Refraction at a Spherical Surface

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Put a glass ball on a dot on your page, and look from directly above. The dot looks bigger, fine. But where does it seem to be? Right there on the page? No, it seems to sit well below the page! How can one curved surface do that? Today's formula answers it, and it is the first step towards lenses, which come up often in JEE, so let's begin.

0:28Sign convention: measure from the pole

So, look at the figure. Light starts at the object, capital O, in a medium of refractive index n one; the small one below the n is a subscript. It crosses one curved surface into a second medium, n two, the shaded part, and the rays meet at the image, capital I, on the far side. Okay? Now the sign convention, the same as for mirrors. All distances are measured from the pole, capital P, the middle of the surface, and distances along the incident light are positive. With me? Now, u is the object distance, from P back to the object; the double-headed arrows under the figure mark u, v and R, the three distances. The light travels to the right. So what sign does u take? Yes, exactly, u is negative! Against the light we use a minus sign, so a real object here has u less than zero; that little sign means less than. And v, the image distance, goes along the light, so here it is positive. Right? Next, capital R, the radius of curvature, from the pole to capital C, the centre of curvature. R is greater than zero when C lies on the side the light is going to. The dashed line from the surface to C is the normal, along a radius. Here C is ahead of the light, inside the shaded medium, so R is positive.

1:51The formula for one curved surface

Clear? Now the main result, also written on the right of the figure. n two over v, minus n one over u, equals n two minus n one, all over R, right? See the pattern: on the left, the new medium goes with v, and the old medium goes with u. And one condition, just like the mirror formula: it holds for paraxial rays, the rays close to the axis. Okay? Next, the magnification, m, which is h prime over h, image height over object height; the little dash is read as prime. And m equals n one times v, over n two times u. Careful, the indices swap here: in m, n one goes with v. Clear? And the sign of m tells you the nature: negative m means an inverted image, and positive m an erect one. Got it?

2:39A flat surface: the apparent depth

Now let us test the formula. What if the surface is flat? What is its radius of curvature then? Yes, it is infinity, the sideways eight on screen. So the right side, n two minus n one over R, becomes zero. With me? So n two over v equals n one over u, and the double arrow means implies: v equals n two over n one, times u. That is exactly the apparent depth result, hidden inside our formula, which is nice, isn't it?

3:11Trap: the sign of R follows the light

Now, the trap in this topic. Many students see a bulge and say R is positive. Be careful here: R takes its sign from the direction of the light, not from the shape you see. Okay? In our figure, light comes from the left, and C lies ahead of it, so R is positive. Right? Now send the light the other way, from the shaded side out to the left. Same surface, same shape. What is the sign of R now? It is negative, that is correct! Now C is behind the light, on the side it came from. So the same surface has R less than zero. Example two uses exactly this. Got it?

3:49Example 1: air to glass, object 30 cm from a convex surface

Now example one, an easy warm up. Here, a point object in air is thirty centimetres from the convex end of a glass rod; the c m means centimetres. The rod has index n equal to one point five, and radius of curvature ten centimetres. Where is the image? Pause the video and try it first. Done? Okay, let's check. Light goes from air into glass, so n one is one, and n two is one point five. Right? The object is in front, against the light, so u is minus thirty centimetres. The centre of curvature is inside the rod, ahead of the light, so R is plus ten. With me? Now we put them in the formula. One point five over v, minus one over minus thirty, equals zero point five over ten; zero point five is n two minus n one. Careful with the two minus signs. When the middle term moves to the right side, what does it become? It becomes minus one over thirty, very good! And zero point five over ten is one over twenty, so one point five over v equals one over twenty, minus one over thirty, which gives one over sixty. Clear? So v equals one point five times sixty, which gives plus ninety centimetres. The image distance is positive, so the image is inside the glass, where the light really goes. It is real, ninety centimetres inside the rod, and that's it!

5:18Example 2: dot under a glass ball, image and magnification

Now example two, our puzzle from the start. Here, a glass ball of diameter twelve centimetres and index one point five rests on a dot on a page. Where does the dot appear from directly above, and with what magnification? Pause the video and try it first. Done? Okay, let's check. Light from the dot goes up through the glass and out of the top into air, so n one is one point five, and n two is one. The dot is a full diameter below the top, against the light, so u is minus twelve centimetres. Okay? Now, the radius is six centimetres, and the top bulges up towards your eye. So R is plus six, right? No, no, wait, that is the trap we just saw! The light goes upwards, and the centre C is six centimetres below the top, on the side the light comes from. So R is minus six centimetres. Got it? Now put everything into the formula. One over v, minus one point five over minus twelve, equals one minus one point five, over minus six. Tidy the signs: one over v, plus one point five over twelve, equals minus zero point five over minus six. With me? One point five over twelve is one over eight, and the right side is one over twelve. So one over v equals one over twelve, minus one over eight, which gives minus one over twenty four, and v is minus twenty four centimetres. Right? The image distance is negative, so the image is on the object's side: it is virtual, twenty four centimetres below the top. The page is only twelve centimetres below the top, so the dot appears twelve centimetres below the page! Clear? Last, the magnification: m equals n one times v, over n two times u. That is one point five times minus twenty four, over one times minus twelve, which gives plus three. It is positive, so the image is erect, and three times larger. So our puzzle is solved: bigger, and below the page!

7:25Quick revision: n2/v - n1/u = (n2 - n1)/R and m = n1v/n2u

So, quick revision. n two over v, minus n one over u, equals n two minus n one, over R, and m equals n one times v, over n two times u, for paraxial rays. Measure from the pole, and R is positive only when C is on the side the light goes to. For a plane surface, R is infinity, giving v equals n two over n one, times u, the apparent depth. Now go and try the practice problems on this page. All the best!