Let Lens 1= a converging lens with focal length f1 = +4 cm. Let Lens 2 =f2 = -12

ID: 2299733 • Letter: L

Question

Let Lens 1= a converging lens with focal length f1 = +4 cm.

Let Lens 2 =f2 = -12 cm)Let an object (represented by an arrow) with a height of 3 cm be placed on the optical axis at x = 0.

Let Lens 1 is placed 8 cm to the left of the object so that the ocation of Lens 1 is x1 = 8 cm.

Let Lens 2 isplaced 12 cm to the left Lens 1 so that the location of Lens 2 is x2 = 20 cm.

a.) Use the Lens Equation to solve for the image distance. Is the image real or virtual?

b.) What is the magnification of the image? What is the height of the image? Is the image upright or

inverted? [Explain how you can tell both from the ray-tracing and from the sign of the magnification.]

C. Where on the x-axis does the image form

for converging lens

u=-8cm

f=+4cm

by lens formula

1/v+1/8=1/4

v=8cm

m=9/-8=-1

h'/3=-1

h'=-3cm(image is real and inverted by converging lens)

now object distance for diverging lens=-(12-8)=-4cm

f=-12cm

by lens formula

1/v+1/4=-1/12

v=6cm

m=v/u=6/-4=-1.5

h''/h'=-1.5

h''=-1.5*-3=4.5cm

a) image distance=6cm to the left of diverging lens.image is real

b) hieght of image=4.5 cm image is upright.

total magnification=4.5/3=1.5

c)16+4+6=26 cm on -x axis

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