Geometrical optics of a light bulb

Transcription

1 Univerza v Ljubljani Fakulteta za matematiko in fiziko Oddelek za Fiziko Seminar - 4. letnik Geometrical optics of a light bulb Avtor: Blaž Zabret Mentor: Prof. dr. Gorazd Planinšič Ljubljana, Abstract: If we submerge a light bulb in the water, an interesting phenomenon can be observed. The outside surface appears to become reflective and the internal surface of the bulb appears to shrink. These phenomena can be explained using geometric optics only. The effect of the light diffusing coating inside the light bulb is examined and also the reason why some apparently coted light bulbs do not show this behaviour. In the conclusion, applicative side of this phenomenon is also discussed.

2 Content Abstract:... 1 Content ) Introduction ) Thin light bulb in water ) Light bulb with thick glass ) Relative size of the inner surface ) Coated light bulbs ) Applications and conclusion ) Literature

3 1.) Introduction When trying to explain the behaviour of nature we are sometimes faced with a daunting problem which might seem unsolvable. But we can help ourselves by starting with a smaller, similar, but understandable problem. Then, by gradually adding to it, we can gain the understanding of the original problem. This is also useful in education when trying to explain a complex phenomenon. By starting with a simple, understandable problem and gradually, step by step, making it more complex, pupils can get a better understanding of the phenomenon. In this paper, we are going to explain what happens to a light bulb when it is submerged in water using this approach. When a light bulb is submerged in water the inner surface of the light bulb appears to shrink and the outside surface appears to become reflective. This can be seen in Figure 1. One might think that this happens because water acts like a lens and by doing so, seemingly shrinks the inner surface of the bulb. But if this would be the case, one might ask oneself why the outer surface appears the same as in the air. The explanation as to why the inner surface of the light bulb appears to shrink when submerged in water is not that simple. This happens because the light rays experience a total reflection on the inner surface of the light bulb [1]. Figure 1: When one immerses an unlit light bulb in water the internal surface appears to shrink and the outside surface appears to become reflective. The light bulb in this figure is coated with the silica powder from the inside [1]. 2.) Thin light bulb in water To see why the inner surface appears to shrink, one must understand what happens to the light passing through the bulb. To simplify our problem we will first consider a thin light bulb. In a thin light bulb a small section of the bulb can be approximated with flat parallel surfaces as shown in Figure 2 and 3. To explain what happens when a thin bulb is submerged in water one can use Snell s law sin φ 1 sin φ 2 = n 2 n 1. Numbers 1 and 2 indicate the medium in which the light rays are travelling. Let us label the angle which is formed by ambient light rays entering the outer surface of the light bulb φ out. The angle that is formed by light rays entering the light bulb at the inner surface will be labelled φ in. We will label the angles which are formed within the glass φ go 3

4 and φ gi, as shown in the Figure 2. When the light bulb is in air, all light entering the bulb will pass through the glass and enter it, because the indexes of refraction outside and inside the bulb are equal; therefore, angles φ out and φ in most be equal as well. This is illustrated in Figure 2. But when the bulb is submerged in water, light entering the bulb can experience a total reflection on the inner surface of the bulb because the index of refraction outside the bulb is larger than that inside. Light ray which strikes outside surface at an angle φ out c leads to a critical angle for a total reflection at bulb s inner surface. This can be seen in Figure 3. Figure 2: Here we can see angles that are formed by the light rays passing through a flat parallel surface with air on both sides [1]. Figure 1: Here we can see angles that are formed by the light rays passing through flat parallel surface with water on one and air the other side [1]. Because indexes of refraction inside and outside the bulb are not equal when the bulb is submerged in water, φ out and φ in are also no longer equal. To determine the relation between the two angles, the Snell s law can be applied with inner as well as with outer surfaces. sin φ out sin φ go = n g n w, (1) sin φ gi sin φ in = n gas n g. (2) In equations (1) and (2) the n g, n w, n gas, stand for indices of refraction of glass, water and diffuse gas inside the light bulb. For a thin bulb the angles φ go and φ gi are equal. We can combine this with equations (1) and (2), and we get the relation sin φ out = n gas sin φ in n w (3) φ out = arcsin n gas sin φ in n w. (4) 4

