2 m

b) 1 m

c) 1 cm

d) 1 yard (yd) - ✔✔b) A 1.00 diopter lens focuses light at 1 m. (Technically, the light entering the
lens must be parallel for this to hold true.)



The range of wavelengths in the visible spectrum is:

a) 100 to 400 nanometers

b) 400 to 800 nanometers

c) 450 to 650 nanometers

d) 400 to 800 meters (m) - ✔✔b) Depending on which reference is consulted, visible light is in
the range of 400 to 800 nanometers. (Be sure to watch the units of measurement given in an
answer. "Meters" in answer d should have indicated that it was incorrect.)



The pointer at which a lens forms an image (whether real or virtual is the:

a) Nodal point

b) Conoid of Sturm

c) Focal length

d) Focal point - ✔✔d) The focal point is the place where the lens focuses incoming light. The focal
point may be real (can be projected onto a screen) or virtual (exists in theory but cannot be projected).



All of the following regarding the optical center of a lens is true except:

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