A3.0-cm-tall object is 50 cm in front of a converging lens that has a 25 cm focal length. you may want to review (pages 589 - 592) . part a calculate the image position. express your answer to two significant figures and include the appropriate units. s′ s ′ = nothing nothing request answer part b calculate the image height. type a positive value if the image is upright and a negative value if it is inverted.

A) 50 cm from the lens

We can use the lens equation:

where

f is the focal length

s is the object distance

s' is the image distance

In this problem, we have

f = 25 cm

s = 50 cm

So using the formula we can find the image distance, s':

and the positive sign means the image is real.

B) -3.0 cm (inverted image)

The image height is given by the magnification equation:

where

y' is the image height

y = 3.0 cm is the object height

s' = 50 cm is the image distance

s = 50 cm is the object distance

Substituting into the formula, we find

And the negative sign means the image is inverted.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2