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3 years ago

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The virtual image produced by a convex mirror is one third the size of the object.(a) If the object is 51 cm in front of the mir

ror, what is the image distance? (Include the sign of each answer.)cm(b) What is the focal length of this mirror?cm

1 answer:

Nimfa-mama [501]3 years ago

4 0

(a) -17 cm

Magnification equation:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o$ is the size of the object

$q$ is the distance of the image from the mirror

$p$ is the distance of the object from the mirror

In this problem, we have:

p = 51 cm (the object is 51 cm in front of the mirror)

$\frac{h_i}{h_o}=\frac{1}{3}$

So, we can find q, the image distance:

$q=-\frac{h_i}{h_o}p=-\frac{1}{3}(51 cm)=-17 cm$

and the negative sign means the image is virtual.

(b) -25.5 cm

Mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

p = 51 cm

q = -17 cm

Substituting and re-arranging the equation, we find the focal length of the mirror:

$\frac{1}{f}=\frac{1}{51 cm}-\frac{1}{17 cm}=-\frac{2}{51 cm}\\f=-\frac{51 cm}{2}=-25.5 cm$

and the negative sign is due to the fact it is a convex mirror.

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