A transparent photographic slide is placed in front of a converging lens with a

Question

A transparent photographic slide is placed in front of a converging lens with a focal length of 2.59 cm. An image of the slide is formed 14.8 cm from the slide.

(a) How far is the lens from the slide if the image is real? (Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.) cm cm
(b) How far is the lens from the slide if the image is virtual? (Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.) cm cm

Explanation / Answer

(a)

If image is real

object distance = s

image distance = s'

s + s' = 14.8 cm

s' = 14.8 - s

focal length f = 2.59 cm

from lens equation

1/f = 1/s + 1/s'

1/2.59 = 1/s + 1/(14.8-s)

s = 3.35 cm , 11.4 cm

object distance s = 3.34 cm , 11.4 cm

======================

(b)

If image is virtual

object distance = s

image distance = -s'

s - s' = 14.8

s' = 14.8 + s

focal length f = 2.59 cm

from lens equation

1/f = 1/s + 1/s'

1/2.59 = 1/s - 1/(14.8+s)

s = 2.25 cm

object distance s = 2.25 cm

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