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# Objective structured practical question (ospe)

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1) I am Dr Md Anisur Rahman Anjum passed MBBS from Dhaka Medical College in 1987. Diploma in Ophthalmology (DO) from the then IPGM&R (now it is Bangabandhu Sheikh Mujib Medical University BSMMU) in 1993. Felllowship in Ophthalmology FCPS from Bangladesh College of Physician and surgeon in 1997. I am now working as associate professor in General Ophthalmology in National Institute of Ophthalmology Dhaka Bangladesh which is the tertiary centre in eye care in Bangladesh.
These OSPE are dedicated to the postgraduate student who are decided to builds their carrier in ophthalmology. I hope that they will be benefitted if they solve these OSPE

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### Objective structured practical question (ospe)

1. 1. Objective Structured Practical Question (OSPE) Topic: Optics According to the course curriculum of Bangladesh College of Physician & Surgeon (BCPS)
2. 2. AUTHOR: Dr Md Anisur Rahman Anjum. MBBS (Dhaka Medical College). DO (Dhaka University) FCPS (EYE) Associate Professor National Institute of Ophthalmology Dhaka, Bangladesh. Chamber: Mojibunnessa Eye Hospital House: 18 Road: 6. Dhanmondi, Dhaka, 1205. Bangladesh. Email: anjumk38dmc@gmail.com Cell: 01711-832397 7/14/2014 2anjumk38dmc@gmail.com
3. 3. OSPE:1 7/14/2014 3anjumk38dmc@gmail.com
4. 4. Question The following are the results of the retinoscopy on a 20 year-old man using a streak retinoscope.(Assuming a working distance of 2/3 metre) OD With the streak in the vertical meridian and sweep horizontally, the retina reflex is neutralized with +0.50D. With the streak in the horizontal meridian and sweep vertically, the retina reflex is neutralized with -1.00D. 7/14/2014 4anjumk38dmc@gmail.com
5. 5. Question  OS With the streak orientated at 30o and sweep at 120o, the retina reflex is neutralized with +1.00D. With the streak orientated at 120o and sweep at 30o, the retina reflex is neutralized with -0.50D Question: 1 Draw the power cross (taking into account the working distance). = 5 Question: 2 Write down the glasses prescription for this patient (using negative cylinder) = 5 7/14/2014 5anjumk38dmc@gmail.com
6. 6. Answer • the power of the cylinder is 900 to the axis • 2/3 meter is equivalent to -1.50 which has to be taken off) 7/14/2014 6anjumk38dmc@gmail.com
7. 7. Answer • 7/14/2014 7anjumk38dmc@gmail.com
8. 8. • OD = -1.00 Dsph/-1.50 Dcyl 90 • OS = -0.50 Dsph/-1.50 Dcyl 120 7/14/2014 8anjumk38dmc@gmail.com
9. 9. OSPE:2 7/14/2014 9anjumk38dmc@gmail.com
10. 10. Question Calculate the spherical equivalent of the following prescriptions and draw the power cross for each. a. +2.50 / -1.50 X 80 b. -4.00 / + 6.00 X 90 c. +1.50 / -3.50 X 45 7/14/2014 10anjumk38dmc@gmail.com
12. 12. Answer a. +2.50 / -1.50 X 80 = +1.75 (spherical equivalent) b. -4.00 / + 6.00 X 90 = -1.00 (spherical equivalent) c. +1.50 / -3.50 X 45 = -0.25 (spherical equivalent) • To draw the power cross remember that the power of the cylinder is 90 degrees to the axis. The spherical equivalent is calculated by adding the value of the sphere and half the value of the cylinder.) 7/14/2014 12anjumk38dmc@gmail.com
13. 13. OSPE:3 7/14/2014 13anjumk38dmc@gmail.com
14. 14. Convert them into spectacle forms with minus cylinder notation • Working distance is 67 cm 7/14/2014 14anjumk38dmc@gmail.com
15. 15. 1) The power crosses translate to spherocylindrical corrections (with minus cylinder) of RE +1.50 / - 4.00 X 135 LE -0.75 / -1.25 X 90 • As the working distance is 2/3 meter, -1.50 D is subtracted from the above giving RE PL / -4.00 X 135 LE -2.25 / -1.25 X 90 7/14/2014 15anjumk38dmc@gmail.com
