Set 2

Transcript

0:00Introduction

Hello everyone! Welcome to Set two of our mixed JEE Main problems on optics: fifteen problems that mix all the chapters of the unit, and thirty minutes to solve them. Here is a puzzle from this set. An object fifty centimetres from a concave mirror of focal length thirty centimetres moves towards it at four centimetres per second. So its image also moves at four centimetres per second, right? Hmm, hold that thought, and we will settle it at the end. So, a warm-up first, then the signs, then two full problems. Let's go.

0:36Warm-up: two parallel plane mirrors

So, the warm-up first, the very first problem of this set. On the left of the figure, the angle of incidence, i, equals the angle of reflection, r, both measured from the dashed normal. On the right, rays from the point object, O, reflect from a plane mirror, and seem to come from the point marked capital I behind it, the virtual image. Both gaps are marked d, because the image is as far behind the mirror as the object is in front. Okay? Now the problem: two parallel plane mirrors are twelve centimetres apart, written c m, and an object is four centimetres from one of them. Take the nearest image in each mirror. How far apart are these two images? Let's see: the object is four centimetres from the first mirror, so its image is four centimetres behind that mirror. It is twelve minus four, that is eight centimetres, from the second mirror, so that image is eight centimetres behind the second mirror. Right? Now go from one image to the other: four, plus the twelve between the mirrors, plus eight. That gives twenty four centimetres. Some students add only four and eight, and forget the gap between the mirrors. Clear?

1:50Signs, the mirror formula and m

Now, before the curved mirrors, the signs. In the left half of the figure, light from the object, the black arrow A B, travels left to right towards a concave mirror, the thick blue curve. P is its pole, C is the centre of curvature, and F is the focus. Every distance is measured from the pole: along the incident light it is positive, the plus sign, and against it, negative, the minus sign. Okay? The object distance is called u, and the focal length is f. For this concave mirror, the focus F lies in front of it, on the object's side. So is f positive or negative? It is negative, very good! From the pole, we go back towards the focus, against the light. The same holds for a real object, so u is less than zero; that is the little less than sign. And f is less than zero for a concave mirror. Got it? Now the mirror formula, with v as the image distance: one over v, plus one over u, equals one over f. It holds for paraxial rays, the rays close to the axis, as the brackets say. With me? And m, the magnification, equals minus v over u, which is also f over, f minus u. This second form is a topper's trick: you do not need v at all. If m comes out negative, the image is real and inverted, just like the dark red arrow in the left half of the figure. On the right, a convex mirror gives a small, erect, virtual image, so there m is positive. Right?

3:22How fast does the image move?

Now the second tool, for our puzzle. When the object moves, u changes, and v changes with it, while f stays fixed. Differentiate the mirror formula: d v over d u, the rate of change of v with u, equals minus v squared over u squared, which is minus m squared. Okay? So for motion along the axis, the image velocity, v sub capital I, equals minus m squared, times the object velocity, v sub capital O, okay? Careful, this v is a velocity, not the image distance. Both velocities are measured relative to the mirror. Clear? Now look at that minus sign. Since m squared is never negative, what does this minus sign tell you about the two motions? Yes, exactly, they always move in opposite directions along the axis! The greater than sign says m squared is positive, so the image velocity has the opposite sign to the object velocity. And the speed is multiplied by m squared, not by m. Got it?

4:22Example 1: focal length and magnification, image on screen

Now, Example one. Here, a concave mirror forms the image of an object on a screen. The screen is one hundred centimetres from the mirror, and the object is twenty five centimetres from it. Find the focal length and the magnification. Pause the video and try it first. Done? Okay, let's check. The image falls on a screen. So is it real or virtual, and what is the sign of v? It is real, of course, because a screen catches only a real image! So the image is in front of the mirror, and v is minus one hundred centimetres. The object is in front too, so u is minus twenty five centimetres. Right? Now the mirror formula: one over f equals one over minus one hundred, plus one over minus twenty five. One over twenty five is four over one hundred, so the sum is minus five over one hundred, which is minus one over twenty. Okay? So f equals minus twenty centimetres. The minus sign only tells us the mirror is concave, so the focal length is twenty centimetres. Clear? Now m equals minus v over u. Minus one hundred over minus twenty five is four, and the minus sign in front makes it minus four. So the image is inverted, and four times the size. Getting plus four is the usual slip, so always put the signs in first. With me?

5:45Example 2: image speed, object at 4 cm/s towards the mirror

Now Example two, and this is our puzzle from the start. An object is fifty centimetres from a concave mirror of focal length thirty centimetres. It moves towards the mirror along the axis at four centimetres per second; c m s to the power minus one, on screen, means centimetres per second. Find the speed of the image. Pause the video and try it first. Done? Okay, let's check. Signs first: the object is in front, so u equals minus fifty centimetres, and the mirror is concave, so f equals minus thirty centimetres. Right? Now the topper's trick: m equals f over, f minus u. That is minus thirty, over minus thirty plus fifty, which is minus thirty over twenty. So m is minus one point five. Okay? Now the speed: the image is one point five times as large, so it moves one point five times as fast, and one point five times four gives six centimetres per second, right? No, no, wait, that is the trap! Along the axis, the speed is multiplied by m squared, not by m. One point five squared is two point two five, and two point two five times four gives nine. So the image moves at nine centimetres per second, more than twice as fast as the object. Clear? Now the direction: object and image move in opposite directions along the axis. The object comes towards the mirror, and this double arrow on screen means so, the image moves away from it. So our puzzle is settled: the image moves at nine centimetres per second, away from the mirror, not four. Got it?

7:27Quick revision: plane mirror, signs, image speed m squared

So, quick revision. First, a plane mirror forms the image as far behind it as the object is in front. Second, put the signs in first: u is negative for a real object, f is negative for a concave mirror, and m equals f over, f minus u. And third, along the axis, the image speed is m squared times the object speed, in the opposite direction. Now set a timer for thirty minutes, and try all fifteen problems of this set on this page. All the best!