TBR: Optics Question

[justadream]

TBR Physics II page 286 #48


"When an object is placed inside the focal length on the right side of a diverging lens, what type of image is formed?"

Answer: Virtual, Upright Image


How do you get this answer? Also, I don't even know how the information that the object is on the "right side" is helpful.

Here's the way I interpreted the scenario

1/f = 1/i + 1/o

Let me make up random numbers:
1/-10 = 1/i + 1/5

Thus, i = -3.33333

This is Upright and virtual. But my reasoning is completely different from TBR's.

TBR's explanation is confusing because it says both i and o are negative (WHAT? I thought object distance is never negative for MCAT purposes).

Here's TBR's explanation (I changed some variable names):

Since the object is to the right of the lens and the lens is diverging, both i and o are negative. If the object is placed inside the focal length, the magnitude of f is greater than the magnitude of o, so the magnitude of 1/o is greater than the magnitude of 1/f. Since 1/o is a larger negative number, then in order to satisfy the lens equation, i must be positive. If i is positive, the magnification equation indicates that that image is upright.

---

Members don't see this ad.

 

[DrDreams]

This is what I have in my notes. Lens Makers Equation
1/f=1/d+1/i

1. f(+)=converging f(-)=Diverging.
2. Object distance: Right of lens=(-) left of lens (+)
3. Real images = (+) Virtual images= (-)

Convex or Diverging mirrors or lenses: Always upright, virtual and smaller.

I hope that helps....

 

[popopopop]

Lol wow, I spent 20 minutes searching for a solution because optics is a weakness of mine too. I think these three pages really cleared things up for me in solving this problem. I think we're dealing with a double lens problem and other sources have just stated that when D.O. is to the right it is assumed to be negative.

Link 3, slide #3 might be what we're dealing with? Distance of Object is negative "for virtual object (only possible if “object” is actually the image formed by another lens)."

1. http://www.thelearningodyssey.com/Graphics/Content/vs/em/Medias/html/a294-diverging-lens.html
2.http://faculty.spokanefalls.edu/InetShare/AutoWebs/AsaB/Phys103/MirrorsThinLens.pdf
3. http://people.virginia.edu/~ral5q/classes/UVa-JLab/summer13/contents/2012 summer/notes/Lecture10.pdf