Diffraction Understanding a solution

[Hebel]

Homework Statement

(see attached problem)

Homework Equations

(see attached problem)

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?

 

[blue_leaf77]

'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?

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

 

[drvrm]

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?

halving its diameter, will decrease its area by a factor of 16.

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.

 

[Hebel]

Why does twice the width mean 4 times the area?

 

[blue_leaf77]

Why does twice the width mean 4 times the area?

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

 

[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?

 

[blue_leaf77]

so we are just assuming that the area of the beam incident on the screen will be circular right?

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