Optics did my professor screw up?

AI Thread Summary
The discussion centers around a physics problem involving a cat in front of a mirror, where a virtual image is formed. The initial calculations for the radius of curvature (R) yield 1.82m, while the answer key states it should be 2.2m. A key point raised is the need to consider the virtual image distance (di) as negative, which affects the calculations. Correctly substituting di as -10.0m leads to the proper application of the mirror formula. The conclusion suggests that the professor's guidance may have overlooked this crucial detail in the problem setup.
Dart82
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Homework Statement


A cat is standing 1m in front of a mirror. a virtual image is formed 10.0m behind the mirror. what is the radius of curvature for the mirror?


Homework Equations


1/f = (1/do)+(1/di)
f=+/-(1/2)*R



The Attempt at a Solution


(1/1m + 1/10m)^-1 = .909
.909*2 = 1.82

According to my answer key, R should be equal to 2.2m. the only way i get that answer is if i take (1/do)+(1/di) and not take its reciprocal...i would just plug it straight into the focal length equation to solve for R. Did he make a mistake??
 
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Dart82 said:

Homework Statement


A cat is standing 1m in front of a mirror. a virtual image is formed 10.0m behind the mirror. what is the radius of curvature for the mirror?


Homework Equations


1/f = (1/do)+(1/di)
f=+/-(1/2)*R



The Attempt at a Solution


(1/1m + 1/10m)^-1 = .909
.909*2 = 1.82

According to my answer key, R should be equal to 2.2m. the only way i get that answer is if i take (1/do)+(1/di) and not take its reciprocal...i would just plug it straight into the focal length equation to solve for R. Did he make a mistake??

The image is virtual so di should be taken to be negative.
 
you miss substitution
di = -10.0 m :biggrin:
 

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