What is the primary focal point (F 1 )?
The primary focal point is the point along the optical axis at which an object must be placed for parallel rays to emerge from the lens. Thus, the image is at infinity ( Fig. 3-1 ).
What is the secondary focal point (F 2 )?
The secondary focal point is the point along the optical axis at which parallel incoming rays are brought into focus. It is equal to 1/lens power in diopters (D). The object is now at infinity ( Fig. 3-2 ).
Where is the secondary focal point for a myopic eye? A hyperopic eye? An emmetropic eye?
The secondary focal point for a myopic eye is anterior to the retina in the vitreous ( Fig. 3-3 , A ). The object must be moved forward from infinity to allow the light rays to focus on the retina. A hyperopic eye has its secondary focal point posterior to the retina ( Fig. 3-3 , B ). An emmetropic eye focuses light rays from infinity onto the retina.
What is the far point of an eye?
The term far point is used only for the optical system of an eye. It is the point at which an object must be placed along the optical axis for the light rays to be focused on the retina when the eye is not accommodating.
Where is the far point for a myopic eye? A hyperopic eye? An emmetropic eye?
The far point for a myopic eye is between the cornea and infinity. A hyperopic eye has its far point beyond infinity or behind the eye. An emmetropic eye has light rays focused on the retina when the object is at infinity.
How do you determine which lens will correct the refractive error of the eye?
A lens with its focal point coincident with the far point of the eye allows the light rays from infinity to be focused on the retina. The image at the far point of the eye now becomes the object for the eye.
What is the near point of an eye?
The near point is the point at which an object will be in focus on the retina when the eye is fully accommodating. Moving the object closer will cause it to blur.
Myopia can be caused in two ways. What are they?
Refractive myopia is caused by too much refractive power owing to steep corneal curvature or high lens power.
Axial myopia is due to an elongated globe. Every millimeter of axial elongation causes about 3 D of myopia.
The power of a proper corrective lens is altered by switching from a contact lens to a spectacle lens or vice versa. Why?
Moving a minus lens closer to the eye increases effective minus power. Thus, myopes have a weaker minus prescription in their contact lenses than in their glasses. Patients near presbyopia may need reading glasses when using their contacts but can read without a bifocal lens in their glasses (see question [CR] ). Moving a plus lens closer to the eye decreases effective plus power. Thus, hyperopes need a stronger plus prescription for their contact lenses than for their glasses. They may defer bifocals for a while. The same principle applies to patients who slide their glasses down their nose and find that they can read more easily. They are adding plus power. This principle works for both hyperopes and myopes.
What is the amplitude of accommodation?
The total number of diopters that an eye can accommodate.
What is the range of accommodation?
The range of clear vision obtainable with accommodation only. For an emmetrope with 10 D of accommodative amplitude, the range of accommodation is infinity–10 cm.
How does a diopter relate to meters?
A diopter is the reciprocal of the distance in meters.
What is the near point of a 4-D hyperope with an amplitude of accommodation of 8?
The far point is 25 cm (¼ D) behind the cornea. The patient must use 4 D of accommodation to overcome hyperopia and focus the image at infinity on the retina. Thus, he or she has 4 D to accommodate to the near point, which is 25 cm (¼ D) anterior to the cornea. However, when wearing a +4.00 lens, he or she has the full amplitude of accommodation available. The near point is now 12.5 cm (⅛ D).
What is the near point of a 4-D myope with an amplitude of accommodation of 8?
The far point is 25 cm (¼ D) in front of the eye. The patient can accommodate 8 D beyond this point. The near point is 12 D, which is 8.3 cm ( <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='112′>112112
D) in front of the cornea.
When a light ray passes from a medium with a lower refractive index to a medium with a higher refractive index, is it bent toward or away from the normal?
It is bent toward the normal ( Fig. 3-4 ).
What is the critical angle?
The critical angle is the incident angle at which the angle of refraction is 90 degrees from normal. The critical angle occurs only when light passes from a more dense to a less dense medium.
What happens if the critical angle is exceeded?
When the critical angle is exceeded, total internal reflection is the result. The angle of incidence equals the angle of reflection ( Fig. 3-5 ).
Give examples of total internal reflection.
Total internal reflection at the tear–air interface prevents a direct view of the anterior chamber. To overcome this limitation, the critical angle must be increased for the tear–air interface by applying a plastic or glass goniolens to the surface. Total internal reflection also occurs in fiber-optic tubes and indirect ophthalmoscopes.
What is the formula for vergence?
<SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='U+P=V,’>U+P=V,U+P=V,where U is the vergence of light entering the lens, P is the power of the lens (the amount of vergence added to the light by the lens), and V is the vergence of light leaving the lens. All are expressed in diopters. By convention, light rays travel left to right. Plus signs indicate anything to the right of the lens, and minus signs indicate points to the left of the lens.
U + P = V ,
What is the vergence of parallel light rays?
The vergence of parallel light rays is zero. Parallel light rays do not converge (which would be positive) or diverge (which would be negative). Light rays from an object at infinity or going to an image at infinity have zero vergence.
What is the image point if an object lies 25 cm to the left of a +5.00 lens?
Everything must be expressed in diopters: 25 cm is 4 D (1/0.25 m). Because the image is to the left of the lens,
<SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='U=−4D’>?=−4?U=−4D
U = − 4 D
<SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='P=+5D’>P=+5DP=+5D
P = + 5D
<SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='−4+5=1′>−4+5=1−4+5=1
− 4 + 5 = 1
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