Just basic, 'ole, retinoscopy!

[haya]

Hope you guys can help me with this!

The spectacle correction of a px is:

R: -1.00/-2.00x90
L: -3.00/-4.00x180

Without any trial lenses in place, what is the ratio of reflex speeds along the two principle meridians of the LEFT eye at a working distance of 50cm?

Thanks, Haya

[Sara]

ratio 1:2 approximately

May be someone will correct me.
Thanks,
Sara
:rolleyes:
optical student

[Darryl Meister]

This isn't the kind of question you see much of over here in the States. Tunnacliffe gives the speed factor as

W / (W + K)

Where W is the dioptric working distance (2.00 D) and K is the ocular refraction.

The speed factor for the -3.00 meridian is -2, while the speed factor for the -4.00 meridian is -1. I don't know if this is correct for astigmats, but dividing the two speed factors of the principal meridians yields a value of 0.5 (or 2, depending upon whether you choose the strongest or weakest meridian). This is consistent with Sara's answer.

Best regards,
Darryl

[haya]

Thanks you guys. That is what i have come up with (honest!) but I really lack the confidence in my ability - and that is not very reassuring considering my exams are precisely two weeks away!

Also, Darryl, I am delighted that a little transatlantic distance has not knocked your talents in the academic department.

And rest assured, I'll be back....... to pick your brains!!!!!!

haya

[Darryl Meister]

But don't break out the champagne just yet until you find out whether or not we got the answer right! ;)

Best regards,
Darryl

[haya]

Thanks, you guys!

Both of you were correct and I am now well into my first week of revision. NO, let me correct myself... in my case, it is one of 'vision'!

a v.happy haya!