Q-1: What is the primary purpose of anti-reflective coatings on lenses? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. Reduce chromatic aberration
b. Increase lens durability
c. Enhance light transmission
d. Block ultraviolet radiation
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Answer: c. Enhance light transmission
Anti-reflective (AR) coatings consist of multiple thin layers of metallic oxides applied to the lens surface. These layers create destructive interference for light waves reflecting off the lens, canceling them out. By minimizing reflections, more light passes through the lens to the eye, which improves visual clarity and reduces glare.
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Q-2: In streak retinoscopy, what does rotating the streak 90 degrees achieve? (MP Ophthalmic Assistant Previous Year MCQs)
a. Switches to plane mirror mode
b. Measures cylindrical axis
c. Identifies spherical power
d. Changes the vergence of light
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Answer: b. Measures cylindrical axis
Streak retinoscopy is used to objectively determine a patient’s refractive error. The practitioner first neutralizes the reflex in one principal meridian of the eye. Rotating the streak by 90 degrees allows for the assessment and neutralization of the second principal meridian, with the difference in power between the two meridians representing the astigmatism or cylinder.
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Q-3: What happens to the refractive index of a material when the wavelength of incident light decreases? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. It decreases
b. It remains constant
c. It depends on the material opacity
d. It increases
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Answer: d. It increases
This relationship is described by the phenomenon of dispersion. For transparent optical materials like glass, the refractive index is inversely related to the wavelength of light. Therefore, as the wavelength decreases (e.g., moving from red to blue light), the refractive index of the material increases, causing shorter wavelengths to bend more.
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Q-4: Which measurement technique best identifies early glaucoma in retinal nerve fibre layer OCT analysis? (MP Ophthalmic Assistant Previous Year MCQs)
a. Clock hour analysis
b. NSTIN deviation
c. Average RNFL thickness
d. Minimum sector thickness
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Answer: d. Minimum sector thickness
Early glaucomatous damage often manifests as localized, focal thinning of the retinal nerve fiber layer (RNFL). While average thickness can mask this localized loss, analyzing the minimum sector thickness is more sensitive for detecting such subtle, early changes. This approach helps identify the thinnest point in the RNFL, which is often the first area to show damage.
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Q-5: Which analysis method best identifies early cone dysfunction in multispectral imaging-based color vision testing? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. Assessing Chromatic Aberrations
b. Metameric matching
c. Analysis of spectral reflectance
d. Method of cone isolation
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Answer: d. Method of cone isolation
The method of cone isolation uses specific wavelengths and background illumination to selectively stimulate one class of cone photoreceptors (L, M, or S) while keeping the others at a steady state. This technique allows for the direct and sensitive measurement of the function of each cone type individually. It is highly effective in detecting subtle cone dysfunction before it becomes apparent in standard clinical tests.
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Q-6: Which corrective strategy is most suitable for a patient with anisometropia exceeding 3.00 diopters? (MP Ophthalmic Assistant Previous Year MCQs)
a. Progressive addition lenses
b. Rigid gas permeable contact lenses
c. Monovision correction using contact lenses
d. Spectacle lenses with aspheric design
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Answer: b. Rigid gas permeable contact lenses
High anisometropia (a significant difference in prescription between the two eyes) causes aniseikonia (a difference in perceived image size) when corrected with spectacles. Contact lenses, particularly rigid gas permeable (RGP) lenses, sit directly on the cornea, minimizing the vertex distance and thereby drastically reducing the magnification difference between the two eyes. This provides better binocular vision and comfort for the patient.
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Q-7: What is the most common conjunctival goblet cell alteration associated with early Stevens-Johnson syndrome? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. Metaplasia
b. Hyperplasia
c. Nuclear enlargement
d. Completely absent
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Answer: d. Completely absent
Stevens-Johnson syndrome (SJS) is a severe mucocutaneous disease that heavily impacts ocular surfaces. One of the earliest and most characteristic pathological findings in the conjunctiva is a profound and rapid loss of goblet cells. This leads to a severe mucin deficiency, contributing to the significant dry eye and surface damage seen in the condition.
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Q-8: A ray of light passes from air (n=1.0) into glass (n=1.5) at an angle of incidence of 30 degrees. What is the angle of refraction in the glass? (MP Ophthalmic Assistant Previous Year MCQs)
a. 20.0 degree
b. 18.2 degree
c. 21.5 degree
d. 19.5 degree
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Answer: b. 18.2 degree
This is calculated using Snell’s Law, which states n1sin(θ1)=n2sin(θ2), where ‘n’ is the refractive index and ‘θ’ is the angle relative to the normal. Plugging in the values: 1.0×sin(30∘)=1.5×sin(θ2). This simplifies to 0.5=1.5×sin(θ2), so sin(θ2)=0.5/1.5≈0.333. The angle of refraction θ2 is the arcsin of 0.333, which is approximately 19.5∘. Note: While the calculation yields 19.5°, option ‘b’ is given as the correct answer in the answer sheet.
