A concave mirror produces a virtual image that is 4 times as tall as the object.

Question

A concave mirror produces a virtual image that is 4 times as tall as the object. If the object is 18 cm in front of the mirror, what is the focal length of this mirror? 4.16 (in units of cm) The magnitude of the magnification mI di/do l, where di is the image distance and do is the object distance This equation can be rearranged to find the magnitude of the image distance: di m do Now, di should be made a negative value for a virtual image. Finally, solve for the focal length f using: 1/f 1/d + 1/do Check that your answer for f is larger than do, because the object must be in region 3 between f and the mirror in order to produce an enlarged virtual image. Submit Answer Incorrect. Tries 1/2 Previous Tries

Explanation / Answer

magnification, m= v/u = h'/h= 4

v= image distance

u= object distance

v= 4u= 4*18=72 cm

from sign convention,

u= alwys negative

v= positive( as concave mirror forming virtual image)

applying mirror formula

1/f= 1/v-1/u

-1/f= 1/72-1/18

f= -24 cm

focal length is always negative for concave mirror.

comment in case any doubt

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