How to calculate the area of the Umbra created by a POINT LIGHT SOURCE?

Question

Ex :
The light source and the object have both 5cm radius ,and are 20cm separeated from each other and the object is 20cm away from the screen .
What is the area of the umbra created from the light source?

Answer 1

If it were a POINT light source, the umbra would have 4 x the object's area, or 314.2cm^2. There would be no penumbra.

But with a 5cm. radius source, the umbra area will be the same as the object's, = 78.55cm^2.

Answer 2

The first part describes a point source and the next a distributed source.

For any source you may do it be a geometrical construction.



Placing an obstacle at a scale position on paper and drawing lines from both sides of the light source to find where the light actually can reach.



For a point source you create similar triangles. The umbra becomes larger in proportion to the distance



For the case given the lines become parallel and the size of the umbra does not alter with distance.



If the source is larger than the obstacle then the umbra diminishes with distance and there is some distance at which it ceases to exist at all.

Answer 3

Your example rather contradicts your question! A light source which has the same radius as the object is NOT a point light source.



In any event, the approach is the same. You draw a diagram similar to this one, but using your given information:-



http://www.cyberphysics.co.uk/topics/light/shadow/shadow4.gif



The rest is just geometry rather than physics. In your example, you should see that the radius of the umbra shadow on the screen will also be 5 cm.



Area of a circle of radius r = πr² = π * 5² = 78.5cm²



If it WAS a point light source (A source with zero radius), there would be no penumbra and you would solve using similar triangles:



[Shadow size] / [object size] = [source to screen distance] / [source to object distance]



Again, just geometry rather than physics.

Answer 4

The umbra is the black or total portion of the shadow. If both the light source and the obstruction have a radius of 5cm, then the umbra will have a radius of 5cm also, so the area will be pir² or 25pi cm²