# Refraction (aperture, apparent depth)

1. Dec 12, 2012

### rishch

1. The problem statement, all variables and given/known data

1: A decrease in the aperture of a lens changes the
(A)Size of the image
(B)Intensity of the image
(C)Portion of the image
(D) All of these

2: A vessel is filled with two different liquids which do not mix. One liquid is 40cm deep and has n=1.6 and the other is 30cm deep and has n=1.5. What is the apparent depth of the vessel when viewed along the normal?

2. Relevant equations

1: None

2: n=real depth/apparent depth
3. The attempt at a solution

1: The first ones answer is supposed to be (B)Intensity of the image. I don't know what intensity of an image is though, and I googled it but nothing came up. My answer would've been (C)Portion of the image, but I don't know what that exactly means either, but I'm assuming it means how much of the image you can see, and since the aperture decreases, that should decrease too.

2: This one is really weird an I have no idea how to solve it. The solution is-

n1 = 1.6, d1 = 40 cm
n2 = 1.5, d2 = 30 cm
Apparent depth d = ?
n = real depth (D)/apparent depth (d)
⇒ d = D/n = d1/n1 + d2/n2
d = (25 + 20) cm = 45 cm

But how do they get d = D/n = d1/n1 + d2/n2? They have a video which goes into a bit more detail but I still don't get it :/ Could someone explain?

2. Dec 12, 2012

### haruspex

Every part of the scene still has straight line access to the centre of the aperture, so the light rays entering the aperture still represent the entire scene. But beyond the aperture is the lens. As the aperture is narrowed, each part of the scene can reach less of the lens and therefore contributes fewer photons to the image. Thus the brightness of the image is reduced.
The refractive index, n, tells you the relative speed of light through the medium, c/n. Light travels by the shortest time route. When light passes through a medium boundary at an angle to it, the shortest time route is different from the shortest space distance; it will be biased toward spending less distance in the slower medium. When you trace the rays back to find their apparent source, you find that the apparent distance from the boundary shrinks in proportion to the refractive index of that medium. So when light travels a distance d through a medium of index n, an observer in air will perceive it has only having travelled d/n.

3. Dec 12, 2012

### rishch

True, but what matters is the light entering our eyes. While the totally, fewer number of photons would seem to come out from the image, but in reality each point on the image is still emitting the same number of photons in each direction, it's just that now they're emitting in less directions, so for an observer the brightness should remain the same. But since the aperture decreased, you could say that the field of vision is narrowed and we can see less of the image so the portion of the image that we could see would reduce, right?

Yes, I know how to calculate the postion of the apparent image if you have two mediums, say a bird in air looking at a fish in water. The two mediums are air and water. But in this question there are three mediums. You have air on the top, you have the medium with refractive index 1.5 on top in the beaker and at the bottom of the beaker, for a depth of 40cms, you have the medium with refractive index 1.6.

So if you were to look at the two mediums separately, in two separate beakers, with your head in air, then you could say the apparent depth of the bottom of the beaker = real depth/refractive index. But when you put the two mediums together, you can't simply add the apparent depths because, for the second medium at the bottom of the beaker, light is travelling from a medium of refractive index 1.6 to one of 1.5! It's not travelling into air.

So here's what I did.

I took the bottom medium and and the upper medium and calculated the relative refractive index of the bottom medium with respect to the first. Now the actual formula if you're observing an object in a denser medium from a rarer medium that need not be air is,

apparent depth= real depth/relative refractive index of denser medium with respect to first.

So I got apparent depth if your head is in the upper medium will be 40/(1.6/1.5)

Here 40 is the real depth of the bottom of the beaker and 1.6/1.5 is the relative refractive index. So this is the apparent depth. As you move your head out into the air this virtual image can be considered as an object in the upper medium. It's depth will be,

30+(40/(1.6/1.5))

You can apply the normal formula here as you are observing from air. Then,

apparent depth= [30+(40/(1.6/1.5))]/1.5
=45

45 is the answer they got. If instead of substituting values you leave them as variables, you get the formula they used, that is,

Apparent depth for the whole thing=d1/n1+d2/n2

However, they used that without giving any explanation, so I was left confused. I still can't intuitively grasp the formula as it doesn't seem very.....natural? Can anyone here get the formuala intuitively?

Last edited: Dec 13, 2012
4. Dec 13, 2012

### rishch

Can anyone explain the first question?

5. Dec 13, 2012

### SammyS

Staff Emeritus
Absolutely the opposite !

Suppose the image falls on a screen. Then fewer photons hitting any part of the screen means that fewer photons are reflected in each and every direction.

Draw a ray diagram. The size of the aperture at the lens (the amount of area of the lens which is allowed to pass light) does not affect the field of view.

6. Dec 13, 2012

### Basic_Physics

The top drawing shows how the lens refracts the light incoming rays to form an image of the point behind it. In the bottom drawing the lens is stopped down (apparture decreased), so the image is formed with less light and thus it has a lower intensity.

#### Attached Files:

• ###### stopped down lens.jpg
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7. Dec 13, 2012

### rishch

Ah yes now it all makes sense. I was imagining a virtual image, in which case I think what I said is correct. If you were to take a convex lens to form a virtual image, like a magnifying glass, the smaller magnifying glass would still make an image of the same brightness, but you wouldn't be able to see as much of the image. In case of a real image where you capture the image on a screen, what you said makes complete sense. I figured it out today morning when I was reading how the iris contracts to control the brightness of the image formed on the retina and I thought "Ohhhhh! So that's what he meant!" Sorry, I was thinking about virtual images. However, in case of virtual images, is what I said correct? That to us, the image would appear just as bright.

8. Dec 13, 2012

### rishch

So what I meant is while in a virtual image the image remains the same, the amount of it that we can see reduces. Am I correct? Oh and can anyone give some feedback on the second question and my solution?