Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a glass block with a tiny air bubble trapped inside it. The bubble bulges out on both sides, just like a convex lens. And a convex lens brings light together, right? So, does this bubble bring light together too? Hmm, keep your guess ready. Today we will see how to use this unit's problem bank, settle this puzzle with one of its concept checks, and solve two of its problems. Questions like these come up often in JEE, so let's begin.

0:34How to use this chapter

So, first, what is in this chapter? Part A has eighteen solved problems, at JEE Main level. And they mix the whole unit, from mirrors and lenses to interference and polarisation. Right? Then Part B has fourteen harder problems, at JEE Advanced level. With me? And Part C has twelve concept checks. They test the ideas behind the formulae, the why questions. After that come twelve practice problems, at the end of Part C, for you to solve alone. Got it? Now tell me, when should you read a solution? Straight away, or only after you have tried the problem yourself? Only after you try, yes, exactly! The course says it clearly: attempt each part as a test, and only then read the solutions. Okay?

1:25Part C: can a convex shape diverge light?

Now, a taste of Part C, and our puzzle. On the left of the figure, a convex lens in air brings the light together, to a real, inverted image; capital F marks a focus, and two F is twice as far. On the right, a concave lens spreads the light out, to a virtual, erect, smaller image. So what decides which one we get? It is the lens maker's formula, for a thin lens in a medium. One over f equals, n L over n m, minus one, times, one over R one, minus one over R two. Here f is the focal length, n with a small L below is the refractive index of the lens, and n with a small m is that of the medium around it. R one and R two, the R with a small one and a small two, are the radii of the two surfaces. Right? For a convex shape, the second bracket is greater than zero; that is the greater-than sign on screen. So for a glass lens in air, the first bracket is positive too, f is positive, and the lens converges. Clear? Now tell me: what happens if the medium around the lens has a higher index than the lens itself? Then n L over n m is less than one, so the first bracket is negative, yes, exactly! So f is negative, and the course says it clearly: if n m is greater than n L, a convex lens diverges. With me? Now back to our bubble. It has a convex shape, but it is made of air, with a lower index than the glass around it. Okay? So the air bubble is a diverging lens; it spreads light out, like the lens on the right. That settles our puzzle: the shape alone does not decide, the indices inside and outside decide. Got it?

3:14Example 1: image of 4 cm object 18 cm from concave mirror

Now, Example one, an easy one from Part A, okay? An object four centimetres high is eighteen centimetres from a concave mirror of focal length twelve centimetres. Find the position and height of the image. The small c m on screen means centimetres. Pause the video and try it first. Done? Okay, let's check. Many students put every number in as positive: one over v, plus one over eighteen, equals one over twelve. That gives v equal to plus thirty six, an image behind the mirror. So, is the image virtual? No, no, wait, we must put the signs in first! The course measures every distance from the pole of the mirror: along the incident light is positive, against it is negative. The object is real and in front, so u, the object distance, is minus eighteen. For a concave mirror the focal length is negative, so f is minus twelve. And the height h is plus four, above the axis. Okay? Now the mirror formula, one over v plus one over u equals one over f, which holds for rays close to the axis. So one over v equals one over f minus one over u: minus one over twelve, plus one over eighteen, which is minus one over thirty six. Right? So v, the image distance, is minus thirty six centimetres. Negative v means a real image, thirty six centimetres in front of the mirror. With me? The magnification m is minus v over u. Here v over u is minus thirty six over minus eighteen, which is plus two, so m is minus two. And a quick check with the course's shortcut, m equals f over, f minus u: minus twelve over, minus twelve plus eighteen, gives minus two again. Clear? Finally, the image height, h prime, the h with a small dash, is m times h: minus two times four, which is minus eight centimetres. The minus sign means the image is inverted, and it is eight centimetres high. Got it?

5:22Example 2: glass slab between object and concave mirror

Now Example two, a hard one from Part B, okay? Here, a point object is thirty centimetres in front of a concave mirror of focal length twenty centimetres. Between them stands a glass slab, six centimetres thick, with refractive index n equal to one point five, perpendicular to the axis. Find the position of the final image. Pause the video and try it first. Done? Okay, let's check. First, the slab. Seen through it, does the object look nearer to the mirror, or farther away? It looks nearer, that is correct, because a slab shifts the image towards the side that receives the light, and here that is the mirror. For viewing close to the normal, the course gives the shift as t times, one minus one over n, whatever the distance of the object. Here t, the thickness, is six, and one minus one over one point five is one third, so the shift is two centimetres. Okay? So, for the mirror, the object now sits twenty eight centimetres away, and u is minus twenty eight. Right? Now the mirror formula with signs, and f is minus twenty. One over v equals minus one over twenty, plus one over twenty eight, which is minus one over seventy. So v is minus seventy centimetres. Clear? So the answer is seventy centimetres in front of the mirror, right? Oh no, careful, that is the trap here! The reflected light travels back through the slab once more before it meets. The slab shifts it by two centimetres again, in the direction the light is now going, so the meeting point moves two centimetres farther from the mirror. With me? So the final image is real, seventy plus two, which is seventy two centimetres from the mirror. Forget the second crossing, and you write seventy and lose the marks. Got it?

7:19Quick revision: signs first, slab shift, air bubble lens

So, quick revision. Part A is at JEE Main level, Part B at JEE Advanced level, and Part C checks the ideas; attempt each as a test, then read the solutions. In mirror problems, put the signs in first: u is negative for a real object, and f is negative for a concave mirror. Next, a slab shifts the image by t times, one minus one over n, every time the light crosses it. And a convex shape diverges light when the medium around it has the higher index, just like our air bubble. Now open Part A, solve it like a test, and finish with the practice problems at the end of Part C, the last part of this chapter. All the best!