Question

A convex thin lens has f= 5cm. if a 3 cm object is placed 10 cm in front of this lens, where is the image formed? How tall is the image? is the image real or virtual? is the image upright or inverted?
Repeat for if object is placed 2.5 cm in front of the lens.

Answer 1

Given that

f = 5 cm

ho = 3 cm

do = 10 cm

We know that

1 / f = 1/ do + 1/di

==> di = do * f / [ do - f]

= 10 * 5 / [ 10 -5]

= 10 cm ( real image)

magnification m = - di / do

= - 10 / 10

= -1 ( image is inverted)

Height of the image

m = hi / ho

==> hi = m * ho = -1 * 3 = - 3 cm ( image is inverted)

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Given that

f = 5 cm

ho = 3 cm

do = 2.5 cm

We know that

1 / f = 1/ do + 1/di

==> di = do * f / [ do - f]

= 2.5 * 5 / [ 2.5 -5]

= - 5 cm ( virtual image)

magnification m = - di / do

= - -5 / 2.5

= 2 ( image is upright)

Height of the image

m = hi / ho

==> hi = m * ho = 2 * 3 = 6 cm ( image is upright)