16. 16. 2) The power crosses translate to spherocylindrical corrections (with positive cylinder) of RE -2.50 / +4.00 X 45 LE -2.00 / +1.25 X 180 • As the working distance is now 1/2 meter, - 2.00D is subtracted from the above RE -4.50 / +4.00 X 45 LE -4.00 / +1.25 X 180 7/14/2014 16anjumk38dmc@gmail.com
17. 17. OSPE:4 7/14/2014 17anjumk38dmc@gmail.com
18. 18. Question A 18 year-old university student presents to the optician one week before his final examination complaining of intermittent double vision and poor distance vision. He was seen one month earlier and the refraction was emmetropic. Q: 1 What is the most likely diagnosis? Q: 2 What treatment would you prescribe? 7/14/2014 18anjumk38dmc@gmail.com
19. 19. Question A 40 year-old myopic woman is recently prescribed soft contact lenses for the first time. She returned two weeks later and complains that her reading vision is not as good as with her glasses. Retest shows her visual acuity to be 6/6 in both eyes with the contact lenses and the lenses were of the right prescription and well-fitted. Q: 3 Why does she have problem reading with her contact lenses but not with her glasses? 7/14/2014 19anjumk38dmc@gmail.com
20. 20. Answer Ans: 1 Accommodation spasm. (When the symptoms occurs the patient is likely to have constricted pupils and esotropia. The spasm can be broken with cyloplegia.) Ans: 2 The following treatment is useful: Frequent looking up from work or Prescription of reading glasses Ans: 3.The patient is pre-presbyopic. Myopes require less accommodation with glasses than contact lenses. In addition, the prismatic effect (base-in prism) offered by the concave glasses assist convergence during reading. • 7/14/2014 20anjumk38dmc@gmail.com
21. 21. OSPE:5 7/14/2014 21anjumk38dmc@gmail.com
22. 22. Question A 15 year-old emmetropic boy has an amplitude of accommodation is 30 D. Q: 1. What is his range of accommodation? Another 20 year-old man has a +10 D hypermetropia with 20 D amplitude of accommodation. Q: 2. What is his near point and the range of accommodation when he is not wearing the spectacle correction? Q:3 What is his near point and the range of accommodation when he is wearing his +10.00 dioptres?7/14/2014 22anjumk38dmc@gmail.com
23. 23. Answer Answer: 1 Infinity to 3.3cm. =3 (The range of accommodation is the linear distance between the far point and the near point. In the absence of accommodation, the boy is in focus at infinity. With maximal accommodation, he is in focus at 1/30 = 0.033m = 3.3cm) Answer:2. 10cm = 3 (To be in focus in infinity, the man needs to use up +10.00D of his accommodation amplitude. As a result, he has only 20 - 10 = 10 D for near vision. His near point is 1/10 = 0.10 m = 10 cm) 7/14/2014 23anjumk38dmc@gmail.com
24. 24. Answer Answer: 3. The near point is 5 cm and the range of accommodation is infinity to 5 cm. = 4 (With hypermetropic correction he does not need to use up any of his accommodation and can therefore use all the 20 D for near vision. The near point is therefore 1/20 = 0.05m = 5cm. The range of accommodation is therefore from infinity to 5 cm) 7/14/2014 24anjumk38dmc@gmail.com
25. 25. OSPE:6 7/14/2014 25anjumk38dmc@gmail.com
26. 26. This is a concave mirror, 5 objects are placed in 5 different location, Q:1 Write down the location of the images. Q: Write down where it will be real and virtual 7/14/2014 26anjumk38dmc@gmail.com
27. 27. Answer • Case 1: Between C and F. REAL IMAGE • Case 2: Image will be at C. REAL IMAGE • Case 3: Image will be beyond C. REAL IMAGE • Case 4: No image will be form • Case 5: the image will always be located somewhere on the opposite side of the mirror. • VIRTUAL IMAGE 7/14/2014 27anjumk38dmc@gmail.com
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