An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. The distance between the lenses is 5.0 cm. Find the location and size of the final image.

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form