Geometric Optics - Magnification

AI Thread Summary
A concave mirror initially forms an image twice the size of an object, and both the object and screen are moved to create a new image three times the size of the object. The screen is moved 75 cm, while the mirror remains stationary. The problem involves determining how far the object moved using the magnification formula, m = image distance / object distance. Participants in the discussion suggest defining variables for initial and final distances and recommend using the magnification equations to derive relationships between these distances. The discussion emphasizes the importance of writing out equations to clarify the solution process.
LANS
Messages
24
Reaction score
0

Homework Statement


A concave mirror forms an image on a screen twice as large as an object. Both object and mirror are then moved such that the new image is 3x the size of the object. If the screen is moved 75cm, how far did the object move?

Homework Equations


m = image distance / object distance

The Attempt at a Solution


Initially, the screen is twice as far from the mirror than the object. The screen and object move so that the screen is three times as far from the mirror as the object. I'm really at a loss and feeling kinda dumb here on solving this - any pointers in the right direction would be much appreciated.

Thank you,
LANS
 
Physics news on Phys.org
Is that exactly how the question was worded? It sounds a bit off because it says the object and mirror are moved, implying the screen stays put, but then it says the screen was moved.
 
vela said:
Is that exactly how the question was worded? It sounds a bit off because it says the object and mirror are moved, implying the screen stays put, but then it says the screen was moved.

Sorry, I miswrote it. The object and screen moved, mirror stays in the same place.
 
LANS said:

Homework Statement


A concave mirror forms an image on a screen twice as large as an object. Both object and mirror are then moved such that the new image is 3x the size of the object. If the screen is moved 75cm, how far did the object move?

Homework Equations


m = image distance / object distance

The Attempt at a Solution


Initially, the screen is twice as far from the mirror than the object. The screen and object move so that the screen is three times as far from the mirror as the object. I'm really at a loss and feeling kinda dumb here on solving this - any pointers in the right direction would be much appreciated.
OK, let's define some variables:
\begin{eqnarray*}
d_\text{obj} &= \text{initial object distance} \\
d_\text{img} &= \text{initial image distance} \\
r_\text{obj} &= \text{final object distance} \\
r_\text{img} &= \text{final image distance} \\
f &= \text{focal length of the mirror}
\end{eqnarray*} So you have five unknowns. You have ##d_\text{img}/d_\text{obj} = 2## and ##r_\text{img}/r_\text{obj} = 3## so far. That's two equations. Can you come up with any more?
 
There's 1/d_I + 1/d_o = 1/f (and same for r_I, r_o), and r_I - d_I = 75, I'm just not sure how to approach it algebraically.
 
OK, good. Since you're looking for how far the object moved, you might want to get equations in terms of the object distances, so I'd use the magnification equations to solve for the image distances in terms of the object distances and then substitute the results into the three remaining equations.

Try writing those out and see if you can see where to go from there. Sometimes you just have to write equations down and try stuff, and then it becomes clear what to do.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top