- Hey, y'all, my name is Prashanth Kumar, or PK,
and I'm an MCAT tutor at Shemmassian Academic Consulting.
Today, we're gonna be doing a passage walkthrough
in the MCAT chem-phys section.
You should be taken roughly
six and a half minutes on this passage.
But today we're gonna go more in depth
so you can see exactly what's going on in my head
as I take on this passage.
Let's get it.
(keys tapping)
All right. Great.
So I've gone ahead and put the passage on the left
and the questions we're going through on the right.
So feel free to take a moment and pause the video.
If not, let's get right into talking strategy.
For me, I like to maintain the same strategy
going into whatever passage I'm gonna take.
The reason for this is it gives me something
to ground myself in.
Whether the passage is super easy or super challenging,
I always have something to come back to.
My strategy looks like reading a paragraph,
writing a quick three to five word summary,
and then going onto the next paragraph.
And so on, and so on, until I'm done.
The reason I like to do this
is that this provides me accountability
in making sure I understand what's going on in a paragraph
before moving on to the next paragraph.
By doing this, I prevent a lot of time from being wasted
when I, instead of having to reread it,
I might not have to reread an entire passage,
I can reread just a paragraph if necessary.
So that little accountability is very, very important to me.
And finally, when I usually look at passages,
I don't wanna spend too much time on the figures,
because I don't know what questions
will be asked about the figures,
but I will need to come back to them eventually
if they do ask questions.
So I'm prioritizing just taking a quick glance
and understanding what's going on,
rather than spending too much time
on the details of experiments or graphs or whatever.
So with that, let's get right into our first paragraph.
"Researchers are interested in setting
a novel optical system to model the human eye.
The experiment to set up includes a lens, object and image.
The object is placed at various positions
relative to the lens in order to produce different images."
All right, so the question I always ask myself
is what's happening.
We're looking for a big picture understanding
of what's being done.
And so here we have an example
of an eye model with a lens object plus image, okay?
So those are the quick summary
of what's going on in the paragraph.
Remember, use abbreviation, use symbols,
use whatever you need to make sure it's simple
and easy to understand for you
and you're not wasting too much time.
No one else is gonna check this little summary you write.
So with that, let's go on to the next paragraph.
"In Experiment 1, the experimental setup appears
as shown in Figure 1.
The object is located at position O,
the image is located at position I,
and the focal length of the lens is shown at position F."
So right away, I can see
that this isn't giving me much, right?
This little blurb doesn't really have that much information.
So we're gonna have to look at the figure.
And this goes against my usual tip, right?
And that's fine. It's a tip.
It's supposed to be flexible.
So here we see a diagram, right?
We see an experimental setup of an object,
a lens, and an image.
And we've seen these types of graphs before
in our content review phase.
And this is called a thin lens,
or anything in optics is dealing
with these types of setups, okay?
So I just want you to pay attention
to a couple of important things,
namely, where's the object and where is the image,
and how far away is that from the lens
with respect to the focal distance, okay?
So here, I'm gonna mark everything.
So this is the focal distance,
this is the object distance,
and this is the image distance, okay?
And so what I wanna focus on is how the object distance
and image distance are effectively related
to the focal distance.
So here, I'm gonna say the object distance
is actually greater than the focal distance.
And here I'm gonna say the image distance
is greater than the focal distance as well, right?
We see the focal distance is pretty small
comparative to both of them.
And that's all I wanna take away from this paragraph.
Next, we'll go on to the second experiment.
"In Experiment 2, the researchers adjusted
the position of the object,
resulting in a different location of the image.
The experimental setup appears as shown in Figure 2."
And then they give us a graph, right,
of this experimental setup.
And what I want you to note
is the same exact thing, right?
This is the similar graph as what we seen in Figure 1.
But what changes, right?
So in this case, our object distance is greater than F,
but our image distance is less than F, okay?
So right away, we have a bunch
of information that we're working on.
And we have the same sort of parameters
that we looked at in Figure 1, compared to Figure 2.
