What is the image distance for a concave mirror using the mirror equation?

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SUMMARY

The image distance for a concave mirror using the mirror equation is calculated as di = (f * do) / (do - f). In this case, with an object distance (do) of 2 cm and a focal length (f) of 4 cm, the resulting image distance (di) is -4 cm. A negative image distance indicates that the image is virtual and located behind the mirror, which aligns with the sign convention applied in optics. This conclusion confirms the correct application of the mirror equation.

PREREQUISITES
  • Understanding of the mirror equation: 1/do + 1/di = 1/f
  • Knowledge of focal length and object distance in optics
  • Familiarity with sign conventions in optics
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the properties of virtual images in concave mirrors
  • Learn about the applications of concave mirrors in real-world scenarios
  • Explore the derivation of the mirror equation in detail
  • Investigate the differences between concave and convex mirrors
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the behavior of concave mirrors and image formation.

toothpaste666
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Homework Statement


mirrorq.png

B) Using the mirror equation, find the image distance.

Homework Equations


1/do + 1/di = 1/f

The Attempt at a Solution


mirrorrays.png
[/B]
B)
do = 2cm f = 4cm
1/do + 1/di = 1/f
1/di = 1/f - 1/do
1/di = do/f(do) - f/f(do)
1/di = (do-f)/(f*do)
di = (f*do)/(do-f) = (4cm * 2cm)/(2cm - 4cm) = (8cm^2)/(-2cm) = -4cm
di = -4cm

 
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is it ok that my image distance came out negative? if i am correct this is fine if the image is behind the mirror
 
It is correct, the distance of a virtual image is negative in the frame of sign convention you applied.
 
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thank you!
 
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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