Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An air bubble trapped in a block of glass is round, shaped like a convex lens. And a convex lens converges light, right? So does this bubble converge light too? Keep your answer in mind; Example one settles it. Today, quick concept checks on optics, from mirrors to diffraction. Such why-questions do appear in JEE, so let's begin.

0:26Check 1: why a convex rear-view mirror?

So, check one. Why is a convex mirror used as the rear-view mirror of a car or a scooter? Look at the right half of the figure; we skip the concave mirror on the left. The blue curve here is a convex mirror, with its pole P, and its focus F and centre of curvature C behind it. The black arrow on the left is the object. The dashed lines extend the reflected rays backwards, and meet in a small upright image, the short brown arrow between the pole and the focus. Wherever the object is, a convex mirror gives a virtual, erect, diminished image. Okay? Since every image is diminished, a wide stretch of the road behind fits into one small mirror: a wide field of view. And the image is always erect. That is the whole answer. Right?

1:17Check 2: a lens in water, a mirror in water

Now check two. Put a glass lens in water, and its focal length changes. Put a mirror in water, and its focal length does not change. So what makes the difference? Start with the lens maker's formula for a thin lens. One over f equals the bracket n L over n m, minus one, times the bracket one over R one, minus one over R two. Here f is the focal length, n L, with a small L below, is the refractive index of the lens, and n m, with a small m below, that of the medium; R one and R two are the radii of the two surfaces. Okay? So refraction depends on the relative index, n L over n m. In water, n m changes, so this ratio and the focal length change too. Clear? But a mirror has f equals capital R over two, with capital R the radius of curvature, so its shape alone fixes the focal length. Reflection does not depend on the medium at all. With me?

2:16Check 3: can a convex-shaped lens diverge light?

Now check three, with the same formula. Can a lens of convex shape ever diverge light? It can, and the formula shows exactly when. Put the lens in a medium of higher index than its own material, so n m is greater than n L; that is the greater than sign on screen. Then n L over n m is less than one, so the first bracket is less than zero. Right? Now the second bracket, with the course's sign convention. All distances are measured from the lens; those in the direction of the incident light are positive, and those against it are negative. So for a convex lens, with light coming from the left, R one is positive and R two is negative, and one over R one, minus one over R two, is positive. With me? So negative times positive is negative: f is less than zero, and a negative focal length means the convex lens diverges light. Strange, but it is just the formula. Got it?

3:15Check 4: total internal reflection from air into glass?

Now check four, a quick one. Can total internal reflection occur when light goes from air into glass? Recall the two conditions on screen. First, light must go from a denser medium to a rarer one; the arrow means going into. Second, the angle of incidence, i, must be greater than theta c, the critical angle, written theta with a small c below. Okay? Air into glass is rarer into denser, so the first condition fails. So the answer is no, at any angle of incidence. Clear?

3:51Check 5: why is the central fringe white?

Now check five. Light up a double slit with white light. The central fringe is white, but the fringes beside it are coloured. So why is the centre white? At the centre of the screen, the two slits are equally far away, so the path difference, capital delta, the little triangle, is zero. And that is true for every wavelength, lambda. Right? Zero path difference means a maximum. So every colour has its bright fringe right at the centre, and all the colours together make white. Clear? Away from the centre, the fringe width beta equals lambda times capital D, over small d, for small angles; capital D is the screen distance, and small d the slit separation. So each colour has its own beta, and the colours separate, violet nearest the centre and red farthest. Got it?

4:45Check 6: one slit closed, or one very wide slit

Now check six. In a double-slit experiment, one slit is closed. What is seen on the screen now? With one slit, there is no second coherent source to interfere with, so the fine interference fringes vanish. Okay? But the screen is not dark. The open slit gives its own broad single-slit diffraction pattern, as in the figure: light through a slit of width small a, a screen at distance capital D, theta, the angle from the centre line, and the pattern drawn beside the screen. Its bright central maximum has width two lambda capital D over small a; we skip the formulas on the right. Right? Now make that slit very wide compared with the wavelength; the double greater than sign means much greater than. Then the central maximum shrinks to the sharp geometrical image of the slit, and diffraction becomes unnoticeable. Clear?

5:42Example 1: air bubble in glass, converging or diverging?

Now, Example one, and it is our puzzle from the start. An air bubble is trapped inside a block of glass. Does it behave as a converging or a diverging lens? Pause the video and try it first. Are you done? Okay, let's check. The bubble is round, so as a lens it is convex-shaped. But the material of this lens is air. Okay? So, convex shape, and a convex lens converges light. Is that our answer? No, no, careful, that is the trap! Around the bubble is glass, of higher index than air, so n m is greater than n L, exactly the case of check three. Right? So n L over n m, minus one, is negative, and the focal length is negative. The bubble is a diverging lens: a convex-shaped region of lower index spreads light out. And that settles our puzzle.

6:37Example 2: YDSE slits made wider, how the fringes change

Now Example two, a harder one. In a double-slit experiment, the width of each slit is slowly increased, keeping the separation fixed. How does the pattern change? Pause the video and try it first. Are you done? Okay, let's check. Many students say that wider slits give wider fringes. Is that right? No, no, careful, that is not right at all. The fringe width beta equals lambda capital D over small d, and the slit separation, small d, is fixed. The slit width is not even in this formula, so beta is unchanged. Right? But each wider slit lets in more light, since intensity is proportional to the slit width. So the fringes get brighter. Okay? Now look at the figure again. Each slit alone makes this single-slit pattern, and the double-slit fringes are seen inside its central maximum, the envelope, of width two lambda capital D over small a. As small a grows, the envelope narrows, but beta does not, so fewer fringes are seen. With me? And when the slits become very wide, the fringes disappear altogether; a clear pattern needs narrow slits. So, same fringe width, brighter, fewer fringes, and finally none. That is the full answer!

7:59Quick revision: lens in water, TIR, white fringe, wide slits

So, quick revision. The convex mirror gives an erect, diminished image and a wide view, and water changes the focal length of a lens, not of a mirror. When n m is greater than n L, a convex shape diverges, like the bubble in glass; and total internal reflection, the TIR on screen, needs denser to rarer, with i greater than theta c. The centre is white as the path difference is zero for every colour, and wider slits keep beta but give brighter, fewer fringes. Now try the practice problems on this page. All the best!