Finally, let's go onto the last paragraph.
"In Experiment 3, the researchers
once again adjust the position of the object.
The experimental setup appears as shown in Figure 3."
And so here we have our final sort of figure, right?
We've had three different experiments
with three different object change, or distance changing.
So in this case, how is the object distance
related to the focal length?
Well, the object distance
is less than the focal length,
and the image distance
is greater than the focal length, okay?
So those are the main things I wanna focus on right now.
Other details you can note
is where the image distance
is related to the object distance.
So here, you could write image opposite,
image opposite, and then image same, right?
You don't have to do that,
but if you wanted to pay attention to that,
that's something we can always work with.
Awesome.
So with that, let's get right into questions.
"Which of the falling determines
the focal length of the lens in Figure 1?"
Is it A, the radius of curvature of the lens,
B, the image distance, the C, the object distance,
or D, the level of lens magnification?
All right. So right away, this is a standalone question.
One that we're not gonna require any prior information for.
So what is their answer gonna be?
Well, we should know a focal length,
this value right here, F,
is an intrinsic property of the lens.
Focal length equals intrinsic property,
so, of the lens.
And so that means if we change the lens,
that is going to change the focal length.
But what is the defining property of the lens?
It's the radius of curvature of the lens.
So A is our answer.
Let's go through B, C and D.
B is the image distance.
That, while it is affected by the lens,
is also affected by the object distance.
And so that does not determine focal length.
C, object distance.
This does not affect focal length.
It's actually independent of focal length.
Because you're changing the distance,
it's not gonna change the focal length at all.
So C is eliminated.
And then D, the level of lens in magnification.
If you remember the formula
and a magnification is equal
to negative image distance over object, right?
And so with that, we can eliminate that as a choice,
because if it's dependent on the image distance
and object distance, which both
are not solely determined by focal length.
So right away, we're one for one,
just from understanding a piece
of information that we already had.
Awesome.
Question number two.
"In Experiment 1, suppose the object distance
is two centimeters,
and the image distance is three centimeters.
What is the focal length of the lens?"
Is it A, 1 over 2 centimeters,
B, 3 over 4 centimeters,
C, 6 over 5 centimeters,
or D, 7 over 9 centimeters?
All right, so right away, we have to note
that we're looking at Experiment 1, okay?
So that's this diagram right here.
And so we have our object distance, our image distance,
and it tells us what exactly we're looking at, right?
It tells us object distance is two centimeters,
image distance is three centimeters,
and the focal length is what we're looking at.
And so with this, we're gonna need off formula, right?
We're gonna need a very important formula
that we should remember from our content review.
And that's the important thin lens equation.
And what is that equation?
Well, that's 1 over object distance
plus 1 over image distance
equals 1 over the focal length.
Okay, so with this,
we have an equation and we're ready to go.
And so we just plug and chuck, right?
We say, what is the object distance? Two centimeters.
So 1 over 2.
And what is the image distance?
1 over 3 equals 1 over focal distance.
So now the question is, do the units match?
That is always what we needed to check.
And it says centimeter, centimeter, centimeter,
centimeters, all the answer choices are in centimeters
and our object distance and image distances
are also in centimeters.
So we're looking good to go,
and we're ready to run through our calculations.
So what is 1/2 plus 1/3?
Well, that is finding the common denominator is six.
We have to multiply this by three and this by three,
and this by two and this by two.
Classic multiplication and addition.
So then we have, oops.
Then we have 3 plus 2 on the numerator
and 6 on the denominator,
which is equal to 5 over 6.
But remember, this is equal
to 1 over the focal length distance.
What is going to be the focal length distance?
It's not asking for 1 over F, it's looking for F.
So with that, we do a little quick flip, right?
If we flip this side, that means we flip this side.
So F is equal to 6 over 5,
and we have our final answer.
And it matches C, right?
So we don't have that matching A, B or D.
And we're good to go.
Right away, our first calculation is dependent
on the thin lens equation.
