Why are magnified images dimmer

Join Here! Already have an account? Log in. Hi, everyone. This is the problem based on image permission by the mirror. In this problem, it is given explain, right magnified image formed by the mirror seems to be dimmer than that off original object here. The explanation. There is less light per unit area but unit area because larger image allow the light toe spread out more, which makes the on large image seemed dimmer, compared toa brightness off normal size. Normal sites. Image, that's all.

Thanks for watching it. If there are two converging lenses in a compound microscope, why is the imag… Magnifying glasses normally are used to produce images that are larger than … But how do we know some exotic combination of lenses can't increase visual image brightness?

Suppose it could. Now, take the smallest possible number of lenses which can combine to increase visual brightness of a scene. Since no single lens can do this, we know the system must have at least two lenses in it. Call the lens closest to the eye -- the one we look through -- the last lens.

Now, take the last lens away. Call the remaining system the reduced system. The last lens was forming an image from an image -- i. Look directly at that image, produced by the reduced system. Observe the brightness of that image. Since the original system used the smallest possible number of lenses to produce brightness enhancement, and the reduced system has fewer lenses, we know that the apparent brightness of the image produced by the reduced system must be no brighter than what you see with the unaided eye.

Now put the last lens back. The last lens is a simple lens, and it must be acting on the image produced by the reduced system via one of the four configurations discussed above. So, the last lens can't increase the apparent brightness of the image beyond what the reduced system produced on its own. And so the entire system must also not be able to produce an apparent image brightness increase.

Holes in the Proof I'm aware of two major holes in the proof. I assumed without proof that the brightness distribution in the cone of light emitted by a lens is the same as the distribution in the cone entering the lens. Without this assumption, the induction step doesn't work. I believe this assertion is true; a geometric argument to prove it seems reasonably straightforward but some of the details may be tricky.

I used small angle approximations everywhere. This makes sense for a single lens used to view a distant scene. It doesn't make much sense for the internal lenses in a complex optical system, however.

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And, the light collected exactly compensates for the magnification, so the surface brightness will appear unchanged from that of the naked eye view. If the objective of the binocular is even bigger, so that at 7x the exit pupil is bigger than the eye's entrance, some of the light collected by the objective spills onto the iris and is wasted. But the usable, effective, central part of the objective is still 7x wider than the eye, and still functions just like the smaller objective of the first binocular: the eye is filled with light, and the brightness of the image is the same as in that case.

There's another common case where surface brightness will appear reduced. If the first binocular's exit pupil exactly matches the eye's pupil at 7x, consider what happens if its magnification is increased to 10x.

All the collected light still enters the eye, since the exit pupil is now smaller than the eye's pupil. But the image, being larger, has the light diluted over a greater area, and the surface brightness will appear reduced. A similar reduction in brightness happens if the magnification was held at 7x, but the size of the objective was reduced. So, increased magnification leads to diminished brightness in the range where the exit pupil is smaller than the eye's entrance.

When the exit pupil is equal or larger than the eye, the surface brightness will be the same as seen by the naked eye. The image cannot be made brighter, since any extra light will only spill outside the eye, ineffectively. I have assumed that no light is lost in passing through the binocular. No coatings are that good, of course, I just wanted to keep it simple, as far as it goes. And, people don't always see in agreement with simple predictions, there are always complicating factors.

But, that's the basic argument. This thread is more than 11 years ago old. It's likely that no further discussion is required, in which case we recommend starting a new thread. If however you feel your response is required you can still do so. I understand this is an old thread, but want to reply anyway.

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