Board Certified Ophthalmologist Optics and Refraction — Questions and Answers
Question 1: An object is placed 50 cm in front of a thin lens. What is the vergence of light from the object arriving at the lens?
- +0.50 D
- +2.00 D
- -2.00 D (Correct answer)
- -0.50 D
Correct answer: -2.00 D
Vergence equals 1 divided by distance in meters (1/0.5 = 2), and light diverging from a real object has negative vergence.
Question 2: According to Knapp's rule, where should a spectacle lens sit so that an eye with axial ametropia has a retinal image the same size as an emmetropic eye's?
- At the corneal plane
- At the nodal point of the eye
- At the eye's center of rotation
- At the anterior focal point of the eye (Correct answer)
Correct answer: At the anterior focal point of the eye
A correcting lens at the anterior focal point (about 15-17 mm in front of the cornea) gives an image the same size as in an emmetropic eye when the ametropia is axial.
Question 3: A patient looks through a -4.00 D lens at a point 5 mm from its optical center. How much prism is induced?
- 20 prism diopters
- 2 prism diopters (Correct answer)
- 1.25 prism diopters
- 0.8 prism diopters
Correct answer: 2 prism diopters
Prentice's rule gives prism = decentration in cm x power in diopters, so 0.5 x 4 = 2 prism diopters.
Question 4: When using a Jackson cross cylinder to refine the cylinder axis, how should the cross cylinder be placed?
- Its handle is perpendicular to the trial cylinder axis
- Its plus axis is aligned with the trial cylinder axis
- Its principal meridians straddle the trial cylinder axis at 45 degrees (Correct answer)
- Its minus axis is aligned with the trial cylinder axis
Correct answer: Its principal meridians straddle the trial cylinder axis at 45 degrees
For axis refinement, the handle lines up with the cylinder axis, which puts the cross cylinder's principal meridians 45 degrees on either side of it.
Question 5: Light passes from air into the cornea at an oblique angle. What happens to the ray?
- It goes straight through without bending
- It bends toward the normal (Correct answer)
- It bends away from the normal
- It is totally internally reflected
Correct answer: It bends toward the normal
By Snell's law, light entering a medium with a higher refractive index slows down and bends toward the normal.
Question 6: Using Hofstetter's average formula, what is the expected amplitude of accommodation in a 40-year-old?
- 8.5 D
- 2.5 D
- 6.5 D (Correct answer)
- 4.0 D
Correct answer: 6.5 D
Hofstetter's average amplitude is 18.5 - 0.3 x age, so 18.5 - 12 = 6.5 D.
Question 7: During the duochrome (red-green) test, the patient says the letters on the red side are clearer. What should the examiner do?
- Add plus power
- Change the cylinder axis
- Add more cylinder
- Add minus power (Correct answer)
Correct answer: Add minus power
If red is clearer, the eye is relatively myopic (focus is in front of the retina), so minus is added. This is the 'RAM-GAP' rule: red add minus, green add plus.
Question 8: What is the spherical equivalent of +2.00 -3.00 x 090?
- -0.50 D
- +0.50 D (Correct answer)
- +3.50 D
- -1.00 D
Correct answer: +0.50 D
Spherical equivalent is the sphere plus half the cylinder: +2.00 + (-1.50) = +0.50 D.
Question 9: What is the correct transposition of -1.00 +2.00 x 180?
- +1.00 -2.00 x 180
- -3.00 -2.00 x 090
- +1.00 -2.00 x 090 (Correct answer)
- +3.00 -2.00 x 090
Correct answer: +1.00 -2.00 x 090
To transpose, add the cylinder to the sphere, reverse the cylinder's sign, and rotate the axis 90 degrees.
Question 10: At a 67 cm working distance, a retinoscopist neutralizes the reflex with +3.00 D. What is the patient's net refractive error?
- +3.00 D
- +2.00 D
- +4.50 D
- +1.50 D (Correct answer)
Correct answer: +1.50 D
A 67 cm working distance adds about 1.50 D, which is subtracted from the gross finding: +3.00 - 1.50 = +1.50 D.
Question 11: During retinoscopy, the examiner sees 'against' motion. What does this mean?
