Question

Two thin lenses with a focal length of magnitude 9.00 cm, the first diverging and the second converging, are located 6.75 cm apart. An object 3.20 mm tall is placed 15.0 cm to the left of the first (diverging) lens. How far from this first lens is the final image formed? Is the final image real or virtual? What is the height of the final image? Is it erect or inverted?

          Two thin lenses with a focal length of magnitude 9.00 cm, the
first diverging and the second converging, are located 6.75 cm
apart. An object 3.20 mm tall is placed 15.0 cm to the left of the
first (diverging) lens. 
How far from this first lens is the final image formed?
Is the final image real or virtual?
What is the height of the final image?
Is it erect or inverted?

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Added by Josefina W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/17/2023

Step 1

First, we need to find the image formed by the first lens. We can use the lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. Since the first lens is diverging, Show more…

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Two thin lenses with a focal length of magnitude 9.00 cm, the first diverging and the second converging, are located 6.75 cm apart. An object 3.20 mm tall is placed 15.0 cm to the left of the first (diverging) lens. How far from this first lens is the final image formed? Is the final image real or virtual? What is the height of the final image? Is it erect or inverted?

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Transcript

00:02 In this question, one diverging lens and one converging lens i .e.

00:11 Two thin lenses are placed at a distance of 6 .75 cm apart.

00:24 A object is placed to the left of diverging lens at 15 cm to the left and the height of the object is 3 .20 mm.

00:39 We have to find how far from the first lens the final image is formed.

00:47 Therefore, the focal length of diverging lens will be minus 9 cm and converging lens will be plus 9 cm.

01:00 We have done with all the information.

01:02 Now we will proceed to the question.

01:04 The lens formula is 1 by v minus 1 by u is equal to 1 by f.

01:12 We have to find 1 by v.

01:14 This is for first lens i .e.

01:20 The concave lens.

01:24 1 by v minus 1 by u is equal to 1 by f.

01:28 1 by v minus 1 by 50 is equal to 1 by minus 9.

01:39 Focal length of diverging lens is negative.

01:42 1 by v plus 1 by 50 is equal to minus 1 by 9.

01:50 Going further up, minus 1 by 9 minus 1 by 50.

01:55 Therefore, v will be 1 by v is equal to minus 45.

02:04 We took the fgm.

02:06 5 plus 3 i .e.

02:09 V is equal to minus 45 divided by 8 i .e...

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