Question

For the last question is the image upright, downward, virtual, or real?

PRACTICE IT Use the worked example above to help you solve this problem. An object 2.56 cm high is placed 20.8 cm from a convex mirror with a focal length of 8.10 cm. (a) Find the position of the image. -5.829 cm (b) Find the magnification of the mirror. 0.28 (c) Find the height of the image. 0.716 cm EXERCISE HINTS: GETTING STARTED I'M STUCK! Suppose the object is moved so it is 4.05 cm from the same mirror. Repeat parts (a) through (c) (а) д %3 cm (Б) М %— (c) h' = cm The image is-Select

Answer 1

Object distance, $o=4.05cm$

Focal length, $f=-8.10cm$

Use formula $\frac{1}{f}=\frac{1}{o}+\frac{1}{i}$

$-\frac{1}{8.10cm}=\frac{1}{4.05cm}+\frac{1}{i}$

$-\frac{1}{8.10}-\frac{1}{4.05}=\frac{1}{i}$

$\frac{-4.05-8.10}{8.10*4.05}=\frac{1}{i}$

$-\frac{12.15}{8.10*4.05}=\frac{1}{i}$

$-\frac{8.10*4.05}{12.15}=i$

(a) ANSWER: ${\color{Red} i=-2.7cm}$

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Magnification, $M=-\frac{i}{o}$

$M=-\frac{-2.7cm}{4.05cm}$

$M=\frac{2.7}{4.05}$

(b) ANSWER: ${\color{Red} M=0.667}$

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Magnification, $M=\frac{h_{i}}{h_{o}}=-\frac{i}{o}$

$\frac{h_{i}}{h_{o}}=-\frac{i}{o}$

$\frac{h_{i}}{2.56cm}=-\frac{-2.7cm}{4.05cm}$

$\frac{h_{i}}{2.56}=\frac{2.7}{4.05}$

$h_{i}=\frac{2.7*2.56}{4.05}$

(c)ANSWER: ${\color{Red} h_{i}=1.707cm}$

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Image distance is negative, so image formed is virtual

Image height is positive, so image formed is upright

ANSWER: Image is virtual and upright.

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