5 When there is water with index of refraction n w = 1,33 on one side and diffuse gas with the index of refraction n gas = 1 on the other side of the glass, the critical angle on the outside surface at which we get total reflection on the inside surface is φ out c = 48,8. Any light ray that strikes outside surface at an angle which is greater than φ out c, will experience total reflection on the inside surface. A distant viewer will see only those light rays which will refract into his or her eyes. These rays can pass through the bulb or experience total reflection at the bulb s inner surface. Rays that pass through the bulb in Figure 4 and exit it on the top or the bottom of the bulb do not refract into distant viewers eyes. The only rays from these portions that he or she can see are those that experience total reflection. Because of that, the light bulb is no longer transparent on these portions and becomes reflective. This reflection leads to seemingly smaller inner surface. Figure 4: Sketch of image formation by tracing some typical rays. Lens in the Figure represents observer s eye. Rays that pass through the bulb and exit it on the outer portions of the bulb do not refract into distant viewers direction. The only rays from the outer portions that he or she can see are those which experience total reflection. Equation (4) suggests that total reflection on the inner surface happens independently of the light bulb s material as long as its index of refraction is larger than that of water. If the effect does not depend on the intermediate material from which the light bulb is made, we can remove the glass and just take a look at a spherical gas bubble in the water. We can apply the Snell s law to the water bubble interface and see that we get the same equation for critical angle of incidence as in equation (4). Thus we can see that we get the same effect with a spherical gas bubble as with a thin wall light bulb if we submerge them in water. 2.1.) Light bulb with thick glass Now that we understand image formation with a thin-wall light bulb let s go a step further and take into account the wall thickness. In this case the surface of the light bulb cannot be approximated with a parallel surface layer any more. We will show later that in this case the 5

6 critical angle for total internal reflection changes. The critical ray that experiences total internal reflection is shown in Figure 5. We will label the angle, at which this ray enters the bulb on the outer surface φ out c, because it differs from that of a light bulb of a negligible thickness. The angles formed inside the glass, φ go and φ gi, are also not equal, as those in the thin light bulb. The φ go is slightly smaller than φ gi. We can apply Snell s law to the both surfaces and get n w sin(φ out ) = n g sin(φ go ) for the outer surface and n g sin(φ gi ) = n gas sin(φ in ) for the inner surface. In Figure 5 we can see that the two radiuses and the light ray within the glass form a triangle. We can apply the law of sine and get sin(φ go ) R t = sin(π φ gi). (5) R To the previous equation we can apply trigonometric identity for sin(π α) and combine it with equations for light rays refracting at the inner and outer glass surface sin(φ go ) = (R t) n gas sin(φ R n in ) (6) g sin(φ out ) = (R t) n gas sin(φ R n in ). (7) w Figure 5: Figure shows a light bulb with exaggerated thickness. The critical light ray at which we get total internal refraction is shown as well [1]. The critical angle at which we get a total internal reflection is φ out c = arcsin { n gas (1 t )}. (8) n w R 3.) Relative size of the inner surface 6

7 The next step we will take in understanding light bulb s behaviour when submerged in water is how small the apparent inside surface is. First we will consider the thin light bulb, again to simplify the problem. If the viewer is above the light bulb, he or she will see only the light rays that will be parallel to a vertical axis after the refraction. If the angle at which the light ray strikes the outer surface of a light bulb is less than φ out c, it will pass through the bulb. But if the angle is larger than φ out c, the light bulb will appear as a mirror surface. For this to happen, the critical angle from the vertical axes must equal φ out c, as shown in Figure 6. There we can also see the outer radius R and the apparent inner radius rapp. We can approximate the ratio of r app R as r app R sin(φ out c) = n gas n w. (9) Figure 6: The light rays that strike the bulb at angles equal or larger than φ out c, will experience total internal reflection. If the angle of incidence is less than φ out c, the light will pass through [1]. We can verify the accuracy of this equation by comparing it to experimental measurements. If we measure the apparent internal and external diameter from Figure 1, we get the ratio r app R = 0,74. If we calculate the ratio from the equation (6), we get the ratio of 0,75. The difference is approximately 2% [1]. If we take glass thickness into account we can calculate a new ratio r app R r app R = n gas (1 t ). (10) n w R In the previous equation we can see that the glass thickens influences the size of the apparent inner surface. Thicker glass results in smaller apparent inner surface. 4.) Coated light bulbs Some light bulbs (so called frosted bulbs) have a diffusive coating on the inside. But not all of them show the apparent shrinkage of the inner surface when submerged in water. This can 7

