Spherometer, Travelling Microscope and Angular Verniers

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Pick up a convex lens. Its curved surface is just a small piece of a big sphere, okay? Now I want the radius of that sphere. But the sphere is not there! Only this small piece is in your hand. So how will you measure the radius of a sphere you cannot even see? With a ruler? No chance. But there is a neat way: one small screw, and one tiny height. Today we meet three instruments: the spherometer, the travelling microscope, and the angular vernier. And note, instrument questions are regular visitors in JEE Main, and these are easy marks if your formulas are clear. So, let's begin.

0:45Spherometer: a screw gauge on three legs

So, first, the spherometer. Honestly, it is nothing new. It is simply a screw gauge, standing on three legs, okay? The three legs sit at the corners of an equilateral triangle, and the side of that triangle we call small l, okay? Now place it on a curved surface, like our lens. The screw measures one small height, h: how high the curved surface rises above the plane of the three legs. Just one small number. Clear? Alright, now tell me. Is h simply the reading you take on the lens? Most students say yes. No, no, careful! This is the trap. First you take a reading on a flat glass plate. That is your zero reading. Then you take the reading on the curved surface. And h is the difference of the two: how much the reading rises above the flat glass reading. Got it? Now the main formula. R, the radius of curvature, equals l squared over six h, plus h over two, okay? Write this down. Notice, in the first term h sits in the denominator, so a tiny h gives a big radius. And never drop the small second term, h by two. We will see why in the example. And how fine can it read? Same as a screw gauge. The least count is the pitch, divided by capital N, the number of divisions on its circular disc, okay? Pitch, remember, is how far the screw moves in one full turn. Nice and simple.

2:14Travelling microscope: a vernier that moves

Next, the second instrument: the travelling microscope. It is a microscope that slides along a main scale, and that main scale is fitted with a vernier, exactly like your vernier callipers, okay? So its least count is also the same as for callipers: one main scale division, minus one vernier scale division. On screen, MSD and VSD are just short forms of these two. Clear? Quick recall from the callipers lesson. Suppose n vernier divisions are equal to n minus one main scale divisions. What is the least count? Yes, exactly! One main scale division, divided by n. But careful: in JEE questions, one main scale division is not always one millimetre. It can be half a millimetre, for example. So always read it from the question, right? And what does it measure? Small lengths. For example, the bore of a capillary tube, the narrow hole inside it. Or the real and apparent depths of a glass slab, okay? From those two depths we get the refractive index of the glass, written mu. Now, which depth goes on top of the fraction? Real on top! Mu equals real depth divided by apparent depth, okay? Remember it like this: real comes first, so real goes on top.

3:35Angular vernier: the same idea in degrees

Now the third one: the angular vernier, which you find on a spectrometer. Same vernier idea, but now the main scale is a circle, marked in degrees, not millimetres. That is the only change. So, what is the least count here? Yes, same rule as before: the value of one main scale division, divided by the number of vernier divisions, okay? Exactly like the travelling microscope, just in degrees. Got it? These least counts are tiny angles, so we write them in minutes and seconds of arc. The small circle on screen means degrees, the single dash means minutes, and the double dash means seconds, okay? One degree is sixty minutes, and one minute is sixty seconds.

4:19Topper's trick: two families, two formulas

Now let me share a trick that toppers use. Callipers, screw gauge, spherometer, travelling microscope, spectrometer. So many instruments! How many least count formulas do you really need? Just two! Family one, the vernier family: callipers, the travelling microscope and the spectrometer. All three have a main scale with a vernier sliding along it, right? For this family, in the usual setup where n vernier divisions equal n minus one main scale divisions, the least count is the value of one main scale division, divided by the number of vernier divisions. Clear? Family two, the screw family: the screw gauge and the spherometer. Least count is the pitch, divided by capital N, the number of circular scale divisions. So in the exam, first ask, vernier or screw? Then the formula is ready. As simple as that.

5:14Example 1: radius of curvature with a spherometer

Now, remember our puzzle from the start? Let's finish it. The legs of a spherometer form a triangle of side four point zero centimetres. On a convex lens, the reading rises by zero point two zero centimetres above the flat glass reading. Find the radius of curvature. Pause the video, and try it first. Done? Okay, let's check. Here l is four, and h is zero point two. First term: l squared is sixteen, and six h is one point two. Sixteen divided by one point two is about thirteen point three centimetres. So the answer is thirteen point three, right? No, no, careful! You forgot something. Which term? Yes, h by two! h by two is zero point two zero divided by two: zero point one zero centimetres. So R equals sixteen over one point two, plus zero point one zero: about thirteen point four centimetres. And there you go, mystery solved! The rise was only zero point two centimetres, just two millimetres, and from it we found the radius of the whole sphere, thirteen point four centimetres.

6:30Example 2: refractive index of a slab and its error

Now a harder one, JEE level. Read it with me. With a travelling microscope of least count zero point zero one millimetre, the real thickness of a glass slab is six point zero zero millimetres, and its apparent thickness is four point zero zero millimetres. Find the refractive index, and its maximum error. Pause and try. Hint: think of the relative errors from the errors chapter. Done? Okay, let's check. Mu equals real over apparent: six point zero zero divided by four point zero zero. One point five zero. Easy part done. Not at all! For a quotient, the relative errors always add, because we want the maximum error. So the relative error, delta mu by mu, equals zero point zero one over six, plus zero point zero one over four. That is about zero point zero zero four two. So the error in mu is zero point zero zero four two? Wait, no! That is only the relative error, delta mu by mu. We want delta mu itself. So, multiply by mu itself. Delta mu equals one point five zero times zero point zero zero four two: about zero point zero zero six. Multiply by mu itself: one point five zero times zero point zero zero four two, about zero point zero zero six. So, mu equals one point five zero zero, plus or minus zero point zero zero six. See how the answer keeps three decimal places, to match the error? Very neat.

8:04Quick revision: spherometer, microscope and least count

So, quick revision. Spherometer: R equals l squared over six h, plus h over two, with h measured from the flat glass reading. Travelling microscope: a vernier least count, and mu equals real depth over apparent depth. And least count: vernier family, one main scale division over n; screw family, pitch over the number of circular scale divisions. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!