IIT-JEE Questions (1978–2006): Lenses

Transcript

0:00Introduction

Hello everyone! Here is something strange. An object creeps towards a convex lens at just one centimetre per second, but at one point, its image moves at nine centimetres per second. How can an image outrun its own object? Today we solve two past IITJEE questions on lenses, from two thousand and five and two thousand and four; the second is exactly this puzzle. Lens questions like these come up often, so let's begin.

0:33Thin lens: sign convention and formula

So, first, let us collect our tools. In the figure: a convex lens on the left, a concave lens on the right; capital F marks a focus, and two F is twice the focal length away. Our sign convention: distances from the lens are plus along the incident light, and minus against it. Okay? Now, u is the object distance, and v is the image distance. The light travels to the right, and the object, the black arrow, stands on the left. So tell me, what sign does u take? Negative, yes, exactly! The object sits against the light, so u is less than zero; that little sign means less than. Right? Now the lens formula says that one over v, minus one over u, equals one over f, where f is the focal length. It holds for thin lenses and paraxial rays, the rays close to the axis. With me? And f has a sign too: greater than zero for a converging lens, like the convex one, and less than zero for a diverging lens, like the concave one. Clear? Next, the magnification, m, equals v over u, and from the lens formula, f over f plus u. Example two runs on this second form. Got it? For a real object, so u less than zero, negative m means real and inverted, like the convex lens image, and positive m means virtual and erect, like the concave lens image. Right?

1:56Lenses in contact: the powers add

Now tool number two: for thin lenses in contact, one over capital F equals one over f one, plus one over f two. The small one and two below label the two lenses, and capital F is the focal length of the combination. Okay? One over f is the power, capital P, and with f in metres, its unit is the dioptre, the capital D in brackets. Right? So in contact, P equals P one plus P two: the powers add. And a converging lens has positive power. With me? Here students often slip: a concave lens diverges, so its focal length and its power carry the minus sign. Forget it, and the answer flips. Clear?

2:40An object moving along the axis

Tool number three, the star of our puzzle. What if the object moves along the axis? The lens is fixed, so f is constant: we differentiate the lens formula with respect to time. Okay? One over v gives minus one over v squared, the little two up top, times d v over d t, the rate at which v changes. One over u gives plus one over u squared, times d u over d t, and the constant gives zero. Right? Rearrange: d v over d t equals v squared over u squared, times d u over d t. But v over u is the magnification, so this is m squared, the magnification squared, times d u over d t. With me? Now, the magnification squared can never be negative. So which way does the image move? Yes, exactly, always in the same direction as the object. And its velocity is the magnification squared times the object's velocity, not just the magnification, okay?

3:36Example 1: focal lengths of lenses in contact (IIT-JEE 2005)

Example one, from IITJEE two thousand and five. Here, a convex lens is in contact with a concave lens. Their focal lengths, in magnitude, are in the ratio two is to three, the colon on screen, and the combination has focal length thirty centimetres; c m means centimetres. Find the individual focal lengths. Pause the video and try it first. Done? Okay, let's check. So which lens gets the two, and which gets the three? The combination's focal length is thirty centimetres, taken as plus thirty, so the pair converges. So the convex lens must be stronger, with the smaller focal length, and it gets the two. Okay? So f one, the convex lens, is plus two x, and f two, the concave lens, is minus three x, minus because it diverges. Right? The powers add, so one over two x, minus one over three x, is one over six x. That equals one over thirty, and the arrow means this gives x equals five centimetres. With me? So the convex lens has focal length ten centimetres, and the concave lens minus fifteen centimetres. Clear? Let us check. One over ten, minus one over fifteen, is three minus two, over thirty, which is one over thirty, and the tick mark says it works. And if you gave the convex lens the three? Then one over three x, minus one over two x, is less than zero: a diverging pair, impossible here. That is the trap, okay?

5:07Example 2: image speed and rate of change of magnification

Example two, our puzzle, from IITJEE two thousand and four. Here, an object approaches a convex lens of focal length zero point three metres along its axis, at zero point zero one metres per second; the m s, with a small minus one up top, means metres per second. When it is zero point four metres from the lens, find the speed of the image, and the magnitude of the rate of change of the lateral magnification. Pause the video and try it first. Done? Okay, let's check. For a real object, u is minus zero point four metres, and for this converging lens, f is plus zero point three. The object moves towards the lens, so is d u over d t positive or negative? It is positive, very good! Moving from minus zero point four towards zero, u is increasing, so d u over d t is plus zero point zero one metres per second. Okay? Now the magnification, f over f plus u. Here f plus u is zero point three minus zero point four, which is minus zero point one, so m is zero point three over minus zero point one, which gives minus three. The image is real, inverted, and three times as tall. Clear? Then surely it moves three times as fast: three times zero point zero one gives zero point zero three metres per second, right? No, no, careful, that is the trap! The speed goes with the magnification squared. Minus three, squared, is nine, and nine times zero point zero one gives zero point zero nine metres per second. That is nine centimetres per second, and our puzzle is solved! Got it? Now the second part. Differentiate m equals f over f plus u with respect to time: d m over d t equals minus f, over f plus u, whole squared, times d u over d t. Here f plus u, whole squared, is zero point zero one, giving minus zero point three over zero point zero one, times zero point zero one. With me? So d m over d t is minus zero point three per second; the s with the minus one up top means per second, as m has no unit. The question asks for the magnitude, shown by the two straight bars: zero point three per second. Clear?

7:22Quick revision: lens formula, powers add, image speed

So, quick revision. The lens formula says one over v, minus one over u, equals one over f, with u less than zero for a real object, and m equals f over f plus u. In contact, the powers add, and a concave lens has negative f. For a moving object, the image velocity is the magnification squared, times the object velocity. Now try the practice problems on this page. All the best!