AIEEE and Earlier JEE Main Questions: Dispersion and Instruments

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You look at a tall tree, far away, through a telescope of magnifying power twenty. Does the tree look twenty times taller, or twenty times nearer? Many students say taller, without a second thought. Today we settle this, and one more question, on colours and the critical angle, both from past JEE Main papers. Questions like these come up often. So, let's begin.

0:29Telescope: magnifying power is about angles

So, first the telescope, in the figure. Light from a far object enters the objective, a lens of long focal length, f o, that is f with a small o below. It forms a real image, the short brown arrow, at its focus. Then the eyepiece, of short focal length f e, with a small e below, magnifies that image. Okay? Now, alpha, the Greek letter on screen, is the angle the object makes at your eye when you look without the telescope, with the unaided eye. Right? And beta, the other Greek letter, is the angle that the final image makes at your eye, when you look through the eyepiece, as in the figure. With me? Now let me ask you this. When we say magnifying power, what are we comparing: the sizes of things, or the angles they make at the eye? Angles, yes, exactly! The magnifying power, capital M, equals beta over alpha. And in normal adjustment, with the final image at infinity, it is f o over f e. Clear? One more tool. For a far object the angle is small, roughly the height over the distance; that wavy equals sign means approximately. So a bigger angle can come from a taller object, or from a nearer one. Got it?

1:49Colour changes the refractive index

Now, colours. In this figure, white light, the grey ray, enters a glass prism and comes out split: red on top, then green, and violet at the bottom. Capital R and capital V on the screen mark red and violet; the formulas at the right of the figure are not needed today. Okay? Why does this happen? The refractive index, n, depends on the wavelength, lambda, the Greek letter on screen. Roughly, n equals small a, plus small b over lambda squared, with small a and small b constants; this is Cauchy's relation. So for transparent media, a longer wavelength gives a smaller refractive index. Right? Now tell me. Red light has a longer wavelength than violet. So in glass, which colour has the bigger refractive index? Yes, it is violet, very good! n V, that is n with a small V below, for violet, is greater than n R, for red; that pointed sign means greater than. So violet bends more, and it is deviated more than red, just as the figure shows. Clear? Now recall the prism at minimum deviation. Here capital A is the prism angle, and delta m, the Greek delta with a small m below, is the angle of minimum deviation. n equals sine of, capital A plus delta m, over two, all over sine of capital A over two. With me? So with the same capital A, a larger n needs a larger sine on top, and so a larger delta m. Violet, with its larger index, has a larger minimum deviation than red through the same prism. Okay?

3:30Critical angle: it depends on colour too

Now the critical angle, with this figure. Below is the denser medium, with index n one, that is n with a small one below; above is the rarer medium, with index n two. The dot marked S is a source, the dashed lines are normals, and the h and the r are not needed here. Right? Going from denser to rarer, the ray bends away from the normal. At one angle of incidence, the refracted ray just grazes the surface, and that angle is the critical angle, theta c, theta with a small c below. Sine theta c equals n two over n one, and for a medium in air, sine theta c equals one over n. Got it? And total internal reflection, TIR on screen, needs two things: the light must go from denser to rarer, and the angle of incidence, i, must be greater than theta c. Then all the light is reflected, like the ray on the right. Clear? Now put the two ideas together. Violet has the larger refractive index. So, does violet have the larger critical angle, or the smaller one? The smaller one, yes, exactly! Sine theta c is one over n, so a bigger n gives a smaller sine, and a smaller angle. So violet has a smaller critical angle than red. Okay? And a question may speak of frequency instead. Lower frequency goes with longer wavelength, like red. So a lower frequency means a smaller n, and a larger critical angle. Example two needs this line. With me?

5:05Example 1: tree through a telescope of magnifying power 20

Now, our puzzle, as it came in JEE Main twenty sixteen. An observer looks at a distant tree of height ten metres, through a telescope of magnifying power twenty. How does the tree appear? Pause the video and try it first. Done? Okay, let's check. Magnifying power is beta over alpha, and that is twenty. So beta equals twenty times alpha: the angle at the eye is twenty times larger. Okay? Now, an angle twenty times larger. So the tree looks twenty times taller, right? No, no, careful, that is the trap! The telescope cannot change the tree; it only changes the angle. And the angle is roughly height over distance. So the same tree, at one twentieth of its distance, would make about this angle. With me? So the tree appears twenty times nearer. And the ten metres? We never used it; it is not needed. Puzzle solved: the angle is twenty times larger, as if the tree were twenty times nearer. Clear?

6:08Example 2: green at critical angle, which colours escape

Now example two, from JEE Main twenty fourteen, a little harder. Green light, travelling in water, meets the surface between water and air at its critical angle. What can be said about visible light of other colours, incident at the same angle? Pause the video and try it first. Done? Okay, let's check. Water to air is denser to rarer, so total internal reflection is possible. The angle of incidence, i, equals the critical angle of green, so green itself just grazes along the surface. Okay? Now the other colours. Violet bends the most, so violet should escape most easily, right? No, no, wait, bending more is exactly the problem! Blue, indigo and violet have a higher frequency than green, so a larger n, and a smaller critical angle. Got it? So for them, i, the critical angle of green, is greater than their own theta c. Both conditions hold, so blue, indigo and violet are totally reflected back into the water. Right? Now the other side. Yellow, orange and red have a lower frequency than green, so a smaller n, and a larger critical angle. Clear? For them, i is less than theta c, and that is the less than sign. So they are not totally reflected; they emerge into the air. With me? So the answer: light of frequency less than that of green emerges into the air, and the rest is totally reflected. Remember it this way: the larger the index, the sooner the light gets trapped.

7:48Quick revision: M as angle ratio, colour and critical angle

Okay? So, quick revision. Telescope: magnifying power is beta over alpha, a ratio of angles, so the object looks nearer. Larger refractive index, like violet: larger minimum deviation, and smaller critical angle. Total internal reflection needs denser to rarer, and i greater than theta c; red, with the largest critical angle, escapes most easily. Now try the third question on this page, on red and blue light through a prism. All the best!