Concave mirror - finding object and image distance

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A concave mirror with a focal length of 27.2 cm has an object-image distance of 50.2 cm, prompting a calculation for both object and image distances. The initial equations used include the mirror formula and the relationship between object and image distances. The user attempts to solve for the object distance but encounters confusion with sign conventions, leading to incorrect negative values. It is emphasized that both distances must be positive, as the object is placed beyond the center of curvature. Clarification on the correct application of the equations is needed to resolve the discrepancies in the calculations.
phys62
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Homework Statement


A concave mirror has a focal length of 27.2 cm. The distance between an object and its image is 50.2 cm. Find (a) the object and (b) the image distance assuming that the object lies beyond the center of curvature.


Homework Equations


f=1/2 R
1/object distance + 1/image distance = 1/f


The Attempt at a Solution


image distance - object distance = 50.2 cm
(image distance + object distance)/(image distance*object distance) = 1/f
50.2/(image distance*object distance)=1/27.2
image distance*object distance=1365.44

image distance=50.2 + object distance

substitute in and get: (50.2 + object distance)object distance = 1365.44
quadratic equation solves for object distance = 69.77 or 19.57 cm


Am I even doing this right?? If so, I'm stuck from here.. Thanks for any help you can give!
 
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phys62 said:

Homework Statement


A concave mirror has a focal length of 27.2 cm. The distance between an object and its image is 50.2 cm. Find (a) the object and (b) the image distance assuming that the object lies beyond the center of curvature.


Homework Equations


f=1/2 R
1/object distance + 1/image distance = 1/f


The Attempt at a Solution


image distance - object distance = 50.2 cm
(image distance + object distance)/(image distance*object distance) = 1/f
Whoops.
image distance - object distance = 50.2 cm
But you substituted 50.2 cm for image distance + object distance

Try substituting for "image distance" in
1/object distance + 1/image distance = 1/f​
 
Can anyone help me figure out if I'm even on the right track?
 
so I tried saying image distance = 50.2 + object distance

substituting that into 1/object distance + 1/image distance = 1/f

I got object distance = -620.65 cm
then image distance would = -620.65 + 50.2 = -570.45 cm

This is incorrect... Have I done this right, and the 620.65 simply needs to be positive? Or am I completely off track?
 
Something is wrong there. Those numbers definitely don't satisfy the equation,

1/object distance + 1/image distance = 1/f

1/(-620.65) + 1/(-570.45) = -0.00336
While 1/f = 1/27.2 = +0.0367

Since f (and also 1/f) is positive, the object and image distances cannot both be negative.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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