8 be seen in Figure 7 (note that both light bulbs are submerged under the water). The main reason for the difference between the two light bulbs is in the coating. Light bulb which is shown in Figure 7a has a smooth inner surface, on which a coating of fine silica powder is deposited. Because the powder is not in optical contact with the inner surface of the bulb [1], the ambient light can experience a total reflection on the inner surface before interacting with the powder and, therefore, creates the apparent shrinkage of the inner surface. One such light ray is shown in the left part of the Figure 8. Figure 7: Here two different frosted light bulbs are submerged in water. A: The bulb with a powder deposit on the inside is shown. Total internal reflection is clearly visible on the edges of the bulb. Because powder is not in optical contact with the inner surface, light can experience total reflection before interacting with the powder. B: The bulb with rough internal surface is shown. Because of thet there is no constant angle at which we get total internal reflection and the outer portions of the bulb do not become reflective. Light which is reflected diffusively by the silica powder is being emitted at all angles. Rays that are emitted at angles 90 or less enter the glass and exit the bulb at angles smaller than φ out c. This can be seen in the right part of the Figure 8. Rays which are emitted with angles larger than 90 are being emitted into the bulb. These rays are than emitted by another particle. Figure 8: Light emitted by the silica particle leaves the bulb in a coned shape. The angle of the light cone is φ out c. Rays which experience total internal refraction and are seen by the distant observer strike the bulb at an angle which is larger than φ out c. But the bulb which is shown in Figure 7b is different. It does not have a coating of a powder on the inside surface. It has been etched (or maybe sanded) to obtain a rough inner surface. Because of that some light rays that strike outer surface at angles greater than φ out c can scatter inside the light bulb and do not experience total internal reflection. This is shown 8

9 in Figure 9. Because of thet there is no constant angle at which we get total internal reflection and the outer portions of the bulb do not become reflective. Figure 9: Light entering the light with rough inner surface can pass into the bulb even if the angle at the outer surface exceeds φ out cr. Because of rough inner surface there is no consistent angle at which we get total internal reflection. The difference between the bulbs which are shown in Figure 7a and 7b is visible if we look at them under a microscope. Powdery coating on 7A and rough surface on 7B can clearly be seen in Figure 10. One might think that the rough surface in Figure 7B is made of more fine and thin layer of powder. But if you take a closer look at the Figure, you will see dotty reflections on the surface. These reflections show that the surface is not coated with a fine powder like in Figure 7A, but, in fact, it is rough. Figure 10: A microscope was used to determine the difference in the coating on the inside of the bulb. A: The bulb is coated with fine powder and a total internal reflection can be seen on the outer portions of the bulb, when immersed in water. B: The bulb with a rough inner surface can be seen. Because of that total internal reflection is not clearly visible. 5.) Applications and conclusion 9

10 In this paper we have shown the steps one might take to understand the phenomena that occur when light bulb is submerged in water. The apparent shrinkage of the inner surface was explained with total internal reflection, because the angle of incidence for light rays that strike the outer portions of the light bulb is larger than the critical angle φ out cr. The phenomenon of a submerged light bulb is already presented in House of experiments with educational purposes. There one can find an experiment called Total reflecting light bulbs. This experiment consists of several different light bulbs that can be immersed in the water as shown in Figure 11. By observing and implementing this experiment people of all ages can learn about total refraction. Figure 11: By rotating the experiment viewer can immerse different light bulbs in water. In the middle of the experiment there are instructions how to do the experiment and qualitative explanations of the observed phenomena [3]. When a person is faced with a hard task, it is very useful to start with a simplified problem and then making it more complex step by step. This does not apply only to the science and laws of nature but also to everyday life. 10

11 6.) Literature [1] M. C. James, Am. J. Phys. 76, 856 (2008) [2] J. Strnad, Fizika 2. Del (DZS, Ljubljana, 1978) [3] House of experiments, 11