Set 3

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture a diver, three metres under a calm lake, looking straight up. The whole sky, from one horizon to the other, is squeezed into one bright circle on the surface. How big is that circle? Keep your guess; Example one will settle it. Today, Set three of the mixed optics problems, in the style of JEE Main. Two quick tools first, then two full problems. Let's go.

0:30Tool 1: signs in the mirror formula

So, tool one: signs in the mirror formula. An object is ten centimetres in front of a concave mirror, and its virtual image is thirty centimetres behind it; the c m on screen means centimetres. Find f, the focal length, okay? First, the sign convention. Distances are measured from the pole: along the incident light they are plus, and against it, minus. So with light from the left, every distance in front of the mirror is negative. And the mirror formula holds for paraxial rays, the rays close to the axis. With me? So the object gives u equal to minus ten centimetres. And the image is thirty centimetres away, so v is minus thirty too, right? No, no, wait, read the question again! The image is behind the mirror, on the side the incident light was heading to, so v is plus thirty centimetres. Got it? Now the mirror formula: one over f equals one over v, plus one over u. One over thirty, minus one over ten, is minus one over fifteen. So f is minus fifteen centimetres: the focal length is fifteen centimetres, and the minus sign says concave. Clear? And a quick check with the magnification m, which is minus v over u. Minus thirty over minus ten gives plus three. Positive m means a virtual, erect image, three times the size. Right?

1:57Tool 2: two wavelengths, one fringe

Now tool two, from wave optics. In a double-slit experiment, the eighth bright fringe of light of wavelength four hundred and eighty nanometres, the n m on screen, falls exactly on the sixth bright fringe of another wavelength, lambda, the Greek letter. Find lambda, okay? In the figure, S one and S two are coherent sources, small d apart, with the screen capital D away. For small angles, the fringe width beta, the Greek letter, is lambda times capital D, over small d. So bright fringe number n sits at y equals n times beta, that is n lambda D over d, measured from the centre O of the screen. Clear? Both wavelengths use the same slits and the same screen, so capital D and small d are the same, and the two fringes meet at the same y. So n one lambda one equals n two lambda two, with the small numbers written below. With me? So, six times four hundred and eighty equals eight times lambda, and lambda is three hundred and sixty nanometres, right? No, no, careful, that is the trap! The eighth fringe belongs to four hundred and eighty, so eight times four hundred and eighty equals six times lambda, which gives six hundred and forty nanometres. It makes sense: the longer wavelength has wider fringes, so fewer fit in the same distance. Got it?

3:14Example 1: radius of the sky circle a diver sees, 3 m deep

Now, Example one: our diver from the start! So, a diver is three metres below the surface of a calm lake, the m on screen means metres, and the refractive index of the water, written n, is four over three. The whole sky looks compressed into a circle on the surface. Find its radius. Pause the video and try it first. Done? Okay, let's check. In the figure, run the rays backwards: the diver's eye is at the point S, at a depth h below the surface. Light from the horizon grazes the surface and bends into the water at theta c, the critical angle, the Greek theta with a small c. Sine theta c is n two over n one, where n one is the refractive index of the denser medium, the water, and n two is that of the rarer medium, the air. Here that is one over four thirds, which is three over four. Okay? So the whole sky arrives inside a cone of half angle theta c. Then the radius is h times sine theta c, and three times three over four gives two point two five metres, right? Oh no, careful, that is the trap! Look at the triangle with its corner at S: the normal at the edge of the circle is parallel to the vertical, so the angle at S, between the vertical and the ray, is also theta c. The depth h is the side next to that angle, and the radius r is the side opposite it. Opposite over adjacent is the tangent, so r equals h times tan theta c. Clear? Sine theta c is three over four, so picture a right triangle with opposite side three and hypotenuse four. The third side is the square root of sixteen minus nine, which is root seven; that is the root sign on screen. So tan theta c is three over root seven. With me? So r is three times three, over root seven, which is nine over root seven, about three point four metres; the wavy equals sign means approximately. Puzzle solved: the whole sky fits in that one circle! Outside it, light from below, going from water, the denser medium, into air at more than theta c, is totally reflected. Right?

5:23Example 2: two convex lenses 40 cm apart, final image and m

And now Example two, a hard one. Two convex lenses, of focal lengths ten centimetres and twenty centimetres, share the same axis and are forty centimetres apart. An object is fifteen centimetres in front of the first lens. Find the final image and the magnification. Pause the video and try it first. Done? Okay, let's check. For lenses, distances are measured from the optical centre, again plus along the incident light. The lens formula, for thin lenses and paraxial rays, is one over v, minus one over u, equals one over f, with a minus sign this time, and m equals v over u. Here a real object has u less than zero, and a convex lens has f greater than zero. Okay? First lens: u is minus fifteen and f is plus ten. One over ten, minus one over fifteen, is one over thirty, so v is thirty centimetres, a real image. And m one, the m with a small one, is thirty over minus fifteen, which is minus two. With me? Now this image becomes the object for the second lens. It is thirty centimetres from the first lens, so u is minus thirty, right? No, no, wait, that is the trap! We measure from the second lens. The lenses are forty centimetres apart and the image is thirty from the first, so it sits ten centimetres in front of the second lens. So u is minus ten centimetres. Got it? Now f is plus twenty. One over twenty, minus one over ten, is minus one over twenty, so v is minus twenty centimetres. Negative v means the image is on the side the light comes from: a virtual image. And m two is minus twenty over minus ten, which is two. Clear? Now the total magnification. Minus two plus two gives zero, right? Not at all, magnifications multiply! m equals m one times m two, so minus two times two, which is minus four. The final image is virtual, twenty centimetres in front of the second lens, and four times the size. And the minus sign in m says it is inverted: virtual, yet inverted, because the first lens had already turned it upside down. With me?

7:42Quick revision: sign of v, fringe overlap, r = h tan theta c

So, quick revision. For a mirror, a virtual image is behind it, so v is positive. When two bright fringes fall on the same point, n one lambda one equals n two lambda two; and the diver's circle has radius h times tan theta c. With two lenses, measure u from the second lens, and multiply the magnifications. Now go and try the other problems of this set on this page. All the best!