When two lenses are used in combination, the first one forms an image that then

Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.60cm -tall object is 51.0cm to the left of a converging lens of focal length 40.0cm . A second converging lens, this one having a focal length of 60.0cm , is located 300cm to the right of the first lens along the same optic axis.

A)

Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0cm .

Enter your answer as two numbers separated with a comma in cm

B)

I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Enter your answer as two numbers separated with a comma in cm

Explanation / Answer

You have no units in your question so I will guess its all supposed to be cm.

Part A)

Use the lens equation...

1/f = 1/p + 1/q

1/40 = 1/51 + 1/q

q = 185.45 cm (So the image is 185.45 cm the the right of this lens)

The height is...

h'/h = -q/p

h'/1.60 = -185.45/51

h' = - 5.82 cm   (The negative means inverted)

Part B)

Since the image for the first is 185.45 cm to the right of the lens, and the second lens is 300 cm to the right, that means that this image, which becomes the object for the second lens, is 114.55 cm to the left of that second lens

1/-60 = 1/114.55 + 1/q

q = -39.38 cm   (This means the image is 39.38cm to the left of this second lens)

h'/h = -q/p

h'/-5.82 = -(-39.38)/(114.55)

h' = 2 cm   (It is upright - re-inverts)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.