Official July 24, 2014 MCAT Thread(because just once wasn't fun enough!)

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HinduHammer said:
Just to reiterate, UV upright virtual and IR inverted real? And it ALWAYS holds true?

I always think I know the optics stuff, then I do it and I get it wrong...grr..
Click to expand...

May Ganesh, Krishna, Shiva, Vishnu, and the rest of the crew be with us on Thursday
 
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Hey does anyone have advice on quick calculations and other ways to get through the PS section. I am also wondering how different the new AAMC CBT FL's are from the old 77q ones. I only have the old 3-9, so that's what I've been using.
 
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dohyh88 said:
yes always true. Knowing that becomes really handy for majority of the optics related questions.
Click to expand...

Can you explain this problem to me, clearly explaining how to manipulate the thin lens equation? I am having trouble manipulating the thin lens equation and don't want to rely on conceptual understanding in case the questions asks me to find the exact value. Thanks!

A person attempts to take a picture of an object, using a lens with a focal length f. If the person then makes a second attempt, using a lens with a focal length 2f and doubles the distance between the lens and the object, the distance to the image during the second attempt will:

a. be reduced to 1/4 of the original
b. be reduced to 1/2 of the original
c. be double that of the original
d. be quadruple that of the original
 
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HinduHammer said:
Can you explain this problem to me, clearly explaining how to manipulate the thin lens equation? I am having trouble manipulating the thin lens equation and don't want to rely on conceptual understanding in case the questions asks me to find the exact value. Thanks!

A person attempts to take a picture of an object, using a lens with a focal length f. If the person then makes a second attempt, using a lens with a focal length 2f and doubles the distance between the lens and the object, the distance to the image during the second attempt will:

a. be reduced to 1/4 of the original
b. be reduced to 1/2 of the original
c. be double that of the original
d. be quadruple that of the original
Click to expand...

Suppose the focal length is f and the object distance is o. Also assume that the object distance is greater than the focal length. The thin lens equation would read 1/I + 1/o = 1/f => 1/I = 1/f - 1/o => 1/I = o-f/of
so then the image is I = of/o-f. Now let f go to 2f and o go to 2o. then we would have, using the same equation, the new image distance as I' = (2o-2f)/(2o)(2f) = 2o-2f/4of = 2(o-f)/4of = o-f/2of = 1/2 I .

so it's reduced to 1/2 of the original.
 
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user12 said:
Suppose the focal length is f and the object distance is o. Also assume that the object distance is greater than the focal length. The thin lens equation would read 1/I + 1/o = 1/f => 1/I = 1/f - 1/o => 1/I = o-f/of
so then the image is I = of/o-f. Now let f go to 2f and o go to 2o. then we would have, using the same equation, the new image distance as I' = (2o-2f)/(2o)(2f) = 2o-2f/4of = 2(o-f)/4of = o-f/2of = 1/2 I .

so it's reduced to 1/2 of the original.
Click to expand...
See i solved the equation for i.... plugged in random values for f & o, then doubled f & o and replugged in..compared values of i to each other... i ended up doubling for me... hmph
 
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It's pretty straight forward if you use the thin lens equation and some inspection of the formula:

So, 1/f = 1/(d_obj) + 1/(d_img)

You are told that the f is doubled and the d_obj is doubled. This implies that f in the above eq is now 2f and d_obj is now 2*d_obj which gives:

1/2f = 1/(2*d_obj) + "something" * 1/(d_img); The "something" has to be 1/2 that way you can factor our half from both sides and you get back to the thin lens equation!

1/(d_img) is now 1/(2*d_img) this implies that the new image is double dist of the old image.

Good luck every one!! Phew never have I studied so hard for a test in my life...one more day. Reviewing some concepts and notes and them I'm going to go watch a movie maybe idk haha!

:)
 
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Terps1011 said:
See i solved the equation for i.... plugged in random values for f & o, then doubled f & o and replugged in..compared values of i to each other... i ended up doubling for me... hmph
Click to expand...

Both methods are good. it's just that user12 flipped the equation it should have been I' = (2o* 2f)/(2o-2f), the answer is (C)
 
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Did you guys put your middle name when registering for the MCAT? I have a middle name on my drivers license but I did not put my middle name when i was registering. Will i be okay?
 
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Okay, So when they see my middle name on my driver's license but not on my registration, they won't deny me?
 
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Eklipz said:
Okay, So when they see my middle name on my driver's license but not on my registration, they won't deny me?
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No, all they care is that your first / last name on your license match exactly your AAMC registration ID first / last name.
 
