Query by tay: Physics aid please! Convex Mirror and Magnification?
1) A convex mirror has a focal length of 60 cm. What is the position of the resulting image if the image is erect and 9 times smaller than
the object? Answer in units of cm.

two)The magnification created by a converging
lens is located to be −2.56 for an object placed .219 m from the lens. What is the focal length of the lens? Answer in units of m.

Please show function. Thanks!

Very best answer:

Answer by kuiperbelt2003
1) we use

1/p + 1/q=1/f

p=object distance
q=image distance
f=focal length (adverse for a convex mirror)

we also use

m=-q/p where m is magnificaiton

we are told that m=+1/9 (the plus follows from the truth the image is erect btw, convex mirrors always type erect virtual images)

if m=1/9, then -q/p=1/9 and p=-9q

substitute into the 1st equation:

-1/9q+1/q=-1/60 (don’t forget the focal length is unfavorable)

8/9q=-1/60 or q=-53.33 which means the image forms 53.33 cm behind the mirror, and the original object was 480cm in front of the mirror

two) we now have a converging lens the focal length is optimistic the sign convention for converging lenses is that image distances behind the lens are optimistic object distances in front of the lens are negative

we are told m=-two.56

this indicates that -q/p=-2.56 so that q=two.56p

we are told that p=.219, from which we know quickly that q=.56 cm (q=2.56p=two.56×0.219)

1/.219+1/.56=1/f
f=.15 cm

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Physics aid please! Convex Mirror and Magnification?
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