Repeat Problem 42 if the eyepiece location is adjusted slightly so that the final image is at th

step 1 In this problem, we have a microscope that produce a final image. The final image that is produc

Repeat Problem 42 if the eyepiece location is adjusted...

Key Concepts

Near Point

The near point is the closest distance at which the human eye can comfortably focus, typically around 25 cm for a normal eye. In optical instruments, having the final image at the near point allows the viewer to see a virtual image with maximum angular size, even though the image is no longer at optical infinity.

Virtual Image

A virtual image is formed when light rays diverge after passing through an optical system, making the image appear to be located at a position from which the rays seem to originate. Virtual images cannot be projected onto a screen but can be conveniently observed by the eye, such as when an image is formed at the near point of the viewer.

Image Formation by Lenses

This concept involves the application of the lens equation to determine the relationship between object distance, image distance, and focal length. Adjusting the eyepiece or secondary lens changes the final image position, and in this case, repositioning the eyepiece yields a final virtual image placed at the viewer’s near point.

Angular Magnification

Angular magnification is the ratio of the angle subtended at the eye by the image to the angle subtended by the object when viewed directly. Optical devices are designed to maximize angular magnification so that the observer perceives an enlarged image, whether the final image is adjusted to be at infinity for relaxed viewing or at the near point for maximum enlargement.

Transcript

00:03 In this problem, we have a microscope that produce a final image.

00:10 The final image that is produced by this microscope is located at a near point.

00:18 And we have some information about this microscope.

00:23 We have the focal length of the objective.

00:28 Also for the ips.

00:32 And we know that the image that is produced by the...

00:36 Objective this is not a final image is the image produced by the objective is at this distance from the objective well in the first part of the problem we want to compute where is the distance between the lenses in order to get a distance between the lenses we need to compute this value pe that is the object distance for the ips this i piece works as a magnifier that produce a final image at a point in or at a near point.

01:20 We're going to use a result from problem 35, where from problem 35 of the same section.

01:36 In that problem we have shown that in this case, when we have a magnifier that produce an image at a near point, the object distance, the object distance here, is just the n times the focal length of this lens, that is the focal length of the i -piece divided by n plus f e.

02:15 So this is one result that we got in problem 35, but we also shown another result.

02:27 We got the angular magnification because these ips.

02:33 So the angular magnification is just n over the focal length plus 1.

02:44 So these two results are very useful in this problem.

02:51 So in the first part of the problem, we want to compute the distance between the lenses from the figure.

02:58 We can see that that distance is just qo plus pe.

03:05 Qo is information from the problem is 0 .65.

03:13 And pe, we can get from this result.

03:18 We have just mentioned is just n, that is remember that.

03:25 The near distance is 0 .25 meters always times the focal length that is 0 .280 and all of this divided by the sum.

03:53 Okay, and all of this, if we compute all of this, this is just 0 .190 mirrors.

04:06 In the next part of the problem, we want to compute what is the total angular magnification...

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