Question

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   

Answer 1

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.