This is a very high yield equation
that will come up over and over again.
So I want you to remember this one.
All right, with that, let's go right on
to question number three.
All right.
"Which of the falling experiments
produces a virtual equation,
a virtual image?"
Is it A, Experiment 1, B, Experiment 2, C, Experiment 3,
or D, none of the experiments produce a virtual image.
All right.
So first to do this, we need to define
what a virtual image is, right?
A virtual image is an image created by what?
Diverging rays of light.
Image created by diverging rays of light.
Awesome.
Okay. So this is a contrast to our real image.
We have real images that are created
by sort of converging rays of light,
and a virtual image is by diverging rays of light.
And so with this, we have to understand what types
of objects result in virtual images, right?
We have convex mirrors,
concave lenses.
They always give us virtual images.
This is guaranteed.
Right?
Then we have to look at the vice versa, right?
We have concave mirrors,
and convex lenses.
And these will not always give us virtual images.
Rather, they'll usually give us real images,
except if the object distance
is less than the focal distance.
If the object distance is less than one focal distance,
this results in a virtual image.
So these are the rules for what gives you a virtual image.
So with that, we have to look
at all three of our experiments
and decide which one's gonna get us a virtual image.
All right, so right away, we see this type of lens,
and we don't know exactly if it's a convex or concave lens,
but we do know the distances, right?
And we've written it so that we have O is greater than F,
O is greater than F, and O is less than F.
So with that, we already see a relationship
between object distance and focal distance.
And remember that regardless of whether a lens
is a concave lens or a convex lens,
if O is less than F, it will be a virtual image, okay?
So remember, if O is less than F in these concave lenses,
it will always be a virtual image.
Even if O is greater than F,
it will be a virtual image in a concave lens.
So I just wanna emphasize that this is always.
But if you're in a convex lens and O is less than F,
then you would have a virtual image.
So the identifying factor of
if it's a convex lens or a concave lens
doesn't really matter.
All that matters is that the object
is within one focal distance of this lens, right?
And so we don't know, remember,
we don't know what the lens is.
We don't know if it's convex or concave lens.
We just know it's a lens.
But with the fact that this has O less than F,
we can eliminate experiments A and B
and just to be left with C, right?
Because we know that is our answer.
That is a guaranteeing factor
that is sufficient to determine
that we will have a virtual image.
So with that, on to question number four.
All right.
Question number four.
"In Experiment 2, suppose the image distance is increased.
What effect would this have on the absolute value
for the magnification of the lens?"
Would it A, increase, B, decrease,
C, stay the same, or D, no change?
All right, so this requires us to understand
what is happening with magnification.
And like we mentioned earlier,
there's a specific formula
I want you to remember for magnification, okay?
Magnification is negative image distance, or I,
over object distance.
All right.
So in this case, our image distance is increasing.
So they're basically saying
if we push this image distance further,
what will happen to our magnification?
And remember, it's saying the absolute value
for the magnification.
So what we do is we're just putting absolute value signs
on both sides.
So regardless of whether it's negative or positive,
the output will be positive.
All right.
So with that, what effect would
an increase on image distance
have on the absolute value of magnification?
Well, if we know fractions, right,
if we increase the numerator
while keeping the denominator same,
this results in an increased value, okay?
You can just think of it as, like,
if I had 1 over 2, that's 0.5,
but if I just increased that to 2 over 2, that's 1.
so we definitely see an increase in value.
So if we increase the image distance,
we would resultantly see an increase in magnification.
So we can say that A is our answer.
If we decrease the image distance
or increase the object distance,
then we would result in a decrease.
So we can eliminate that.
Staying the same, we would either
not have to change anything,
or increase image distance
and object distance at the same rate.
So we can eliminate C.
And then no change, same thing,
we can eliminate D as well.
All right.
All right. Great work, team.
We crushed another physics passage.
Kudos for you for getting it done,
dealing with thin lens equation,
magnification, optics, all of that.
So great job reviewing all of that material.
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