Light reflections in concave/convex mirrors

  • Thread starter Thread starter freepancakes
  • Start date Start date
  • Tags Tags
    Light Mirrors
AI Thread Summary
The discussion revolves around solving a physics problem involving light reflections in a hubcap that acts as a concave and convex mirror. The key questions include determining the distance of the man's face from the hubcap, the radius of curvature, magnification for each image, and the nature of the images (real or virtual) as well as their orientation (upright or inverted). The equations provided for solving the problem include the mirror formula and magnification formula. Participants suggest using the same object distance for both cases to derive the necessary values, emphasizing that the images are virtual and upright. The conversation highlights the importance of understanding the relationships between object distance, image distance, and focal length in mirror optics.
freepancakes
Messages
5
Reaction score
0

Homework Statement



A man polishes the insided and outside of a hubcap that is a section of a sphere. When he looks into one side of the hubcap, he sees an image of his face 30cm behind the hubcap. He then turns the hubcap over and sees another image of his face 10 cm behind the hubcap.
a)How far is his face from the hubcap?
b)What is the radius of curvature of the hubcap?
c)What is the magnification for each image?
d)Are the images real or virtual?
e)Are the images upright or inverted?


Homework Equations



1/Di + 1/Do = 1/f Di= distance of image Do=distance of object f= focal point

M= Hi/Ho = -Di/Do M= magnification Hi= height of image Ho= height of object


The Attempt at a Solution



a) i believe you have to use the first equation but i am unsure of how to use the given data in it, could you possibly do... 1/-30 + 1/-10 + 2/Di = 2/f ?

b)find the focal point and double it i believe

c)use M= -Di/Do for both but i don't know the Do from part "a"

d)virtual because the image is behind the mirror

e)upright because all virtual images are upright?

-- i think i am on the right track, but not sure--
 
Physics news on Phys.org
Keeping Do same, write two equation for two cases. Since R is the same fore both the case, f is the same, but sign will change. From these equation you can find Do. Once you find Do, other thing are easy.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top