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azor ahai said:
Hey does anyone have advice on quick calculations and other ways to get through the PS section. I am also wondering how different the new AAMC CBT FL's are from the old 77q ones. I only have the old 3-9, so that's what I've been using.
Click to expand...

You can get away with heavily rounding everything to simplify the math and convert everything to scientific notation to minimize mistakes.
 
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Doing this a second time has made me so much more relaxed.... being on vacation is also helpful haha
 
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Just a little equation review and a nice solid workout tonight for me before the big day. What is everyone else up to?
 
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vik66 said:
Just a little equation review and a nice solid workout tonight for me before the big day. What is everyone else up to?
Click to expand...
Watching Greys anatomy, baking, and working out are on my list of to dos :)
 
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July 12th MCATer here guys, good luck tomorrow! Just remember to enjoy and relax today so that you can have peaceful and great state of mind tomorrow!
 
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Does anyone know of any good links for all the formulas we need to know? Just want to make sure I'm not missing anything. Thanks!
 
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read that site. maybe I missed it, but it appears they didn't include

G = -RTlnKeq

might also want to know the double slit equation. im not saying it's going to be on there, but it wouldn't hurt to know. dy = nlambaD, where d is the distance between slits, y is distance between fringes, D is distance to screen, and lambda is wavelength.
 
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last summer i didnt really look at single slit, just kinda glanced over bc they say its low yield. FIRST passage on physical was a tough single slit passage. didnt help with my confidence. lesson learned : DO NOT DISCOUNT ANYTHING
 
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no way man, the slit experiments are a huge
vik66 said:
last summer i didnt really look at single slit, just kinda glanced over bc they say its low yield. FIRST passage on physical was a tough single slit passage. didnt help with my confidence. lesson learned : DO NOT DISCOUNT ANYTHING
Click to expand...

nah nothing is low yield on the mcat. the more you know the better. the slit experiments were a HUGE deal in demonstrating the wave properties of light. any intro physics course should discuss them (and any mcat test prep company should too)
 
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user12 said:
Suppose the focal length is f and the object distance is o. Also assume that the object distance is greater than the focal length. The thin lens equation would read 1/I + 1/o = 1/f => 1/I = 1/f - 1/o => 1/I = o-f/of
so then the image is I = of/o-f. Now let f go to 2f and o go to 2o. then we would have, using the same equation, the new image distance as I' = (2o-2f)/(2o)(2f) = 2o-2f/4of = 2(o-f)/4of = o-f/2of = 1/2 I .

so it's reduced to 1/2 of the original.
Click to expand...

Terps1011 said:
See i solved the equation for i.... plugged in random values for f & o, then doubled f & o and replugged in..compared values of i to each other... i ended up doubling for me... hmph
Click to expand...

RocketSurgeon said:
It's pretty straight forward if you use the thin lens equation and some inspection of the formula:

So, 1/f = 1/(d_obj) + 1/(d_img)

You are told that the f is doubled and the d_obj is doubled. This implies that f in the above eq is now 2f and d_obj is now 2*d_obj which gives:

1/2f = 1/(2*d_obj) + "something" * 1/(d_img); The "something" has to be 1/2 that way you can factor our half from both sides and you get back to the thin lens equation!

1/(d_img) is now 1/(2*d_img) this implies that the new image is double dist of the old image.
Click to expand...

Thanks to @Terps1011 and @RocketSurgeon @IamZee @user12 for the replies! I like both ways Terps and Rocket suggested....first plug in numbers and see what works! Duh! So much easier lol. But I really understood what you said Rocket about how to solve it using the thin lens equation. I think I just needed my coffee or something this morning lol, but thanks to all of you for helping!!

I am taking it August 2 and studying like a fiend, but good luck to you all tomorrow!!!!!
 
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@kenjixshadow we're thinking the same thoughts lol

I was going to review some equations and stuff tonight, but I ended up spending the last 4 hours eating and playing games at Dave n Busters. I'll just review a little in the morning, take this thing, and enjoy my 1 month of summer vacation! :D Hopefully I won't have to retake in October!
 
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We got this guys! I hope to hear from you all after the exam!
 
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About to get on the Ferry. Yes my commute includes a friggin' ferry. So I couldn't sleep in haha. GOOD LUCK EVERYONE!!!
 
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Today is the 3rd exam registered for in '14, noshowed for the first two b/c didn't feel prepared enough for noticeable improvement (and was being indecisive). Can't test 2day bc of documented illness (at same univ. as my pre-health committee). could It be possible to reschedule for Aug/Sep? It would be pointless for me to reapply without a new mcat score but obviously can't risk testing when ill. ;={ Advice?
 
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