Kaplan Error or Math dunce? Lenses.

[tshank]

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So, can someone point out how this ends up being D and not B? I know its math, so I should logically be able to see this but I can't. How did the 2i show up? and not 2/i? Thanks!

In the case of contact lenses, the cornea and the correcting lens are actually touching and act together as a single lens. If the focal length of both the cornea and the contact lens are doubled, then the image distance i for a distant object would:

A - be 1/4 the old value.
B - be 1/2 the old value.
C - be the same as the old value.
D - be twice the old value.


This question again involves the equation given in the passage for a two-lens system:

In the present case, we have a simplification. Since we are dealing with a lens in contact with the cornea, the distance x between the lenses is zero. The equation then reduces to:

Therefore, if the focal length of the cornea doubles and the focal length of the contact lens doubles, the image distance would also double:

(A) Miscalculation.
(B) Miscalculation. The new value for the image distance is half of 1/i, not half of i.
(C) Miscalculation.

 

[Sammy1024]

I didn't solve it, but mentally started it. Usually in the equation you use, if you increase the bottom part of the fraction, you have to figure out a way to bring d (distance) to the top, which would flip the fraction, and instead of 1/2, you would get 2/1.

 

[tshank]

@Sammy1024 , thanks for the response. I edited the post, their solution processing is now posted. It wasn't before. What do you make of it?

 

[Sammy1024]

I still feel the same way. If you look at the equation you notice how the f's and i's are all on the bottom of the fraction. To get the proper value of i, the fraction would have i on top, instead of at the bottom, so to get that you would flip the fraction giving you a 2 instead of 1/2.

It's like when you calculate the resistance for electrical circuits in parallel. The equation is 1/R1 + 1/R2 = 1/Req but lets say you have 2 resistors of 2 and 3 ohms, you would solve it the following way:

1/Req = 1/R1 + 1/R2
1/Req = 1/2 + 1/3
1/Req = 5/6
Req = 6/5

You had to flip the fraction at the end because 1/Req does not equal the resistance, it's 1/the total resistance.

The same thing is applying here. Hope it made sense!

 

[tshank]

I always thought the best way to get rid of a 1/2 on a side was to multiply both sides by 2. If this is done then it should be 2/i, which when flipped as you were saying, you get i/2 = f + f not i2

Why is there suddenly a 1/2 on the right side and not a 2/1?

I can do circuits fine, and that makes sense. But I just don't see the mathematical step here.