- #1

- 6

- 0

## Main Question or Discussion Point

hey quick question about Fraunhoffer diffraction:

RESTULT 1

for a rectanguar apperture, we have the following equation for irradiance I:

I = I(0)(sin(B)/B)^2(sin(A)/A)^2 = I(0)[sinc(B)]^2[sinc(A)]^2

where, B = bu/2; A = av/2

where b = apperture breadth (x-axis); a = apperture height (y-axis)

and u = ksin(theta_x) v = ksin(theta_y)

and where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie at theta_x = 0, and theta_y = 0.

RESULT 2

Now, I also have the equation for single slit diffraction:

I = I(0)(sin(B)/B)^2 = I(0)[sinc(B)]^2

also here, B = bu/2;

where b = apperture breadth (x-axis);

and u = ksin(theta_x)

where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie where theta_x = 0.

QUESTION:

RESULT 1 and 2 can be derived independently of each other using double intregral formulas. Now my question is as follows:

is it satisfactory to think of RESULT 1 reducing to RESULT 2 in the limit that (say) the height (y-axis dimension) of the rectuangular apperture approaches zero?

In this case, the apperture height, a ----> 0

thus A-----> 0

thus sinc(A)----> 1

thus (from RESULT 1) I----> I(0)(sin(B)/B)^2 x 1 = I(0)sinc(B)^2 as required.

This seems to make sense to me. and what I get is a result where by a single (infinitely thin) slit of length b in the x-dimenstion, produces a diffraction pattern on the screen in teh SAME dimension (ie the x-dimension).

WHY I ASK.........is that in one form of the derivation of RESULT 2 for the single slit I have come across, the following is stated:

'consider a slit with width b and assume the hight is infinite'.

FYI: In this derivation, the width is still along the x-axis, and the height is along the y-axis.

the working then proceeds to claim that SINCE we have assumed the slit is infinitely high, we need not consider integrating across the y-variable, and so the integration problem is simplifed to a single integration from -b/2 to +b/2 in the x-variable. RESULT 2 is then achieved.

YOU MAY HAVE SEEN MY PROBLEM AT THIS POINT (or maybe I have missed the point):

(i) in the derivation I just related to you, we end up with RESULT 2, but we do so by applying the fraunhoffer double integral to a slit of finite width b (along the x-axis) and infinite height a (along the y-axis)

WHEREAS

(ii) when I make RESULT 1 reduce to RESULT 2, I do so by assuming that our slit has finite width b (along the x-axis) but ZERO hieght (a = 0) along the y-axis.

Could you please comment on my thoughts, and point out my error if one exists. THANK YOU!

N.B.

when I refer to the fraunhoffer double integral, I refer to the following:

Electric Field At point P on the Screen =

constant x {double integral accross apperture of:}(e^-i(ux+vy))dxdy

where u and v are defined in the section called RESULT 1 above.

RESTULT 1

for a rectanguar apperture, we have the following equation for irradiance I:

I = I(0)(sin(B)/B)^2(sin(A)/A)^2 = I(0)[sinc(B)]^2[sinc(A)]^2

where, B = bu/2; A = av/2

where b = apperture breadth (x-axis); a = apperture height (y-axis)

and u = ksin(theta_x) v = ksin(theta_y)

and where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie at theta_x = 0, and theta_y = 0.

RESULT 2

Now, I also have the equation for single slit diffraction:

I = I(0)(sin(B)/B)^2 = I(0)[sinc(B)]^2

also here, B = bu/2;

where b = apperture breadth (x-axis);

and u = ksin(theta_x)

where I(0) is the irradiance evaluated along the central axis between the apperture and the screen, ie where theta_x = 0.

QUESTION:

RESULT 1 and 2 can be derived independently of each other using double intregral formulas. Now my question is as follows:

is it satisfactory to think of RESULT 1 reducing to RESULT 2 in the limit that (say) the height (y-axis dimension) of the rectuangular apperture approaches zero?

In this case, the apperture height, a ----> 0

thus A-----> 0

thus sinc(A)----> 1

thus (from RESULT 1) I----> I(0)(sin(B)/B)^2 x 1 = I(0)sinc(B)^2 as required.

This seems to make sense to me. and what I get is a result where by a single (infinitely thin) slit of length b in the x-dimenstion, produces a diffraction pattern on the screen in teh SAME dimension (ie the x-dimension).

WHY I ASK.........is that in one form of the derivation of RESULT 2 for the single slit I have come across, the following is stated:

'consider a slit with width b and assume the hight is infinite'.

FYI: In this derivation, the width is still along the x-axis, and the height is along the y-axis.

the working then proceeds to claim that SINCE we have assumed the slit is infinitely high, we need not consider integrating across the y-variable, and so the integration problem is simplifed to a single integration from -b/2 to +b/2 in the x-variable. RESULT 2 is then achieved.

YOU MAY HAVE SEEN MY PROBLEM AT THIS POINT (or maybe I have missed the point):

(i) in the derivation I just related to you, we end up with RESULT 2, but we do so by applying the fraunhoffer double integral to a slit of finite width b (along the x-axis) and infinite height a (along the y-axis)

WHEREAS

(ii) when I make RESULT 1 reduce to RESULT 2, I do so by assuming that our slit has finite width b (along the x-axis) but ZERO hieght (a = 0) along the y-axis.

Could you please comment on my thoughts, and point out my error if one exists. THANK YOU!

N.B.

when I refer to the fraunhoffer double integral, I refer to the following:

Electric Field At point P on the Screen =

constant x {double integral accross apperture of:}(e^-i(ux+vy))dxdy

where u and v are defined in the section called RESULT 1 above.