Question

In: Physics

A photographic slide is to the left of a lens. The lens projects an image of...

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 7.00 m to the right of the slide. The image is 73.5times the size of the slide.

(a) How far is the slide from the lens?
________ m

(b) Is the image erect or inverted?

erectinverted

(c) What is the focal length of the lens?
_______ m

(d) Is the lens converging or diverging?

convergingdiverging

Expert Solution

given

image distance di = 7.00 m

= 700.0 cm

magnification m = - 73.5 ( as image is real)

object distance = do

focal length = f

a.

magnification m = - di/do

- 73.0 = - 700.0 / do

slide distance do = 700.0 / 73.5

= 9.52 cm = 0.0952

b.

Image is inverted as magnification is negative.

c.

1/f = 1/di - 1/do

= 1/700.0 - 1/( - 9.52)

( do is taken as - ve as the object lies to the left of lens)

focal length f = 700.0 * 9.52 / (700 + 9.52)

= 9.39 cm

= 0.0939 m

d.

Since focal length is + ve, lens is converging.

genius_generous answered 1 year ago

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