Laws of Reflection and Plane Mirrors

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. You are one point eight metres tall, and want to see yourself from head to toe in a wall mirror. How tall must the mirror be? One point eight metres, surely? And if you step back, will a smaller mirror do? Example two will settle this. Plane mirror questions come up often in JEE, so let's begin.

0:24The laws of reflection, on any smooth surface

So, look at the left half of the figure. Light comes in along the incident ray and leaves along the reflected ray. The dashed line, perpendicular to the mirror where the ray hits, is the normal. The first law says that all three lie in one plane. With me? Now, small i on the figure is the angle of incidence, and small r is the angle of reflection. Let me ask you this: are they measured from the mirror, or from the normal? From the normal, yes, exactly! And the second law says i equals r: the angle of incidence equals the angle of reflection. As the heading says, these laws hold on any smooth surface, plane or curved. Clear?

1:08Deviation, and turning the mirror

Now, the angle through which the ray is turned is the deviation, delta, the Greek letter on screen. Delta equals one hundred and eighty degrees minus two times i, and the small circle means degrees. So a ray along the normal, with i equal to zero, comes straight back. Okay? Now here is a useful shortcut. The glancing angle, between the ray and the mirror, is ninety degrees minus i, and twice that is delta. So the deviation is two times the glancing angle; the cross means times. Right? Next, keep the incident ray fixed and turn the mirror through an angle theta, the Greek letter you see here. Through what angle does the reflected ray turn? Two theta, that is correct! The normal turns through theta, so the angle of incidence changes by theta, and the angle between the two rays, which is two times i, changes by two theta. The double arrow on screen means implies. Got it?

2:07Two mirrors at an angle

Now this figure: mirrors M one and M two at an angle theta, where the small one and two are just labels. After one reflection at each, the total deviation is three hundred and sixty degrees minus two theta, for any angle of incidence. Okay? Now tell me, if the two mirrors are at right angles, so theta is ninety degrees, what is the total deviation? One hundred and eighty degrees, yes, exactly! So the ray is sent back parallel to itself, going the opposite way. Clear?

2:39The image in a plane mirror

Now, the right half of the figure. Rays from the point object, capital O, reflect off the mirror; traced backwards along the dashed lines, they seem to come from capital I, behind the mirror. No light really reaches there, so the image is virtual. It is also erect, the same size, and laterally inverted, with left and right swapped. Okay? And the two distances marked small d are equal: the image is as far behind the mirror as the object is in front. Right? Now keep the object still, and move the mirror away along its normal through a distance, call it x. How far does the image move? Two times x, very good! The object is now that much farther from the mirror, and the image that much farther behind it. The gap grows by x plus x, and the object is still, so the image moves two times x. Okay? Here v stands for velocity. For motion along the normal, the velocity of the image equals two times the velocity of the mirror, minus the velocity of the object. The small words below each v say whose velocity it is. Got it? And now, the answer to our puzzle. To see your full height, you need a mirror only half your height, at any distance. Example two will prove it. With me?

3:57Images between two inclined mirrors

Now, put an object between two mirrors at an angle theta, and let us count its images. First find n, three hundred and sixty degrees over theta. If n is even, there are n minus one images. Okay? So let me ask you this: if the mirrors are at sixty degrees, how many images do you see? Five images is correct, because sixty degrees gives n equal to six, which is even. If you said six, oh no, careful, you must subtract one when n is even. Right? If n is odd, the answer depends on where the object sits. The bisector cuts the angle between the mirrors in half. On the bisector there are n minus one images; anywhere else, n images. Clear? And if n is not an integer, the count is its integer part, or one more, depending on where the object sits. So taking just the integer part is only a rough rule. Okay? And parallel mirrors give infinitely many images, like facing mirrors in a salon. Got it?

4:58Example 1: 35 degrees to the mirror, find r and deviation

Now example one, an easy one to warm up. Here, a ray strikes a plane mirror, making thirty five degrees with the mirror surface. Find the angle of reflection and the deviation. Pause the video and try it first. Done? Okay, let's check. The ray makes thirty five degrees, and i equals r, so the angle of reflection is thirty five degrees, right? No, no, careful, that is the trap! Thirty five degrees is from the mirror surface, so it is the glancing angle. So the angle of incidence is ninety minus thirty five, which gives fifty five degrees, and the angle of reflection is fifty five degrees too. Okay? The deviation is one hundred and eighty minus two times fifty five, which gives seventy degrees. Or take two times the glancing angle: two times thirty five, again seventy degrees. Both ways give seventy, clear?

5:53Example 2: shortest mirror for a full view of a 1.8 m man

Now example two, our puzzle from the start. Here, a man one point eight metres tall has his eyes ten centimetres below the top of his head. Find the shortest mirror in which he sees himself completely, and the height of its lower edge. Pause the video and try it first. Done? Okay, let's check. The small m on screen means metres, and ten centimetres is zero point one zero metres. So his eyes are at one point eight zero minus zero point one zero, which gives one point seven zero metres. With me? Light from his feet must reflect off the mirror into his eyes. At what height does this ray hit the mirror? Use i equals r. Midway in height between his feet and his eyes, yes! Because i equals r, the path up to the mirror and the path on to his eyes are mirror images. So the ray hits at one point seven zero over two, which gives zero point eight five metres. Okay? Same idea for the top of his head: midway between his eyes and the top. One point seven zero plus one point eight zero, all over two, gives one point seven five metres. Right? So the mirror needs a length of one point seven five minus zero point eight five, which gives zero point nine zero metres, exactly half his height. Its lower edge is zero point eight five metres above the floor. His distance from the mirror never entered the working, so our puzzle is solved. Got it?

7:24Quick revision: i = r, deviation, images, half-height mirror

So, quick revision. Measure angles from the normal: i equals r, the deviation is one hundred and eighty degrees minus two times i, and a mirror turned through theta turns the ray through two theta. The plane mirror image is virtual, erect, same size, and as far behind as the object is in front. For two mirrors, n is three hundred and sixty degrees over theta, an even n gives n minus one images, and a full view needs half your height. You have done really well today. Now try the practice problems on this page. All the best!