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Problem 25 Easy Difficulty

A leaf of length $h$ is positioned 71.0 cm in front of a converg- ing lens with a focal length of 39.0 cm. An observer views the image of the leaf from a position 1.26 m behind the lens, as shown in Figure P25.25. (a) What is the magnitude of the lateral magnification (the ratio of the image size to the object size) produced by the lens? (b) What angular magnification is achieved by viewing the image of the leaf rather than viewing the leaf directly?

Answer

a. -1.22
b. no answer available

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Top Physics 103 Educators
Elyse G.

Cornell University

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Video Transcript

{'transcript': "Okay, so we're doing Chapter 34 problems. 59 here. So it says the focal length of an eyepiece microscope is 1.8 centimeters, and the focal length of the objectives 0.87 years separated a distance of 19.77 years. As all shown in this figure over here. So treating all the Windsors as Stine lenses part ay asks what is the distance from the objective to the object be viewed. So first, we need to realize that for this microscope system that final images at a distance infinity because that's how we define is that so if we're working backward from the park the second one's case, if we plug in our equation, we see that the object distance is equal of the folk. Oi! So the object distance is 1.8 centimeters to the left of eyepiece. So this is object is also the image formed from the objective so we can set up next things. Let's write this down. So I'm gonna call this image one who is not. The final images is the image between them. So we're gonna call this right here. This this is Mitch Warden 1.8 centimeters away to the left of the second winds. And that makes it 17.9 centimeters to the right of the other leads. So this being object distance for the second case on this object distance for the second case on this, the image distance for the first case. So now you know where the image is formed from the first lens. We working backwards again. So we're trying to get to where the object is so we can use our Finland's equation. Rearrange for the object. So we're trying to sell for this, which is s prime f over s prime this Beth. So we can put that in. So 17 point times 0.8, Teen Court one Ms is equal to 8.377 Oops. Sorry. That's not centimeters that this actually we're doing in centimeters. So we did the rest incident here, so keep that in. This is 0.837 seven years or 8.37 No meters. Cool. So that's the distance. It is around 8.37 millimeters to the left of the first. So part B asks what is the, uh what is angular magnification of side. Wrong question. What is the magnitude of the linear magnification produced by the objective? So to do that, we obviously know we can remedy the image distance over the object distance here, So that is negative. 17 point now over point A 37 So we plugged that in, and that becomes a magnification negative. 21.4. Awesome negative because it is inverted. So part see asks what is the overall back angular magnification of the microscope? So from this we can we can use the equation 30 or 23 relating the total too linear of the objective in one. Call it in 12 times into where into is given as 25 centimeters over left. So school unclog Madden 25 centimeters over 1.8 centimeters. So this is 13.9. So that means our overall magnification is negative. 21.4 times 13 point. Not so if we plug that in, we get a negative 297 for a magnification. So yeah, this microscope magnifies roughly 300 times. It's very, very good"}

University of Oregon
Top Physics 103 Educators
Elyse G.

Cornell University

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg