The diffracted becomes the diffractor

The resultant intensity in the single slit diffraction experiment, assuming the far field limit, can be expressed in the form of the well known Sinc function:

I ( θ ) = A ( sin π a sin θ λ π a sin θ λ ) 2 I(\theta )=A{ \left( \frac { \sin { \frac { \pi a\sin { \theta } }{ \lambda } } }{ \frac { \pi a\sin { \theta } }{ \lambda } } \right) }^{ 2 }

If I carry out an experiment where in instead of a single slit I use a sheet of material that is partially permeable to the incoming light and with the special quality that at a distance x, from a chosen centre, along the sheet, the 'permeability' of the sheet is defined as ( sin ( a x ) ( a x ) ) 2 { \left( \frac { \sin { \left( ax \right) } }{ \left( ax \right) } \right) }^{ 2 } , where a is a constant chosen appropriately ( 10 3 ) \left( \approx { 10 }^{ 3 } \right) so that the permeability falls off rather quickly with increasing x, so as to preserve the validity of the far field assumption, then the resulting intensity pattern at a screen kept at a large distance D away with the same point chosen as the centre contains one maximum and no minimum.

If the length of the pattern on the screen, without making a small angle approximation, can be expressed as:

2 D a λ π p a q λ r \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \boxed { \frac { 2Da\lambda }{ \sqrt { { \pi }^{ p }-{ a }^{ q }{ \lambda }^{ r } } } }

Where λ \lambda is the wavelength of light used ( 10 3 m ) \left( \approx { 10 }^{ -3 }m \right)

Then find p + q r \frac { p+q }{ r } to the nearest integer


The answer is 2.

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