The focal length of a simple magnifier is 8.00 cm . Assume the magnifier is a thin lens placed v

step 1 So we have a focal length of a magnifier is 8 centimeters. So I'm going to draw our magnifier he

The focal length of a simple magnifier is 8.00 cm . Assume...

The focal length of a simple magnifier is 8.00 $\mathrm{cm} .$ Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 $\mathrm{cm}$ in front of her eye? (b) If the object is 1.00 $\mathrm{mm}$ high, what is the height of its image formed by the magnifier?

Key Concepts

Virtual Image Formation

Virtual image formation occurs when the image produced by an optical device cannot be projected onto a screen because it appears to be located on the same side of the lens as the object. Magnifiers use this principle to create enlarged, upright images that are positioned at the observer's near point, thereby enhancing the detail of the view.

Near Point

The near point is the minimum distance at which a person can comfortably focus on an object, typically around 25 cm for a normal human eye. In optical instruments like magnifiers, placing the image at the near point maximizes the angular size and makes viewing easier and more comfortable.

Thin Lens Equation

The thin lens equation (1/f = 1/do + 1/di) is fundamental in optics, relating the focal length of a lens to the distances of the object (do) and the image (di) from the lens. This relationship is used to determine either the position of an image for a given object placement or the necessary object distance to obtain an image at a predetermined location.

Magnification

Magnification in lens systems describes how much larger or smaller an image is compared to the object. It is defined as the ratio of the image height to the object height and, in thin lens systems, is given by m = -di/do. This concept helps in assessing the size change and orientation of an image relative to the original object.

Transcript

00:01 So we have a focal length of a magnifier is 8 centimeters.

00:06 So i'm going to draw our magnifier here.

00:08 It's a lens with a focal length of 8 centimeters.

00:12 Here is our focus for each of these 8 centimeters.

00:19 And we're saying this is a converging lens because it's a magnifier.

00:23 I assume it's a thin lens.

00:24 Lens is very close to the eye.

00:27 How far in front of the magnifier should an object be placed if...

00:30 Okay, so first let's draw the eye real quick.

00:33 Here is someone's eye.

00:36 I'm going to draw their pupil and we're going to assume they have green eyes because that's the color that i have handy.

00:44 How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25 centimeters in front of the eye? let's see.

00:52 So over here, this is the focus.

00:58 Here, 25 centimeters is going to be the image.

01:02 You want to find the object.

01:03 Well, 1 over the focal length is equal to 1 over the object plus 1 over the image position.

01:10 So if we solve for the object here, let's see, 1 over f minus 1 over i equals 1 over o.

01:21 The object should equal 1 over f plus 1 over i inverse...

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