Kaplan Optics Question

[sparkleys]

---

Members do not see this ad.

The near point, or nearest point at which an object can be seen clearly, of one of your eyes is 100 cm. You wish to see your friend's face clearly when she stands 50 cm in front of you. If you use a contact lens to adjust your eyesight, what must be the focal length and the power of the contact lens?

Do = 50cm
Di = 100cm
f = ?

Answer is 100cm. How is this possible according to the 1/f = 1/Do + 1/Di formula. The answer key states 1/Do = 1/Di = 1/f where 1/50 = 1/100 = 1/100. I'm not sure how they got this?

 

[Gauss44]

The near point, or nearest point at which an object can be seen clearly, of one of your eyes is 100 cm. You wish to see your friend's face clearly when she stands 50 cm in front of you. If you use a contact lens to adjust your eyesight, what must be the focal length and the power of the contact lens?

Do = 50cm
Di = 100cm
f = ?

Answer is 100cm. How is this possible according to the 1/f = 1/Do + 1/Di formula. The answer key states 1/Do = 1/Di = 1/f where 1/50 = 1/100 = 1/100. I'm not sure how they got this?

Did they mean to make the image distance negative?

http://cnx.org/content/m42484/latest/#import-auto-id1867992

 

[sparkleys]

I don't think so

 

[7502]

Did anyone figure this out?

 

[DrDashInd]

Ok so if you can see an image, it shows up on your retina, right?
So in the 1st case...
when he can see his friend, the image shows up on the retina.
thus:
1/f = 1/i + 1/o
1/f = 1/retina dist. + 1/100

Now if you want to see something, the image must show up on your retina so the purpose of the correctional lenses is to make it so that closer objects also show up on your retina.
So, I derived the retina dist. : 1/f - 1/100= 1/retina dist.

New case. obj. dist = 50cm
Thus
1/fcorrected = 1/i + 1/o
1/fcorrected = 1/i + 1/50 // want image to = retina dist so plug in...
1/fcorrected = (1/f - 1/100) + 1/50

Now 1/fcorrected = 1/f old + 1/f contacts //http://www.citycollegiate.com/geometricaloptics3.htm
thus
1/f + 1/fcontacts = 1/f - 1/100 + 1/50
You can take it from here.

Hope that helped!

 

[Laureatebarrel]

Remember that this is a lens, so if you want the image to appear in front of you, the di has to be negative.