Physics, 12.07.2019 22:20 Arielledt10

Areal object with height of 3.20 cm is placed to the left of a converging lens whose focal length is 90cm. the image is on the right of the lens and 4.50cm tall and inverted. where is the object? where is the image? is the image real or virtual?

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Answer from: annamcda301

$d_{o}$ = 154 cm

$d_{i}$ = 216.6 cm

The image is real

Explanation:

$h_{o}$ = height of the object = 3.20 cm

$h_{i}$ = height of the image = 4.50 cm

f = focal length of the converging lens = 90 cm

$d_{o}$ = object distance from the lens = ?

$d_{i}$ = image distance from the lens = ?

using the equation for magnification

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

$\frac{4.50}{3.20}= \frac{d_{i}}{d_{o}}$

$d_{i}$ = 1.40625 $d_{o}$ eq-1

using the lens equation

$\frac{1}{d_{i}} + \frac{1}{d_{o}} = \frac{1}{f}$

using eq-1

$\frac{1}{( 1.40625)d_{o}} + \frac{1}{d_{o}} = \frac{1}{90}$

$d_{o}$ = 154 cm

Using eq-1

$d_{i}$ = 1.40625 $d_{o}$

$d_{i}$ = 1.40625 (154)

$d_{i}$ = 216.6 cm

The image is real

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Answer from: Quest

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Areal object with height of 3.20 cm is placed to the left of a converging lens whose focal length is...

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