Problem 24.41: Combination of lenses I. When two lenses are used in combination,

Question

Problem 24.41: Combination of lenses I. 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. A2.00 cm -tall object is 55.0 cm to the left of a converging lens of focal length 40.0 cm A second converging lens, this one having a focal length of 60.0 cm is located 300 cm to the right of the first lens along the same optic axis. Part A Find the location and height of the image (ca it I1) formed by the lens with a focal length of 40.0 Enter your answer as two numbers separated with a comma My Answers Give Up Submit Part 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. S2, lyre Submit My Answers Give Up

Explanation / Answer

For first lens,

As the lens formula is given by

-1/p + 1/q = 1/f

Where

                p = object position

                q = image position

                f = focal length of lens

And for the given case as

                p = -55 cm

                q = x

                f = 40 cm

So as per

-1/p + 1/q = 1/f

q can be calculated as

                1/q = 1/p + 1/f

So

                q = 146.67 cm

Also as magnitude of magnification, m = 146.67/55 = 2.67

Hence magnitude of height, h = 2.00x2.67 = 5.33 cm

Hence location = 146.67 cm, height = 5.33 cm.

For second lens,

As the lens formula is given by

-1/p + 1/q = 1/f

Where

                p = object position

                q = image position

                f = focal length of lens

And for the given case as

                p = -153.33 cm (as separation between the two lenses is 300 cm and image location for 1st was 146.67 cm)

                q = x

                f = 60 cm

So as per

-1/p + 1/q = 1/f

q can be calculated as

                1/q = 1/p + 1/f

So

                q = 98.7 cm

Also as magnitude of magnification, m’ = 98.57/153.33 = 0.643

Hence magnitude of height, h’ = 8.00x0.643 = 5.143 cm

Hence location = 98.7 cm, height = 5.143 cm.

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