6 Volt 2 watt solar cell panel from adafruit. With chalk, mark off 2, 4, and 6 meters on the floor.Īs a quantitative experiment, this demonstration works remarkably well, even with the room lights on, as long as the light source is not near a wall or other reflecting surfaces.ġ. The distance is varied by moving the light source on a cart. Setting it up: The light detector remains fixed in position. A video camera is needed to show the meter's display to the audience. This divides the current by two and the meter reads 4.0, 1.0, and 0.45 mA at the prescribed distances. A diagram of the experiment is given below. ![]() It works remarkably well! A 5 mA analog meter is used (because we don't have a 10 mA meter) with a 3.3 ohm shunt across its input. A simple experiment to carry out with a lamp, a light meter and a ruler is to measure the effect of distance on the irradiance from the lamp as it is moved away from the meter. ![]() (The current can also be measured by a digital meter or computer.) Measured currents are 8.0, 2.0, and 0.9 mA at distances of 2, 4, and 6 meters, respectively. ![]() 1 The output current is directly proportional to the intensity of the light falling on the panel and the current is displayed on an analog milliameter. How it works: For the point light source, we use a 1500 watt clear light bulb. What it shows: The intensity of light from a point source decreases as 1/r 2, where r is the distance from the source. According to the inverse square law, the intensity, I, of the radiation from a point source depends on the distance, x, from the source Intensity is proportional to the corrected count rate, C, so Comparing this to the equation of a straight line, y mx. Watch out for simple ratios like this it can speed up the calculation.Inverse square law, luminosity We can see that the distance has halved so 1/2 2 = 1/4, therefore the light has 4 times the power The light irradiance of a 40 W lamp is measured by a pupil with a light level meter to be 1.6 Wm -2 at a distance 1.5 m from the lamp.Ĭalculate the light irradiance at a point 0.75 m from the lamp Calculate the Irradianceof the light at a distance of 25 mfrom the source. In the real world, the incident light is very rarely normal to a surface nearly always light. In the case of constant light source intensity I, it can be said that: E 2 /E 1 r 12 /r 22 (r 1 /r 2) 2 Equation. The Irradiance of light from a point source is 10 Wm -2 at a distance of 5 m from the source. It is evident that the illumination is inversely proportional to the square of the measured distance from the light source. This also leads on to being able to compare two results directly using: I 1 d 1 2 = I 2 d 2 2 Questions Where k is the gradient of the straight line (or the constant of proportionality). This graph allows us to develop the other two relationships required in the course. Thus, at three times the original distance, the intensity of the light passing through a single square will be 1/9 of the original intensity. The graph of the above data is shown below: Therefore irradiance is inversely proportional to the square of the distance from a point source. When the irradiance is plotted against 1/d 2 the graph shows direct proportionality. ![]() On plotting these results we get a curve showing an inverse relationship between irradiance and distance The surface of the desk should be darkened to avoid reflection and the room darkened, with no other sources of light to determine the relationship.Ī table of results is given below Distance to meter (m) Calculate the irradiance of the lamp.Ī simple experiment to carry out with a lamp, a light meter and a ruler is to measure the effect of distance on the irradiance from the lamp as it is moved away from the meter. The power from a lamp is measured as 0.25 W across an area of 100 cm 2. The units are therefore Watts per metre squared (W m -2). This is given by the following equation: I = P Aįrom the relationship we can see that the definition of irradiance is the power per unit area of light incident on a surface. When looking at the power output from a point source of light (simple filament lamps provide a good approximation) there is a relationship to power of the source and the amount of light recieved per unit area. Professor Brian Cox shows in the video how this was worked out by John Herschel in the early 1800's. The Sun is radiating energy in all directions and here on Earth we can work out how much energy is received on the surface per square metre.
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