What's the surface power and refractive index?

**Optometry student** · 10-03-2011, 07:58 AM

Need help in solving this.

The fixed outer legs of a lens measure calibrated for spectacle crown glass, n=1.523, are separated by 18mm when the instrument is new, but in wear, become separated by 20mm. What readings will this instrument give when placed upon a spectacle crown glass surface of power +5.00D?
The instrument will give correct readings when placed on glass of a certain refractive index. Find this index.

**Wes** · 10-03-2011, 08:32 AM

Start thinking about sagittal depth formula.

After you calculate the reading from the first problem, you have three numbers (+5.00, and the calculated reading; and the original index) and should be able to calculate the fourth.

**Darryl Meister** · 10-03-2011, 12:25 PM

In addition to Wes's comments, remember that all of these calculations will rely on the following formula:

$s=\frac{\left(\frac{1}{2}d\right)^2\times F}{2000(n-1)}$

where d represents the separation of the lens measure pins, F is the measured surface power, and n represents the refractive index for which the lens measure has been calibrated. Basically, you are just substituting values into this expression and rearranging as necessary to solve the two problems.

Best regards,
Darryl