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 1.00 cm -tall object is 60.0 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. A-Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 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.

          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 1.00 cm -tall
object is 60.0 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.
A-Find the location and height of the image (call it I1)
formed by the lens with a focal length of 40.0 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.

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Added by Christopher C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

07/20/2022

Step 1

Solving this equation gives us \( t_1 = -24 \, \text{cm} \), meaning the first image is formed 24 cm to the left of the first lens. Show more…

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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 1.00 cm -tall object is 60.0 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. A-Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 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.

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Transcript

00:02 Hi, in the first sub -part of this question, using lens formula, we can write 1x t i minus 1 by t o equal to 1 by f.

00:20 From the question object distance is minus 60 centimeters and focal length is minus 40 centimeters for the first lens.

00:35 So we can we can write 1x t i minus 1 by minus 60 is equal to 1 by minus 40 from this we can get t i equal to minus 24 centimeters so the first image is formed at minus 24 centimeters to the left of first line.

01:23 Now using the equation for magnification we can write m is equal to minus di i .d .o that is equal to h i by h o and we have h o as one centimeters so we can write this as minus of minus 24 by minus 50 is equal to is equal to to h .i .1.

02:02 From this we will get h .i.

02:06 Is equal to minus 0 .48 centimeters as the value for the image height.

02:17 So that is the required answer for the first subpart of this question.

02:24 Now in the second subpart for the second lens, the object distance is, minus of a given length of 300 cm plus 24 centimeters which we obtained the image distance in the first subpart.

02:53 So this will be equal to minus 2 76 centimeters.

02:59 So now using the lens formula we can write 1 by image distance we will take as ti dash minus object distance is minus 2 76.

03:12 Which is equal to focal length in this case is 60 so it will be 1 by 60 from this we can get the i dash is equal to 76 .7 centimeters so second image is formed at 76 .7 centimeters to the right of second lens so again we'll use the equation for magnification, that is m2 is equal to t .i...

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