A leaf of length h is positioned 71.0 cm in front of a...

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?

Key Concepts

Angular Magnification

Angular magnification refers to the ratio of the angular size of the image as seen by the observer to the angular size of the object when viewed directly. It quantifies how much larger (or smaller) the image appears to the eye, taking into account not only the size of the image but also the distance from the observer, making it a key concept in instruments like magnifiers and telescopes.

Thin Lens Equation

This concept involves the relationship 1/s? + 1/s? = 1/f, where s? is the object distance, s? is the image distance, and f is the focal length of the lens. It is a fundamental tool in geometric optics used to determine the location and nature (real or virtual) of the image formed by a thin lens.

Lateral (Linear) Magnification

Lateral magnification is defined as the ratio of the image height to the object height, often given by m = -s?/s? for thin lenses. The negative sign indicates that a real image is inverted relative to the object. This concept is essential for comparing the sizes of the object and its image in optical systems.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that a leaf of length age is positioned it 71 centimeters in front of a converging lens with a focal length of 39 centimeters.

00:14 An observer views the image of the leaf from a position 1 .26 meters behind the lens, as shown.

00:24 So for part a of this problem, we are asked, what is the magnitude of the length? of the lateral magnification produced by the lens.

00:34 So from the thin lens equation, we know that a real inverted image is formed at the image distance that we can determine by that equation, so that is going to be the prorotic between the object distance, the focal length divided by the difference between these two values.

00:54 So we now substitute those values that we are given.

00:57 The position of the object is and we are given that that is 71 centimeters, and the focal length is also given.

01:06 That is 39 centimeters.

01:09 So we have this divided by the difference between these two values.

01:16 So from here we obtain that the image distance is a positive value of 86 .5 centimeters.

01:26 Now, for part b of the, well, so we need to keep it going, because we need the lateral magnification.

01:33 And the lateral magnification can be obtained from the image size and the size of the real object, that we can also write as minus the image distance over the object distance.

01:47 So we're going to substitute the values that we have for the image distance, which is 86 .5 centimeters divided by the object distance, which is 71 centimeters.

01:58 So from this, we obtain a magnificence.

02:00 Of minus 1 .22.

02:05 Now, if we want the magnitude of this magnification, we just take the absolute value and we obtain 1 .22.

02:13 So that's a solution for this part of the problem...

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