An object 10.0 cm tall is placed at x=0.0 cm and a converging lens of focal length 30.0 cm is placed at x=40.0 cm. A diverging lens of focal length 30.0 cm is then placed at x=140.0 cm. (a) Find the location of the image (along the x-axis) and the height of the image formed by the converging lens. (b) Find the location of the image (along the x-axis) and the height of the image formed by the diverging lens.

Question

contestada

Respuesta :

Xema8 Xema8 · Answer 1 · 2019-08-29T18:13:05+02:00

Answer:

a ) 120 cm

size = 30 cm

b ) 12 cm from diverging lens

size = 12 cm

Explanation:

For converging lens

object distance u = - 40 cm

Focal length f = 30 cm

From lens formula

[tex]\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

frac{1}{v}+\frac{1}{40}=\frac{1}{30}

v = 120 cm[/tex]

m=[tex]\frac{v}{u}[/tex]

m=[tex]\frac{120}{40}[/tex]

=3

Size of image = size of object x Magnification

= 3 x 10 = 30cm

For diverging lens

object distance u = 140 - 120 = 20 cm

focal length f = -30 cm

frac{1}{v}+\frac{1}{20}=-\frac{1}{30}

v = -12 cm

m = 12/20 = 0.6

size of final image = .6 x 30 = 18 cm