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Light Intensity - help10

A Physics formula is more than just an algebraic recipe for solving word problems. It is a guide to thinking about how a change in one variable would alter another variable. The Illuminance (rate at which light from a source falls on a given surface area some distance away) depends on two variables. Illuminance is directly proportional to the power of the light bulb. And the illuminance is inversely proportional to the square of the separation distance between the light source and the surface.

There are three similar versions of this question. Here is one of those versions:

Version 1:

A 60-Watt bulb illuminates a surface a distance d away with an illuminance of 1800 lux. What would be the illuminance on a surface a distance 2•d away (i.e., twice as far away) when illuminated by a 30-Watt bulb?

Illuminance is the rate at which light from a source lands on a given area of surface some distance away. Illuminance depends on two variables - the power (in Watts) of the light bulb and the distance between the bulb and the surface. In this question, both variables are changed. So you will need to make two changes to the original illuminance value. One change will take into account the change in power. The second change will take into account the change in distance. To do so, get yourself organized and be prepared to make two changes to the original force.

Power Change
The value for illuminance is directly proportional to the power of the light source. You are provided the illuminance value on a surface due to a 60-Watt bulb. You are then asked to find the illuminance for a 30-Watt bulb. This is a halving of the light bulb power. So the first change you will need make is to halve the illuminance value (i.e., divide it by 2).

Distance Change
The value for illuminance is inversely proportional to the square of the distance between the light source and the surface. So if the distance increases, the illuminance value decreases. And the factor by which the illuminance decreases is the square of the factor by which the distance increases. So if the distance increases by a factor of 2 (i.e., is doubled), then the illuminance decreases by a factor of 4 (2²). In this question the distance is doubled (2•d). So the second change you will need to make is to quarter the illuminance value. To quarter means to divide by four.

These two changes, when made to the original illuminance value, will provide the accurate answer for the new illuminance value.

Try these links to The Physics Classroom Tutorial for more help with the topic of illuminance:

Sorry. This topic is currently not addressed at The Physics Classroom Tutorial.