Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An object is sixty centimetres in front of a convex mirror, and its image is fifteen centimetres behind it. What is the focal length? Many students write one over fifteen, plus one over sixty, which gives one over twelve. So it is twelve centimetres, right? Hold that answer; one small sign will decide it. This is part A of our optics problem bank, at JEE Main level. So, let's begin.

0:31Signs first: the mirror formula

So, every mirror and lens problem here starts with signs. In the figure, light comes from the left, onto a concave mirror, and on the right, a convex mirror. We measure every distance from P, the pole, the centre of the mirror's surface; F marks the focus, C the centre of curvature, and the dark red arrows are the images. Along the incident light is positive, the plus sign; against it is negative, the minus sign. Okay? Now, u is the object distance, and the object, the black arrow from A up to B, sits on the left. So, is u positive or negative? It is negative, yes, exactly! We measure back to the object, against the light, so u is less than zero; that is the less-than sign on screen. Now f, the focal length: for a concave mirror, the focus is in front, so f is less than zero; for a convex mirror, it is behind, so f is greater than zero, the greater-than sign. Right? Now the mirror formula, with v the image distance. One over v, plus one over u, equals one over f. It holds for paraxial rays, the rays close to the axis. Clear? And m, the magnification, equals minus v over u. Negative m means an inverted image, and positive m means an erect one. With me?

1:57Warm-up: a concave mirror

Let us warm up with a concave mirror. An object four centimetres high is eighteen centimetres from a concave mirror of focal length twelve centimetres; c m means centimetres. Find the position and height of the image, okay? Let us put the signs in first. The object is in front, so u is minus eighteen centimetres. The mirror is concave, so f is minus twelve centimetres. Right? From the mirror formula, one over v equals one over f, minus one over u. That is minus one over twelve, plus one over eighteen. Taking thirty six as the common denominator, this gives minus one over thirty six. Got it? So v is minus thirty six centimetres. Negative v means the image is in front of the mirror, so it is real, thirty six centimetres from the mirror. Clear? Next, m equals minus v over u. Minus thirty six over minus eighteen is two, and the minus sign in front makes it minus two. Okay? Now the image height, h prime, the h with the small dash, equals m times h, the object height; the cross is the times sign. Minus two times four gives minus eight centimetres. So the image is eight centimetres high, and the minus sign says inverted. With me?

3:20The puzzle: a convex mirror

And now, our puzzle from the start. An object sixty centimetres from a convex mirror has its image fifteen centimetres behind the mirror. Find the focal length. So, is it twelve centimetres, as many students write? No, no, careful, that twelve forgot the signs! The object is in front, against the light, so u is minus sixty centimetres. The image is behind the mirror, along the light, so v is plus fifteen centimetres. Right? Now, one over f equals one over v plus one over u, which is one over fifteen, minus one over sixty. Taking sixty as the common denominator, this gives three over sixty, which is one over twenty. Clear? So f is plus twenty centimetres, not twelve. And it came out positive, exactly as a convex mirror needs, a nice built-in check. So one sign solved our puzzle. Got it?

4:16Lenses: same signs, a new formula

Now let us move on to lenses; the sign rules stay exactly the same. The lens formula is one over v, minus one over u, equals one over f, for a thin lens and paraxial rays. Careful with that minus sign; the mirror formula had a plus. Okay? In the figure, the lens on the left is convex, a converging lens, so f is greater than zero. The F marks the focus, and two F sits at twice the focal length. The one on the right is concave, a diverging lens, so f is less than zero. Right? When v is greater than zero, the image is real, on the far side of the lens. When v is less than zero, it is virtual, on the object's side. Got it? And for a lens, m is just v over u, with no minus sign this time. Negative m still means inverted, and positive m means erect. Clear?

5:11Example 1: concave lens, focal length 12 cm, object at 24 cm

Now, example one. An object is twenty four centimetres from a concave lens of focal length twelve centimetres. Find the image distance and the magnification. Pause the video and try it first. Done? Okay, let's check. The focal length is twelve centimetres, so f is plus twelve, right? No, careful, that is the trap! Our lens is concave, a diverging lens, so f is minus twelve centimetres. And the object is in front, so u is minus twenty four centimetres. With me? From the lens formula, one over v equals one over f, plus one over u. That is minus one over twelve, minus one over twenty four. Taking twenty four as the common denominator, this gives minus three over twenty four, which is minus one over eight. Okay? So v is minus eight centimetres. It is negative, so the image is virtual, on the object's side. Right? And m equals v over u, so minus eight over minus twenty four gives one third. It is positive, so the image is erect, and one third the size: virtual, erect and smaller, as a concave lens always gives. If you caught that minus sign on f, very good!

6:32Example 2: convex lens image on a screen 1 m away, find f, m

And now, example two, a little harder. Here, a convex lens forms on a screen one metre from it, the image of an object twenty five centimetres from it; the upright m there is the metre. Find the focal length and the magnification. Pause the video and try it first. Done? Okay, let's check. The screen is one metre away, but the object is twenty five centimetres away. So, can we put v equals one and u equals minus twenty five straight into the formula? No, no, wait, the units are mixed! Put everything in centimetres first: one metre is one hundred centimetres. Clear? Now let us fix the signs. An image on a screen is real, so it is on the far side, and v is plus one hundred centimetres. The object is in front, so u is minus twenty five centimetres. Right? Then, one over f equals one over v, minus one over u, which is one over one hundred, plus one over twenty five. Taking one hundred as the common denominator, this gives five over one hundred, which is one over twenty. Okay? So f is plus twenty centimetres; positive, as a converging lens should be, a nice check again. With me? Finally, m equals v over u, so one hundred over minus twenty five gives minus four. So the image is real, inverted, and four times the size of the object. Got minus four, sign and all? Then you are doing really well!

8:07Quick revision: signs, mirror vs lens formula, m, same units

So, quick revision. Signs first: u is negative for a real object, and f is negative for a concave mirror and a diverging lens. For a mirror, one over v plus one over u equals one over f, and m is minus v over u. For a lens, one over v minus one over u equals one over f, and m is just v over u. Keep the same units throughout, and a negative m means an inverted image. Now attempt the rest of part A on this page as a test, before reading the solutions. All the best!