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AAMC Practice Test FLE 1
Keira_6444
#1 Posted : Wednesday, July 14, 2021 1:35:11 AM
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Stand Alone #11: Are you able to explain the math here? I am having a little trouble understanding how they got to 51D and the explanation is a little confusing.

Stand Alone #45: Another lens equation question equation that I am a little confused by. I'm not following their explanation with the formula we were given in class so if you are able to explain this as well that would be great thanks!

INSTR_Katrina_128
#2 Posted : Sunday, July 18, 2021 5:56:04 PM
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Hi Keira!

Q11: We're lucky because they give us the formula for this question.

O is the distance from the lens to the object and I is the distance from the lens to the image. Both are given. Make sure to convert to meters.

Then,

S = 1/O + 1/I = 1/1 + 1/0.02 = 1 + 50 = 51 D (or diopters)

To figure out the 1/0.02 (which is kind of wonky), we can do a little fraction trick.

Since 0.02 is the same think as 2/100, we can rewrite this as

1/0.02 = 1/(2/100) = 1*(100/2) = 100/2 = 50

HA. Take that, math.

Q45: I will attempt an explanation! 😤

First, let's use the thin-lens equation:

1/f = 1/o + 1/i

We are TOLD that the object is at 3f, so o = 3f.

We will use this information to solve for i. Note that we cannot solve for a specific VALUE of i. Instead, we will solve for i as a function of the focal length, f.

1/f = 1/(3f) + 1/i

1/i = 1/f - 1/(3f)

Now we need a common denominator. Multiply the first term by 3/3.

1/i = 3/(3f) - 1/(3f)

Subtract.

1/i = (3 - 1)/(3f)

1/i = 2/(3f)

Flip both sides.

i = (3f)/2

Now we can use the magnification equation.

m = height of image / height of object = -i/o = -(3f)/2 / (3f)

3f's cancel, and we get -1/2.

We don't care about the negative sign. That just tells us that the image is INVERTED. What we care about is the MAGNITUDE: 1/2.

This tells us that the image is 50% the size of the object. Ooooh. 😀

The best answer is B.
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