Here's something to tickle your painstaking calculation-bone with

  • Thread starter Thread starter KillaMarcilla
  • Start date Start date
AI Thread Summary
The discussion revolves around a complex optical problem involving a convex lens and a concave mirror, requiring multiple calculations to determine the final image location. Participants emphasize the importance of understanding optical principles and suggest using diagrams to track coordinate changes throughout the imaging process. One user expresses frustration with the calculations, especially at a late hour, while another encourages perseverance and highlights the value of attempting challenging problems. Ultimately, it is revealed that the original poster mistakenly worked on the wrong assignment, alleviating some pressure. The conversation underscores the challenges of optical calculations and the supportive nature of the community.
KillaMarcilla
Messages
56
Reaction score
0
I don't know if it qualifies for "funny" bone, necessarily..

Anyway, this problem is really off the charts for me at my current 3 AM IQ level (likely 40 points below what it was four hours ago..)

This one is just for fun, if you think you can solve it.. even if someone digs up an answer somehow, it'll only get me 70% of 5/80 points on an assignment, so don't feel like I need you (and if there's someone who really needs help soon, you oughtta move along)

(insert picture of an arrow, then 105 cm, then a convex-on-each-side lens with a focal length of 31 cm, then 51 cm, then a concave mirror with a focal length of 41cm)

Problem:
"An optical system consists of a convex lens of focal length f1 = 31 cm and a concave mirror of focal length f2 = 41 cm. The lens and mirror are co-axial and are spaced a distance a = 51 cm apart.

An upright object is located a distance d = 105 cm to the left of the lens. The image of this object formed by the lens serves in turn as the object for the mirror. The subsequent image formed by the mirror serves as the object for the lens as it acts on the (left moving) rays that have been reflected from the mirror. The lens forms the final image.

(a) Calculate the location of the final image, as measured from the point where the lens cuts the optical axis.

d' = cm
-79.08 NO

HELP: The second paragraph of the problem statement gives you the essential part of the overall strategy for finding the answer. What's involved is a three-step application of the imaging equation: step 1, the lens equation; step 2 the mirror equation; and step 3, the lens equation, again. The only aspects of this problem that make it more complicated than three independent imaging calculations is that you must make a coordinate change - add or subtract the spacing between two optical elements - when you convert a computed image distance from optical element into an object distance for the next optical element.
HELP: To keep track of the coordinate changes, the orientation of images, and the reality or virtuality of images, diagrams are needed. These should be the three Principal Ray Diagrams corresponding to each optical element. Draw these sequentially after you know from step 1 what kind of image is formed and where it is located, etc. Note: the three principal rays for optical element 1 (the lens) will generally not be principal rays for optical element 2 (the mirror), etc. No problem. Just start over in each step treating the image formed by the previous step as the new object. Sometimes the object will be "virtual", but that may or may not occur in this problem.

Computer systems used in optical design work differently. They trace rays through the entire optical system. This has many advantages including the ability to calculate image intensities and optical aberrations. Your textbook has a complete discussion of the latter very important topic. The advent of computer design tools for optical systems has contributed greatly to improvements in quality and reductions in costs."

I dunno, I just can't get the numbers to work out right.. does anyone maybe just have some ray-tracing software? Oddly enough, I have a copy of 3dsmax, but I think that calculating the necessary index of refraction for a virtual lens that I make and the proper curvature for a virtual mirror would prove to be much more difficult than spending a few hours and working this problem out correctly
 
Physics news on Phys.org
Hey, guys, never mind about this problem

It turns out that I was working on the regular section's homework, and the homework for the honors section (which I'm in) doesn't have this part due until next week
 
.


Wow, that is quite a complex problem! It definitely requires a lot of careful calculations and a good understanding of optical principles. I can see why you're struggling with it at 3 AM. But don't worry, I'm sure with some focus and determination, you can solve it. And even if you don't, it's just a fun problem, so don't stress too much about it.

I don't have any ray-tracing software, but I agree that it would probably be more difficult to use than actually working through the problem. Plus, being able to solve it on your own will give you a great sense of accomplishment.

I think the advice given in the problem statement is really helpful. Drawing diagrams and keeping track of coordinate changes can definitely make it easier to visualize and solve the problem. And who knows, maybe you'll even discover some new optical principles along the way.

Just remember, don't get too discouraged if you can't solve it right away. Take breaks, ask for help if needed, and keep trying. And even if you don't get full points on the assignment, just the fact that you attempted such a challenging problem at 3 AM is impressive in itself. Best of luck to you!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top