Question

A converging lens of focal length 8.030 cm is 20.4 cm to the left of a diverging lens of focal length -6.73 cm . A coin is placed 12.2 cm to the left of the converging lens.

Find the location of the coin's final image. Answer: to the right of the diverging lens

PART B: Express your answer using two significant figures.

di= __________ cm to the right of the diverging lens

PART C: Find the magnification of the coin's final image.

m=___________

Answer 1

The image of the first lens becomes the object of the second So 1/s'1 + 1/s1 = 1/f.1.....=>1/s'1 = 1/f 1- 1/s1 = 1/8.03 - 1/12.2 = 0.04256 Therefore s'1 = 23.496cm This means the object for the second lens is 23.496 - 20.4 = 3.096 cm behind the diverging lens...so s = -3.096 (a virtual object) Now for the second lens we have 1/s'2 + 1/s2 = 1/f2 So 1/s'2 = 1/f 2- 1/s2 = 1/(-6.73) - 1/(-3.096) = 0.1744 So s'2 = 5.734cm So the image forms 5.734 cm to the right of the diverging lens Final magnification = m1*m2 = s'1/s1*s'2/s2 = 23.496/12.2*5.734/3.096 = 3.72