# Diffraction Understanding a solution

1. Mar 9, 2016

### Hebel

1. The problem statement, all variables and given/known data
(see attached problem)

2. Relevant equations
(see attached problem)

3. The attempt at a solution
So assuming the area of the hole is circular, halving its diameter, will decrease its area by a factor of 16. Therefore LI will be I/16? However in the solution (see attached solution) it says: 'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?

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2. Mar 9, 2016

### blue_leaf77

Because $P=IA$, therefore any change in the scaling of $A$ will result in the same amount of scaling in $P$.

3. Mar 9, 2016

### drvrm

perhaps some error is there- Area is proportional to (d/2)^2 !

i think one should remember 'as given' that halving the diameter will decrease the area of hole by a factor of four - but at the same time the diffracted beam increases its width by a factor of 2 - so it leads to a further spread of the same power in four times the area (decrease by a factor of four )so one gets a net 1/16 of the power.

4. Mar 9, 2016

### Hebel

Why does twice the width mean 4 times the area?

5. Mar 9, 2016

### blue_leaf77

$A_1=\frac{1}{4}\pi D_1^2$. Now if you set $D_2=2D_1$, what is $A_2=\frac{1}{4}\pi D_2^2$ in terms of $A_1$?

6. Mar 9, 2016

### Hebel

So the new area is 4 times the old one I see, so we are just assuming that the area of the beam incident on the screen will be circular right?

7. Mar 9, 2016

### blue_leaf77

You are supposed to assume that the illumination is of plane wave from the left.