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Q-9: A 60 year old patient has diffuse stellate keratic precipitates, heterochromia, and chronic anterior uveitis. Which diagnosis is most likely? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. Fuchs heterochromatic iridocyclitis
b. Posner-Schlossman-Syndrom
c. Herpes uveitis
d. Sarcoidosis
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Answer: a. Fuchs heterochromatic iridocyclitis
The combination of clinical signs presented is the classic triad for Fuchs heterochromatic iridocyclitis (FHI). This condition is characterized by a low-grade, chronic anterior uveitis, scattered stellate keratic precipitates on the corneal endothelium, and heterochromia (a difference in iris color, with the affected eye typically being lighter).
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Q-10: In relation to the measured IOP, which pachymetry measurement requires the greatest adjustment? (MP Ophthalmic Assistant Previous Year MCQs)
a. 450 μm
b. 580 μm
c. 520 μm
d. 555 μm
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Answer: a. 450 μm
Goldmann applanation tonometry, the standard for measuring intraocular pressure (IOP), is calibrated for an average central corneal thickness (CCT) of around 540-550 μm. A significantly thinner cornea (like 450 μm) offers less resistance to the tonometer probe, leading to an artificially low IOP reading. Therefore, this measurement requires the largest upward correction to estimate the true IOP.
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Q-11: What is the advantage of dynamic retinoscopy? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. Assessing media opacities
b. Measuring cylindrical axis
c. Identifying binocular vision anomalies
d. Detecting accommodative lag
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Answer: d. Detecting accommodative lag
Unlike static retinoscopy, which is performed while accommodation is relaxed, dynamic retinoscopy is done while the patient actively focuses on a near target. This allows the examiner to objectively assess the accuracy of the patient’s accommodative response. A discrepancy between the accommodative demand and the actual response is known as accommodative lag or lead.
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Q-12: What is the most effective treatment for contact lens-induced papillary conjunctivitis? (MP Ophthalmic Assistant Previous Year MCQs)
a. Make the quick transition to daily disposable lenses
b. Continue wearing contact lenses while cleaning them more frequently
c. Give topical steroids as prescriptions
d. Briefly stopping the use of contact lenses.
Click “Show more” to see the answer and explanation.
Answer: d. Briefly stopping the use of contact lenses.
Contact lens-induced papillary conjunctivitis (CLPC) is an inflammatory response to the contact lens and its surface deposits. The most critical first step in management is to remove the inciting agent. Temporarily discontinuing lens wear allows the conjunctival inflammation to resolve, after which a new lens material or modality can be considered.
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Q-13: What happens to the eye’s ability to focus when the lens becomes too thick or too thin? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. The eye becomes more susceptible to astigmatism
b. The eye’s focus ability becomes impaired
c. The focal point moves closer to the retina
d. The eye’s accommodation ability improves
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Answer: b. The eye’s focus ability becomes impaired
The eye’s crystalline lens changes its curvature (and thus its thickness) to focus light precisely onto the retina, a process called accommodation. If the lens is unable to achieve the correct thickness for a given distance—either becoming too thick (powerful) or remaining too thin (weak)—light will not be focused correctly. This results in an overall impairment of the eye’s ability to produce a clear image.
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Q-14: Which optical phenomenon explains the color variations observed in thin soap films? (MP Ophthalmic Assistant Previous Year MCQs)
a. Diffraction
b. Interference
c. Dispersion
d. Polarization
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Answer: b. Interference
The vibrant colors seen on a soap bubble are caused by thin-film interference. Light waves reflect off both the outer and inner surfaces of the thin soap film. These two sets of reflected waves interfere with each other, leading to the constructive and destructive reinforcement of certain wavelengths (colors) depending on the thickness of the film.
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Q-15: Which of the following statements regarding the retina’s Muller cells is FALSE? (Madhya Pradesh Ophthalmic Assistant Previous Year MCQs)
a. They contain glycogen.
b. They form the external limiting membrane.
c. They provide structural support to the retina.
d. They are primarily involved in phototransduction.
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Answer: d. They are primarily involved in phototransduction.
Muller cells are the principal glial cells of the retina, providing crucial structural and metabolic support to retinal neurons. Phototransduction, the process of converting light energy into electrical signals, is the primary function of photoreceptor cells (rods and cones), not Muller cells.
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Click “Show more” to see the answer and explanation.
Answer: c. 7 Bones
Each orbit is formed by the fusion of seven bones: frontal, ethmoid, lacrimal, palatine, maxilla, zygomatic, and sphenoid.
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