- The far point is behind the examiner; add plus
- The patient's far point is between the examiner and the patient; add minus (Correct answer)
- The patient is hyperopic; add plus
- The patient is at neutrality
Correct answer: The patient's far point is between the examiner and the patient; add minus
Against motion means the far point lies in front of the peephole, so minus lens power is needed to reach neutrality.
Question 12: About what pinhole diameter gives the best acuity when screening for uncorrected refractive error?
- About 6.0 mm
- About 0.3 mm
- About 4.0 mm
- About 1.2 mm (Correct answer)
Correct answer: About 1.2 mm
A pinhole of about 1.2 mm balances a smaller blur circle against increasing diffraction and reduced light.
Question 13: What happens when a high-plus spectacle lens is moved farther from the eye?
- It becomes a minus lens
- Its effective power does not change
- Its effective power at the cornea increases (Correct answer)
- Its effective power at the cornea decreases
Correct answer: Its effective power at the cornea increases
Moving a plus lens away from the eye increases its effective power, which is why aphakic patients slide glasses down their nose to read.
Question 14: What is the conventional magnification of a +20.00 D simple hand magnifier?
- 20x
- 10x
- 5x (Correct answer)
- 2x
Correct answer: 5x
Conventional magnification of a simple magnifier is D/4, so 20/4 = 5x.
Question 15: Why can't the anterior chamber angle be seen without a gonioscopy lens?
- The iris blocks all light coming from the angle
- The sclera absorbs light coming from the angle
- Corneal spherical aberration scatters the light
- Light from the angle is totally internally reflected at the tear-air interface (Correct answer)
Correct answer: Light from the angle is totally internally reflected at the tear-air interface
Light from the angle hits the tear-air interface at more than the critical angle (about 46 degrees), so it is totally internally reflected. A goniolens removes that interface.
Question 16: Where is the circle of least confusion in an astigmatic conoid of Sturm?
- At the dioptric midpoint between the two focal lines (Correct answer)
- At the anterior focal line
- At the posterior focal line
- Always on the retina
Correct answer: At the dioptric midpoint between the two focal lines
The circle of least confusion is at the dioptric midpoint of the interval of Sturm, which corresponds to the spherical equivalent.
Question 17: Which refraction represents mixed astigmatism?
- +1.00 -2.00 x 180 (Correct answer)
- -2.00 +2.00 x 090
- -1.00 -1.00 x 090
- +2.00 -1.00 x 090
Correct answer: +1.00 -2.00 x 180
In +1.00 -2.00 x 180, one meridian is +1.00 and the other is -1.00, so one focal line is in front of the retina and the other is behind it.
Question 18: A myope wears -10.00 D spectacles at a 12 mm vertex distance. About what contact lens power is needed?
- -8.00 D
- -9.00 D (Correct answer)
- -11.25 D
- -10.00 D
Correct answer: -9.00 D
Effective power = F/(1 - dF) = -10/(1 + 0.12), which is about -8.93 D, so roughly -9.00 D.
Question 19: A patient looks at an object through a prism. Which way does the image appear to move?
- Perpendicular to the base-apex line
- Toward the apex of the prism (Correct answer)
- It does not move
- Toward the base of the prism
Correct answer: Toward the apex of the prism
A prism bends light toward its base, so the image appears shifted toward the apex.
Question 20: A Galilean telescope uses a +5.00 D objective and a -20.00 D eyepiece. What is its magnification?
- 0.25x, upright image
- 4x, inverted image
- 25x, upright image
- 4x, upright image (Correct answer)
Correct answer: 4x, upright image
Telescope magnification is -Feyepiece/Fobjective = 20/5 = 4x, and the minus eyepiece of a Galilean design gives an upright image.
Question 21: With the SRK formula, about how large an IOL power error results from a 1 mm error in axial length?
- About 2.5 D (Correct answer)
- About 0.9 D
- About 0.25 D
- About 6 D
Correct answer: About 2.5 D
In P = A - 2.5L - 0.9K, the axial length coefficient is 2.5, so a 1 mm error changes the IOL power by about 2.5 D.
