Light and Optics Light and Optics Physics Help Forum Jun 29th 2009, 06:37 AM #1 Member   Join Date: Jun 2009 Posts: 79 Correction of defects in vision Q) The diameter of an eyeball of a certain person is 24mm,but the normal diameter of an eyeball should be 23mm. What is the power of the lens to be used inorder to correct this defect? He seems to be having a larger eye than normal... so the image formed of an object would in this case be at 23mm,from the eyelens,which, would mean that he would have to use a diverging lens to correct this defect,to form an image at 24mm. As for trying to find the focal length of the correction lens,I have no idea how to do it. I know that, 1/f = 1/v - 1/u , and that power = 1/f, but I can't figure out what values need to be substituted. Thanks in advance   Jun 29th 2009, 06:24 PM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 You can take the object at infinity. Then use the formula 1/f1 + 1/f2 = 1/F Here f1 is the focal length of the normal eye and F is the focal length of defective eye. And f2 is the focal length of the correcting lens.   Jun 29th 2009, 11:37 PM   #3
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 Originally Posted by sa-ri-ga-ma You can take the object at infinity. Then use the formula 1/f1 + 1/f2 = 1/F Here f1 is the focal length of the normal eye and F is the focal length of defective eye. And f2 is the focal length of the correcting lens.
So taking the object at infinity,for the defective eye,and applying sign convention(taking the distances measured in the direction of the rays from the object as negative,while those measured in the direction opposite to the rays as positive),

1/F = -1/2.3 - 1/infinity
1/F = -1/2.3cm = -0.435cm

taking the object at infinity,for the normal eye,
1/f1 = -1/2.4 - 1/infinity
1/f1 = -0.417cm

Then using the formula suggested,
1/f2 = 1/F - 1/f1
1/f2 = -0.435 -(-0.417)
I get f2 = -55.56cm,
since the focal length is negative ,wouldn't the correcting lens be a converging lens?   Jun 30th 2009, 04:11 AM   #4
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 So taking the object at infinity,for the defective eye,and applying sign convention(taking the distances measured in the direction of the rays from the object as negative,while those measured in the direction opposite to the rays as positive), 1/F = -1/2.3 - 1/infinity 1/F = -1/2.3cm = -0.435cm taking the object at infinity,for the normal eye, 1/f1 = -1/2.4 - 1/infinity 1/f1 = -0.417cm Then using the formula suggested, 1/f2 = 1/F - 1/f1 1/f2 = -0.435 -(-0.417) I get f2 = -55.56cm, since the focal length is negative ,wouldn't the correcting lens be a converging lens?
No. It is not correct.
For normal eye focal length of the eye lens is 2.3 cm. You have to increase it to 2.4 cm for the defective eye by keeping a diverging lens of focal length f2 in front of the eye lens. Find the value of f2.   Jun 30th 2009, 06:48 AM   #5
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Thank you so much,sir.
 No. It is not correct. For normal eye focal length of the eye lens is 2.3 cm. You have to increase it to 2.4 cm for the defective eye by keeping a diverging lens of focal length f2 in front of the eye lens. Find the value of f2.
Yes,I've made another careless mistake.Thanks for pointing it out.

correction : 1/f2 = 1/F - 1/f1
= -1/2.4 - (-1/2.3)
f2= 55.56cm
and P= 1.8D
I hope it's correct now?

Last edited by hana1; Jun 30th 2009 at 07:33 AM.   Jan 23rd 2011, 11:03 AM #6 Junior Member   Join Date: Jan 2011 Posts: 1 Hello evrybody. Here what i found -> vision correction  Tags correction, defects, vision ,

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# sign convention used in the defects of eyes

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