GS-2 Physical Sciences Question 3

[sharra.26428]

Could you please explain question number 3 because I am having trouble understanding what the question is asking and so the answer is not making sense to me either.
thanks

[admin]

http://www.mcat-prep.com/forum/viewtopic.php?t=75&start=0

[JoonSKim1557]

I agree, this question made me pretty upset.
This question hasn't mentioned that the the image is virtual. How am I suppose to make an assumption that the image formed by the eyepiece would be on the same side as the image formed by the objective?

[tjk447963]

The link posted above doesn't seem to work. Can you re-post the explanation as I am having trouble understanding why the image magnification gets smaller as the objective distance decreases.

[sleepyzboy2302]

the explanation is confusing.. can anyone explain it a little better? thanks

[jellywing_2058]

Object distance for the Objective = u
Image distance for the Objective = v
Magnification for the Objective = Mo = - v/u

If the distance between the Objective and the Eyepiece = d
Object distance for the Eyepiece, which is the distance of the Image formed by the Objective and the Eyepiece = d – v

Image distance for the Eyepiece = v’
Magnification for the Eyepiece
Me = - v’/ (d – v)

Given in the Question:
Mo = Me
-v/ u = - v’/(d – v)
v/ u = v’/(d – v)
v d – v^2 = u v’
v’ = (v d – v^2)/ u

Overall magnification:
M = - v’/ u = (- v d + v^2)/ u^2
M = (- v d/ u^2) + (v^2/ u^2)

Because A gave an overall magnification MA that was greater than that given by B, MB, it could only be because M is given by:
M = (- v d/u^2) + (v^2/u^2),

Therefore
MA = (v^2/u^2) – (v d A/ u^2)
And,
MB = (v^2/u^2) – (v d B/u^2)

If,
MA > MB
(v^2/u^2) – (v d A/u^2) > (v^2/u^2) – (v d B/u^2)
(v d A/u^2) > (v d B/u^2)
dA > dB

That is, Choice A: The distance between objective and eyepiece in A, dA is greater than the corresponding distance in B, dB.

[kim.dangal6935]

Is it simple mathematics?

Ma = -ia/0a

Mb = -ib/0b

Since Ma > Mb then Mb has to be more negative.

For Mb to be more negative in value, 0b < Oa.

So that means "the distance between the objective and the eyepiece in A is greater than the distance between the objective and eyepiece in B"

Right?

[klaw19867026]

I used the previous post's reasoning also, however I got a different conclusion. If we want magnification of A to be larger, then don't we decrease Oa so that the magnitude of magnification is larger?

Ma = -ia/0a

Mb = -ib/0b

so having 0a < Ob will make A's magnification larger. Decreasing the distance between then two lenses should have this effect, so wouldn't this make choice B true?

[btdumford5156]

Can someone please explain this question? I choose B, Because it seemed intuitively more accurate, but I now see that im incorrect, please help!

[sanji2896161]

[jeffreyt.y8164]

thanks jellywing_2058!