Resolving Power

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. At night, a car comes towards you on a long, straight road. Far away, you see just one light. Then, as it comes closer, that light splits into two headlights! The car had two headlights all along, so why did your eye see only one? Is something wrong with your eyes? No, it is a limit that every eye, every telescope and every microscope lives with, and it does come up in JEE papers. So, let's begin.

0:33Why a point looks like a small disc

So, light from a point passes through a circular opening, like the objective lens of a telescope, or the pupil of your eye. This opening is called the aperture. Because of diffraction, the image of the point is not a point at all, but a small disc. Okay? Now take two points very close together, say two stars, each making its own small disc. If the two discs overlap a lot, what do you see, two points, or just one? Just one, yes, exactly! The two discs blur into a single patch. So we need a clear rule for the boundary between one and two. Right? That rule is called Rayleigh's criterion. Two points are just resolved, that is, just seen as two, when the central maximum of one falls on the first minimum of the other. Any closer, and they merge into one. Clear?

1:25Telescope: the least angle it can resolve

Now, the telescope. Its objective has diameter capital D, and the light has wavelength lambda, the Greek letter on screen. Careful, this capital D is the diameter of the objective, not the screen distance from Young's experiment, okay? For stars, we measure separation as an angle. The least angle it can resolve is written capital delta theta: the little triangle is the Greek capital delta, and the letter after it is the Greek letter theta. It equals one point two two times lambda, over capital D, and we call it the telescope's limit of resolution. With me? Now, here is a quick question. One telescope has a bigger delta theta than another. Is it the better telescope? No, no, it is the worse one! With a bigger delta theta, the stars must be farther apart before it can separate them. So a smaller delta theta means finer detail, and a better telescope. Clear? That is why the resolving power is defined as the reciprocal, one over delta theta, which is capital D, over one point two two times lambda. Now bigger is better, so never write it upside down. Okay?

2:37Microscope: the least distance it can resolve

Now, the microscope. Here we want the least distance between two points of the object, written d min, a small d with min below. It equals one point two two times lambda, over two times n, times sine beta. Okay? Here n is the refractive index of the medium between the object and the objective. Right? And beta, the Greek letter on screen, is the half-angle of the cone of light entering the objective. Careful, this beta is an angle, not the fringe width from interference. Clear? The product n times sine beta is called the numerical aperture. Remember this name. Got it? The resolving power is again the reciprocal: two times n, times sine beta, over one point two two times lambda. With me? Now, some microscopes put a drop of oil between the object and the objective. Look at the formula. Why would that help?

3:33Finer detail: aperture and wavelength

Yes, exactly! Oil has a higher refractive index than air, so the numerical aperture goes up, and so does the resolving power. So a larger aperture helps: a bigger capital D for a telescope, and a bigger n times sine beta for a microscope. Okay? Now the wavelength. To see finer detail, should we use light of a shorter wavelength, or a longer one? The answer is shorter, very good! Lambda sits in the denominator of the resolving power. Right? That is the idea behind the electron microscope. Its electrons have a wavelength thousands of times smaller than that of light. Clear? Now, a topper's trick. For the same instrument, the resolving power is proportional to one over lambda. The sign on screen, like an open loop, means is proportional to. Got it? What is the ratio of the resolving powers of a telescope, for light of four thousand angstrom and six thousand angstrom? The capital A with a small ring on top is the angstrom, ten to the power minus ten metre. So the ratio is two to three, right? Oh no, careful, it is flipped! Resolving power goes as one over lambda, so the ratio is six thousand to four thousand, which is three to two. With me?

4:52Example 1: limit of resolution, 10 cm telescope at 550 nm

Now let us use it. Example one: find the limit of resolution of a telescope with an objective of diameter ten centimetres, for light of wavelength five hundred and fifty nanometres. Pause the video and try it first. Done? Okay, let's check. The c m on screen is centimetres, so ten centimetres is zero point one metre. And n m is nanometres, so five hundred and fifty nanometres is five point five times ten to the power minus seven metre. Right? Delta theta equals one point two two times lambda, over capital D, which is one point two two times five point five times ten to the power minus seven, over zero point one. Okay? One point two two times five point five gives six point seven one. Divide by zero point one, and we get six point seven one times ten to the power minus six. The wavy equals sign means approximately, and r a d means radian, so the answer is about six point seven times ten to the power minus six radian. Oh no, and here is the trap: leave capital D as ten, in centimetres, and your answer comes out a hundred times too small. Clear?

6:07Example 2: how far a 3 mm pupil can resolve two headlights

Now a harder one, and it is our puzzle from the start. Example two: the headlights of a car are one point two metres apart. From how far can an eye with a pupil of three millimetres see them as separate, for light of five hundred and fifty nanometres? Pause the video and try it, and find the angle first. Done? Alright, let's go. The pupil is the aperture of the eye, so capital D is three millimetres, the m m on screen, which is three times ten to the power minus three metre. Okay? Delta theta equals one point two two times five point five times ten to the power minus seven, over three times ten to the power minus three. Six point seven one over three gives about two point two four, so delta theta is two point two four times ten to the power minus four radian. Right? Now, we have an angle, but the question asks for a distance. How do we connect the two? Through the small angle idea, yes! For a small angle in radians, the angle equals the separation over the distance. With me? So the distance is one point two, over two point two four times ten to the power minus four. That is about five point four times ten to the power three metres, and k m means kilometres, so about five point four kilometres. Any farther, and the two headlights merge into one light. It was diffraction at your pupil, not a weak eye!

7:39Quick revision: Rayleigh, 1.22 lambda/D, microscope d min

Clear? So, quick revision. Rayleigh's criterion: two points are just resolved when the central maximum of one falls on the first minimum of the other. Telescope: delta theta equals one point two two times lambda, over capital D, and microscope: d min equals one point two two times lambda, over two times n, times sine beta. Resolving power is the reciprocal: it grows with a larger aperture and a shorter wavelength. Now go and try the practice problems on this page. You know exactly what to do now. All the best!