Image Distance - Chad Quiz Question

[Gauss44]

---

Members don't see this ad.

The radius of curvature of a diverging lens is 10cm. If an object is placed 20cm in front of the lens then where will the image appear?

Chad's Ans: 4cm in front of the lens

Anyone?

Edit: I clarified that the above answer is the one from the answer key. I think the answer key's wrong and agree with N7chell about the negative sign. I will report that question.

 

[N7chell]

Ok, first of all, a diverging lens is a concave lens and looks like this: )(. You need to know that a diverging lens produces a virtual image, which means that the image will appear on the opposite side of the observer. Draw a diagram if it helps you.

For the calculation, you need to use the lensmaker's equation: 1/i + 1/o = 1/f (i/o/f = distance of image/object/focal point). O is always positive, and i/f are positive when they are on the same side as the observer. In this case. i and f are negative. You can get f by halving the radius of curvature. Now plug in the numbers and solve for i.

1/i = 1/f - 1/o
1/i = 1/-5 - 1/20 = -4/20 - 1/20 = -5/20
i = -4cm

Since the observer is always behind the lens, the negative sign tells us that the image will be on the opposite side. Hope that helps!

 

[gettheleadout]

Ok, first of all, a diverging lens is a concave lens and looks like this: )(. You need to know that a diverging lens produces a virtual image, which means that the image will appear on the opposite side of the observer. Draw a diagram if it helps you.

For the calculation, you need to use the lensmaker's equation: 1/i + 1/o = 1/f (i/o/f = distance of image/object/focal point). O is always positive, and i/f are positive when they are on the same side as the observer. In this case. i and f are negative. You can get f by halving the radius of curvature. Now plug in the numbers and solve for i.

1/i = 1/f - 1/o
1/i = 1/-5 - 1/20 = -4/20 - 1/20 = -5/20
i = -4cm

Since the observer is always behind the lens, the negative sign tells us that the image will be on the opposite side. Hope that helps!

No you cannot. f = C/2 is the mirror maker's equation, and does not hold for a diverging lens. Also what you've called the lens maker's equation is not, it's the thin lens equation. I know what TBR calls it; TBR is wrong.

Edit: By the way, Chad's answer is correct because he specifies that the image is in front of the lens, which is correct.

 

[GomerPyle]

Ok, first of all, a diverging lens is a concave lens and looks like this: )(. You need to know that a diverging lens produces a virtual image, which means that the image will appear on the opposite side of the observer. Draw a diagram if it helps you.

For the calculation, you need to use the lensmaker's equation: 1/i + 1/o = 1/f (i/o/f = distance of image/object/focal point). O is always positive, and i/f are positive when they are on the same side as the observer. In this case. i and f are negative. You can get f by halving the radius of curvature. Now plug in the numbers and solve for i.

1/i = 1/f - 1/o
1/i = 1/-5 - 1/20 = -4/20 - 1/20 = -5/20
i = -4cm

Since the observer is always behind the lens, the negative sign tells us that the image will be on the opposite side. Hope that helps!

So the object is on the side of the lens that the image is formed (which is negative and which is also the side that the focal point is)...shouldn't the object distance be -20 as well?

 

[gettheleadout]

So the object is on the side of the lens that the image is formed (which is negative and which is also the side that the focal point is)...shouldn't the object distance be -20 as well?

By convention object distance is always positive regardless of the orientation of the optical axis.

 

[GomerPyle]

By convention object distance is always positive regardless of the orientation of the optical axis.

makes sense

 

[N7chell]

No you cannot. f = C/2 is the mirror maker's equation, and does not hold for a diverging lens. Also what you've called the lens maker's equation is not, it's the thin lens equation. I know what TBR calls it; TBR is wrong.

Edit: By the way, Chad's answer is correct because he specifies that the image is in front of the lens, which is correct.

So how do you calculate f in this case? Does it end up being C/2 since you get the same answer? I admit I'm not an expert in lenses, so I might have gotten the name wrong. Dispite that though, is my calculation still correct?

And OP, I never said Chad's answer is wrong. His answer is right because as the above person said, he specified that the image is in front, which is basically what the negative sign means.

 

[Gauss44]

Edit: By the way, Chad's answer is correct because he specifies that the image is in front of the lens, which is correct.

In the question he doesn't. Sure he does in the answer key, but you don't read that until the quiz is over.

 

[gettheleadout]

So how do you calculate f in this case? Does it end up being C/2 since you get the same answer? I admit I'm not an expert in lenses, so I might have gotten the name wrong. Dispite that though, is my calculation still correct?

And OP, I never said Chad's answer is wrong. His answer is right because as the above person said, he specified that the image is in front, which is basically what the negative sign means.

I don't know. For a diverging lens, which has two radii of curvature (differing at least in sign) we have to know both and the index of refraction of the lens material to find f using the lens maker's equation:

I don't know if additional information was given in the problem set, or if Chad is forcing the student to incorrectly apply an equation, or if I'm missing something (but I've checked multiple sources on this).

In the question he doesn't. Sure he does in the answer key, but you don't read that until the quiz is over.

In the question he states that the object is in front of the lens, and by drawing the ray diagram for a diverging lens, you can see that the image will be virtual and thus on the same side of the lens as the object. Since the object is in front of the lens, the image will also be in front of the lens.

 

[N7chell]

I don't know if additional information was given in the problem set, or if Chad is forcing the student to incorrectly apply an equation, or if I'm missing something (but I've checked multiple sources on this).

It's most likely the second case. He probably expected people to use f=C/2 to simplify the concept. My TPR course instructor did the same thing. If TPR didn't bother explaining the lens focal point issue, I'm assuming the MCAT doesn't expect you to know that much? Thanks for the clarification though!

 

[gettheleadout]

It's most likely the second case. He probably expected people to use f=C/2 to simplify the concept. My TPR course instructor did the same thing. If TPR didn't bother explaining the lens focal point issue, I'm assuming the MCAT doesn't expect you to know that much? Thanks for the clarification though!

That's incredibly frustrating though, because it's wrong. I can't believe (well...) that a TPR course instructor would do the same thing. I just finished a calculus-based physics course that ended with optics, and I've checked my notes from the course, TBR, and Wikipremed. The mirror maker's equation does not apply to lenses. In fact, after our last test I was chatting with one of the grad student TA's, and he mentioned that this exact misuse was a common error people made on the exam.

Chad's wrong if he's having students apply it to the lens in this problem. It's bad physics and bad teaching.

That said, TBR doesn't even give the real lens maker's equation, so I would assume you're correct that it isn't needed for the MCAT. I assume for a lens problem they would simply give you the focal length, or give you both object and image distances to start off with, so you would only need to use the thin lens equation.

 

[milski]

From a practical point of view, everything I've seen from AAMC related to lenses seems to be relatively simple and straightforward. Nothing anywhere near having to deal with an asymmetrical lens.

1/i+1/o=1/f is nice and quick and gives you the solution without much effort but I ended up drawing ray diagrams for all the optics questions. The equation relies on never getting the signs wrong by accident and remembering what the sign means in each case. That's a source of silly mistakes, at least for me.

TBR does misuse the name for lens maker's equation, there is at least one more thread discussing it somewhere around here. The are absolutely wonderful for physics for the MCAT but they may be a bit off when getting into more complicated topics. That should not worry you at all, since AAMC will not make you deal with the complicated stuff anyway.