AIEEE Questions: Reflection

Transcript

0:00Introduction

Hello everyone! Picture this: another car is overtaking yours at fifteen metres per second. Now look at its image in your side-view mirror: it moves much, much slower than the car. So why is it so slow, and how much slower is it? By the end of this video, you will calculate it yourself. Today we solve two A I triple E questions on reflection, and questions like these come up often in JEE, so let's begin.

0:28Two mirrors at an angle: how many images?

First, a tool from the chapter's first lesson. In the figure, two plane mirrors, M one and M two, with small number labels, meet at an angle theta, the Greek letter between them. As the note says, two reflections turn the red ray through three hundred and sixty degrees minus two theta; the small circle means degrees. Today, though, we count the images of an object placed between them. With me? Step one is to find n, which is three hundred and sixty degrees over theta. What happens next depends on what kind of number n is. Okay? If n is even, the number of images is n minus one, wherever the object sits between the mirrors. Right? If n is odd, there are two cases. With the object on the bisector, the line that cuts theta exactly in half, you get n minus one images. Off the bisector, you get n images. Clear? And if n is not a whole number, just take its integer part, the whole number before the decimal point. Got it? Now a quick check, from a solved example in that lesson. The mirrors are at seventy two degrees, with the object on the bisector. How many images will you see? If you said four, very good! Three hundred and sixty over seventy two is five, which is odd, and the object is on the bisector, so n minus one gives four images. Move it off the bisector, and you see five. Okay? And at fifty degrees, n is seven point two, not a whole number, so the integer part gives seven images.

1:59Sign convention and the magnification shortcut

Right? Now look at the figure: the left mirror is concave, and the right one is convex, bulging towards the object, the black arrow labelled A B, and the brown arrows are its images. P is the pole, F the focus, and C the centre of curvature. Here the concave mirror gives a real, inverted image, and the convex one a virtual, erect, diminished image, as the captions say. Okay? First, the sign convention of our course. All distances are measured from the pole: along the incident light they get a plus sign, and against it a minus sign. Right? Now, light travels from the object towards the mirror. So tell me, is the object distance, u, measured from the pole back to the object, positive or negative? Negative, yes, exactly! It is measured against the light, so every real object has negative u. For a convex mirror, the focus lies behind the mirror, along the light, so f, the focal length, is positive. Got it? The mirror formula: one over v, plus one over u, equals one over f, where v is the image distance. It holds for paraxial rays, which means rays close to the principal axis. Clear? From it, the magnification, m, is minus v over u, which also equals f over, f minus u. This last form is the shortcut, because you never need v. Put in f and u with their signs. For a real object, negative m means real and inverted; positive m means virtual and erect. With me? And a convex mirror always gives a virtual, erect, smaller image behind the mirror, between P and F, as in the figure. That is our side-view mirror. Right?

3:38Speed of the image along the axis

Now the last tool, the speed of the image. Differentiate the mirror formula to get d v over d u, the rate at which v changes with u. It equals minus v squared over u squared; the small two means squared. And v over u is minus m, so this is minus m squared. Okay? So, if the object moves along the axis, how fast does the image move? Yes, m squared times as fast! The velocity of the image, v with the word image written small below, is minus m squared times the velocity of the object, for motion along the axis, relative to the mirror. Here v means a velocity, not a distance. Clear? And the minus sign tells us that the object and the image always move in opposite directions along the axis. Got it? Now, here is a very common slip in the exam. Along the axis, the speed goes with m squared, but for motion across the axis, it goes with just m. Keep these two apart.

4:37Example 1: number of images, two mirrors at 60 degrees

Okay? Now Example one, the easier one, from A I triple E two thousand and two. Two plane mirrors are inclined at sixty degrees to each other. How many images of an object placed between them are formed? Pause the video and try it first. Done? Okay, let's check. n equals three hundred and sixty degrees over sixty degrees, which is six. Right? So the answer is six images, correct? No, no, wait, six is the trap here! Six is even, so the number of images is n minus one, which gives five, wherever the object sits between the mirrors. So the answer is five images. Easy, isn't it?

5:22Example 2: image speed in a convex side-view mirror

And now Example two, from A I triple E twenty eleven. The car's convex side-view mirror has a focal length of twenty centimetres, the c m on screen. Another car, two point eight metres behind, is overtaking at a relative speed of fifteen metres per second, the m s to the power minus one on screen. Find the speed of its image in the mirror. Here is a hint: the shortcut for the magnification helps. Pause the video and try it first. Done? Okay, let's check. The mirror is convex, so f is plus twenty centimetres. The second car is a real object, so u is negative. And watch the units: two point eight metres is two hundred and eighty centimetres, so u is minus two hundred and eighty centimetres. Clear? Now you tell me: using f over, f minus u, what is the magnification? It is twenty over, twenty plus two hundred and eighty, since minus a negative u gives plus. That is twenty over three hundred, which gives one fifteenth. It is positive, so the image is virtual and erect, one fifteenth of the size of the car. Right? So the image speed is fifteen times one fifteenth, which is exactly one metre per second, correct? Oh no, careful, that is the very slip we warned about! The car moves along the axis, so the speed goes with m squared, not m. So m squared is one over two hundred and twenty five, times fifteen, the car's speed relative to the mirror. With me? Fifteen over two hundred and twenty five gives one over fifteen. So the image moves at just one fifteenth of a metre per second; m has no unit, so the answer stays in metres per second. That solves our opening puzzle: the car rushes past at fifteen metres per second, but its image crawls at one fifteenth. That is m squared at work!

7:19Quick revision: n = 360/theta, signs, m = f/(f-u), m squared

So, quick revision. Two mirrors at angle theta: n is three hundred and sixty degrees over theta, and an even n gives n minus one images. Signs: a real object has negative u, a convex mirror has positive f, and m equals f over, f minus u. Along the axis, the image speed is m squared times the object speed. Now try the practice problems on this page. All the best!