The Prism: Deviation against Incidence

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. We send a thin ray of light into a glass prism, change the angle at which it strikes, and measure how much the ray turns. Now here is the strange part. Two quite different angles of incidence can give exactly the same deviation. How is that possible? By the end, you will answer it in one line. This is part of experimental skills in JEE, and questions on this graph do come up. So, let's begin.

0:32Deviation by a prism

So, look at the left part of the figure, where the triangle is our prism. The orange ray strikes the first face at small i, the angle of incidence. It leaves the second face at small e, the angle of emergence. Okay? The top corner, capital A, is the angle of the prism. And the Greek letter delta, where the two dashed lines cross, is the deviation: the angle between the incoming ray, carried straight on, and the ray that comes out. Right? The deviation is given in one line, written just below the prism. Delta equals i plus e, minus capital A, so any three of these angles give the fourth. Clear?

1:11Deviation against incidence: the curve

So how do we get this graph? We trace the ray with pins for several angles of incidence, usually thirty degrees to sixty degrees, and find delta each time. The little circle after each number means degrees. Then we plot delta against i, like the graph in the figure. Okay? So let me ask you this. As the angle of incidence increases, what happens to the deviation? The orange curve shows it does not keep rising. As i increases, delta first falls, reaches a minimum, and then rises again. That lowest value is delta m, the Greek letter delta with a small m below it, for minimum deviation. With me? Now, the curve looks like a neat U shape. So it must be the same on both sides of the minimum, right? That would be a mistake, and it is a common trap. The curve is not symmetrical. It is steeper on the side of small i, and gentler on the side of large i, got it?

2:11At minimum deviation

Now, the lowest point is special. The label under it in the figure says i equals e, so at minimum deviation the ray goes in and comes out at the same angle. Clear? Here is a question for you. Put e equal to i, and delta equal to delta m, in our deviation formula. What do you get for the angle of incidence? Yes, exactly, delta m equals two times i, minus capital A, the prism angle. The double arrow means therefore, so i equals e equals capital A plus delta m, all over two. Right? Now, inside the glass, the ray makes angles r one and r two with the normals at the two faces, small r with subscripts one and two, and r one plus r two equals the prism angle. At minimum deviation, r one equals r two equals capital A over two, so the ray passes symmetrically through the prism. Okay? Now the result we want. The refractive index, the Greek letter mu, equals the sine of half the sum, capital A plus delta m, divided by the sine of capital A over two. So read delta m from the lowest point of your graph, and this one line gives you mu. With me?

3:19One deviation, two angles of incidence

Now, back to our puzzle. Look at the upper dashed line in the graph. It cuts the curve twice, at i one and i two, that is small i with subscripts one and two. So any deviation greater than delta m occurs at two angles of incidence. Right? But tell me, why should two different angles of incidence give exactly the same deviation? The answer is really simple. The two angles are the i and the e of the same path. Send the light in at the old angle of emergence, and it retraces the path, coming out at the old angle of incidence. Got it? So put them into delta equals i plus e, minus capital A, and you get i one plus i two equals capital A plus delta. So our puzzle is solved, in one line. Clear?

4:08Tracing the ray with pins

Now, one word on the pins. The pins are fixed vertically, at least eight centimetres apart, and we view the feet of the pins. Okay? Keeping them far apart matters, because two points far apart fix the direction of the ray more accurately. Right? And upright pins, viewed at their feet, mark points in the plane of the paper, where the ray is drawn. With me?

4:32Example 1: second angle of incidence for the same deviation

Now Example one, a quick one to start. On the graph of delta against i for a sixty degree prism, delta is forty four degrees when i is thirty six degrees. At what other angle of incidence is the deviation forty four degrees? Pause the video and try it first. Are you done? Okay, let us check it together. Capital A is sixty degrees, delta is forty four degrees, and the first angle, i one, is thirty six degrees. Okay? Now notice, the question gives no minimum deviation and no refractive index. Can we still solve it? Yes, we can, because the two angles for the same deviation, like i one and i two in the figure, always add up to capital A plus delta. Sixty plus forty four gives one hundred and four degrees. Right? So i two equals one hundred and four, minus thirty six, which gives sixty eight degrees. Very good if you used that sum rule straight away.

5:33Example 2: minimum deviation and mu from i-delta data

Now Example two, a little harder. For a sixty degree prism, a student records pairs in brackets, i first, then delta, all in degrees. Thirty with forty seven, then forty with thirty eight point five, then fifty with thirty seven, then sixty with thirty nine, and seventy with forty three. Estimate delta m and mu. Pause the video and try it first. Are you done? Okay, let us check it together. Read the deviations in order, and each little arrow means next. They fall from forty seven to thirty seven, then rise to forty three, exactly the shape of our curve. With me? Now draw a smooth curve through these points. Its lowest point is near i equal to forty nine degrees, so delta m is about thirty seven degrees, and the wavy equals sign means approximately. Got it? Now let us find mu. Capital A plus delta m is ninety seven degrees, so mu is the sine of ninety seven degrees over the sine of sixty degrees, right? That is the trap, so be careful, because both angles in the formula are halved. Capital A plus delta m, over two, is sixty plus thirty seven, over two, which gives forty eight point five degrees. That is also the angle of incidence at the minimum, and it agrees with the forty nine from the curve. Clear? So mu equals the sine of forty eight point five degrees, over the sine of thirty degrees, which is half of sixty. That is zero point seven four nine over zero point five, which gives about one point five zero. Very good if you halved both angles.

7:10Quick revision: i-delta curve, mu and i1 + i2 = A + delta

So, quick revision. Delta equals i plus e, minus capital A, and it falls to delta m, then rises, and it is steeper when i is small. At delta m, i equals e equals capital A plus delta m, over two, and the sine formula gives mu. Any deviation greater than delta m comes from two angles, with i one plus i two equals capital A plus delta. Now go and try the practice problems on this page, and all the very best to you.