Apparent Depth and the Glass Slab

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. There is a fish one point two metres below the surface of a lake, and a bird zero point nine metres above the surface, right above the fish. But the fish sees the bird two point four metres away, and the bird sees the fish just one point eight metres away! They look along the same straight line, so how can they disagree? By the end of this video, you will work out both numbers yourself. Apparent depth comes up often in JEE, so let's begin.

0:33Apparent depth: looking straight down

First, let us look at the course figure. On the left is Snell's law: n one sine i equals n two sine r, where n one and n two, with small number labels, are the refractive indices, and i and r are the angles with the normal, the dashed line. On the right, the object, O, sits under a liquid. Its rays bend away from the normal into air and seem to come from I, the image, higher up; the grey dashed lines trace them back. With me? So, the object is at a real depth d, in a medium of refractive index n, and we look from air, close to the normal. Okay? Now let me ask you this. Will the object look deeper than d, or shallower? Shallower, yes, exactly! The apparent depth is d over n; the bar is the fraction line, read as over. Water and glass have n bigger than one, so d over n is less than d. The label under the figure says the same thing: apparent depth equals real depth over n. Clear? So the object appears raised by d minus d over n, which is d times, in brackets, one minus one over n. Right? Quick check, from the course. Say a pool is two metres deep, the m on screen, and water has n equal to four thirds. Looking straight down, how deep does it look? One point five metres, very good! Two over four thirds gives one point five; the slanted line is just another way to write a fraction. Half a metre has vanished! That is why a pool looks shallower than it is. Got it? Now, careful, here is a common trap. d over n is correct only for viewing along the normal. Look at the water at a slant, and the object appears even shallower, and its image is no longer vertically above it. Okay?

2:28Layers, and looking up from the water

Next, an object under several layers, say glass with water on top. The apparent depth is t one over n one, plus t two over n two, and so on; the dots mean more terms like these. The small one and two written below t and n number the layers: t one and t two are the thicknesses, and n one and n two the indices, so each layer is divided by its own index. With me? Now let us flip the situation around. The object is in air, and the observer is in the water, like a fish looking up at a bird. Does the bird look higher than it really is, or lower? Higher, that is correct! An object in air at height h, seen from inside the water, appears at a height of n times h. And n is bigger than one, so the bird looks farther away. Right? So here is a topper's trick, one rule for both cases. Apparent distance from the surface equals real distance times n observer over n object; the cross is the times sign, and the word written small below each n names its medium. Observer in air and object in water, the factor is one over n; observer in water and object in air, it is n over one. Clear? And speeds along the normal scale by the same factor, so a fish at rest sees a diving bird approach n times faster. Got it?

3:53The glass slab: normal shift and lateral shift

Now the glass slab. In the figure, the slab has parallel faces, thickness t, and refractive index n. The red ray hits the top face at angle i, bends towards the normal to angle r inside, and comes out at the bottom; the dashed red line just continues the incident ray. Okay? First, look at an object through the slab, close to the normal. Its image shifts by t times, in brackets, one minus one over n, the same form as before. Right? Now here is something to think about. Suppose the object sits far behind the slab. Does the shift change with the distance of the object? No, it does not, very good! The image shifts towards the observer by the same amount, whatever the distance of the object. With me? Another quick check from the course. There is a coin under a glass slab six centimetres thick, the c m on screen, with index one point five. How much does it appear raised? Two centimetres, correct! One over one point five is two thirds, so the bracket is one third, and six times one third gives two centimetres. Got it? Finally, the slanting ray. It comes out parallel to the incident ray, as the note says, but shifted sideways by d, the lateral shift; this d is not a depth. The figure gives d equals t times sine of, i minus r, over cos r. Clear?

5:25Example 1: mark under 6 cm of glass and 10 cm of water

Now Example one. There is a mark at the bottom of a glass block six centimetres thick, of index one point five, with ten centimetres of water on top, of index four thirds. Find the apparent depth of the mark, seen from above. Pause the video and try it first. Done? Okay, let's check. The total depth is six plus ten, which is sixteen centimetres. So we just divide sixteen by one refractive index, right? No, no, careful, there are two different media here! Each layer is divided by its own index, so the glass gives six over one point five, which is four centimetres. Got it? The water gives ten over four thirds, which is seven point five centimetres. Okay? Add them, and four plus seven point five gives eleven point five centimetres. The real depth is sixteen, so the mark looks closer, as it should. Clear?

6:24Example 2: fish and bird, how far each sees the other

Now Example two, our opening puzzle. The fish is one point two metres deep in a lake of index four thirds, and the bird is zero point nine metres above the surface, right above the fish. Find how far each one sees the other. Pause the video and try it first. Done? Okay, the fish first. The observer is in water, and the object is in air. So, using our one rule, do we multiply by n, or divide by n? We multiply, yes, exactly! n observer over n object is four thirds over one, so the bird appears four thirds times zero point nine, which is one point two metres above the surface. With me? The fish itself is really one point two metres below the surface, so it sees the bird one point two plus one point two, which gives two point four metres away. Right? Now the bird looks down, from air into water, so this time we divide. One point two over four thirds gives zero point nine metres below the surface. Okay? Add the bird's own height of zero point nine metres, and the bird sees the fish one point eight metres away. Clear? The real distance is one point two plus zero point nine, which is two point one metres. So the puzzle is solved: both are right, because each one sees the image of the other! Got it?

7:50Quick revision: apparent depth rule and slab shift t(1-1/n)

So, quick revision. Near the normal, apparent distance equals real distance times n observer over n object. From air, that is d over n, and under layers, add t over n for every layer. And a slab shifts the image towards the observer by t times, one minus one over n, for any object distance. Now go and try the practice problems on this page. All the best!