Past JEE Questions: Optics Experiments

Transcript

0:00Introduction

Hello everyone, a small question to start. An optical bench scale has four divisions in each centimetre, so it reads to a quarter of a centimetre. You read two marks and subtract them to get a distance. So the error in that distance is also a quarter of a centimetre, right? Hold on to your answer, because we will test it later today. We solve two past JEE questions on optics experiments, both on the error in a focal length. So, let's begin.

0:30Recall: the parallax method

First, a quick recall of the parallax method. In the figure, the curved line is a concave mirror with its pole, capital P, and the orange arrow pointing up, capital O, is the object pin. Now, a JEE Main twenty twenty six question asked this: where should the object always be placed? Yes, exactly, at any point beyond the focus of the mirror. Then the mirror forms a real image, and we need a real image so that the image pin can coincide with it. Okay? The image pin, capital I, is the dark arrow pointing down. We move it until it shows no parallax with the image of the object pin, as the note says, and then it marks the image. Right? Small u and small v, the double arrows at the bottom, are the distances from the pole, as magnitudes. Then f, the focal length, equals u times v, over u plus v. Clear? For a convex lens in the u v method, the same formula works, with distances from the lens. The two graphs on the right are from the earlier lesson, and we skip them today. With me?

1:36The error in the focal length

Now the main tool, the error in the focal length. The little triangle is capital delta, and delta f means the error in f. Delta f over f squared equals delta u over u squared, plus delta v over v squared, and the little two means squared. Okay? Multiply by f squared, and delta f equals f squared, times the same bracket. That is the error itself, in centimetres. Got it? But some questions ask for the percentage error, and for that we need delta f over f. So how do we get it from the first line? Yes, exactly, we multiply the first line by f, just once. So delta f over f, the relative error, is f times the same bracket. Times a hundred, with the cross for the times sign, it gives the per cent error. Clear?

2:26Where do Δu and Δv come from?

Now, where do delta u and delta v come from? The rule from earlier in this unit says a single reading is uncertain by the least count of the instrument. Okay? But on an optical bench, we read two marks and subtract them. So, do the errors of the two readings cancel? They do not cancel, because in a sum or a difference, the absolute errors add. For two readings, capital A and capital B, delta of A plus or minus B equals delta A plus delta B, and the plus or minus sign covers both cases, right? So a distance found from two marks on the bench carries two least counts of error. Keep this in mind for Example two. With me?

3:09Example 1: concave mirror, u 40 cm, v 120 cm, error 1/K

Now Example one, the easier one, from JEE Main twenty twenty three. In an experiment to estimate the focal length of a converging mirror, the image of an object placed forty centimetres from the pole is formed one hundred and twenty centimetres from the pole. These distances are measured with a modified scale having twenty small divisions in one centimetre. The error in the focal length is one over capital K centimetres, so find capital K, okay? Pause the video and try it first. Are you done? Okay, let's check. Converging means concave, so small u is forty and small v is one hundred and twenty. Then f equals forty times one hundred and twenty, over one hundred and sixty, which gives thirty centimetres. Right? Now the errors. The scale has twenty small divisions in one centimetre, so what is its least count? It is one small division, one over twenty centimetre, which is zero point zero five centimetre. Here each distance is taken as a single reading on this scale, measured from the pole, so delta u and delta v are each one least count, zero point zero five. With me? Now use delta f equals f squared, times the bracket, with f squared as nine hundred and the error as zero point zero five. In the bracket, u squared is one thousand six hundred, and v squared is fourteen thousand four hundred. Okay? Nine hundred times zero point zero five gives forty five. One over one thousand six hundred equals nine over fourteen thousand four hundred, so the bracket is nine plus one, over fourteen thousand four hundred. Forty five times ten gives four hundred and fifty, and that over fourteen thousand four hundred is exactly one over thirty two centimetre. Clear? So capital K equals thirty two. Very good if you got it on your own!

5:05Example 2: convex lens on optical bench, per cent error in f

Now Example two, the harder one, from JEE Advanced twenty nineteen. An optical bench has a one point five metre long scale, with four equal divisions in each centimetre. While measuring the focal length of a convex lens, the lens is kept at the seventy five centimetre mark, and the object pin at the forty five centimetre mark. The image of the object pin, on the other side of the lens, overlaps an image pin kept at the one hundred and thirty five centimetre mark. Find the percentage error in the measurement of the focal length. Pause the video and try it first. Are you done? Okay, let's check. Small u, from the pin to the lens, is seventy five minus forty five, which is thirty centimetres. Small v, from the lens to the image pin, is one hundred and thirty five minus seventy five, which is sixty centimetres. Okay? Then f equals thirty times sixty, over ninety, which gives twenty centimetres. Right? There are four divisions in each centimetre, so the least count is one quarter of a centimetre, which is zero point two five centimetre. With me? So delta u is just zero point two five centimetre, one least count, like in Example one, right? That is exactly the trap, and it is our puzzle from the start. Each distance here is a difference of two bench readings, so the errors add. Delta u and delta v are both two times zero point two five, which is zero point five centimetre. Got it? Now the relative error is f, which is twenty, times the bracket. Zero point five over nine hundred equals two over three thousand six hundred, so the bracket is two point five over three thousand six hundred. Twenty times two point five is fifty, and fifty over three thousand six hundred is one over seventy two. Clear? Times a hundred, the per cent error is one hundred over seventy two, which is about one point three nine per cent, so about one point four per cent. The slash with two small circles is the per cent sign, and the wavy equals sign means approximately. If you doubled the least count, very well done!

7:12Quick revision: error formula, least counts, parallax

So, quick revision. Focal length f equals u times v, over u plus v, and delta f over f squared equals delta u over u squared, plus delta v over v squared. One reading carries one least count of error, and a difference of two readings carries two. In the parallax method, the object goes beyond the focus, so that the image is real. Two more past questions wait on this page, on the travelling microscope and the prism. Try them first, and all the very best!