For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?

Explanation

Note : Explanation about "Fresnel distance".
Consider Fraunhofer diffraction of monochromatic light produced by single slit.


Here for first order minimum, path difference is,
`dsintheta_(m)=mlamda`
`:.dsintheta_(1)=lamda" "("":.m=1)`
`:.sintheta_(1)=(lamda)/(d)`
( `:.` when `theta_(1)` is very small and measured in radian `sintheta_(1)=theta_(1)` )
`implies` Angular width of central maximum (in both the figures given above)
`=2theta_(1)=(2lamda)/(d)`
(i) In figure-1, linear width of central maximum is `(2lamdaD_(1))/(d)` and the width of beam incident on the screen is `A_(1)B_(1)`.
(ii) In figure-2, linear width of central maximum is `(2lamdaD_(2))/(d)` and the width of beam incident on the screen is `A_(2)B_(2)`.
When linear width of beam (AB) made incident on the screen is greater than linear width of slit (d), light rays are said to be diffracted (or bent) by appreciable amount. When this width (AB) is of the order of width of the slit (d), amount of bending would be quite small and so light rays can be considered to be propagating almost on rectilinear paths and hence "Ray Optics" can be applied.
Here `D_(2)ltD_(1)and A_(2)B_(2)ltA_(1)B_(1)`.
Now, suppose distance between slit and screer is Z at which half of linear width of centra maximum is `(1)/(2)((2lamdaZ)/(d))=(lamdaZ)/(d)` and suppose it is equal to linear width of slit (d).
Then,
`(lamdaZ)/(d)=d`
`:.Z=(d^(2))/(lamda)`
Above distance is called ..Fresnel distance.. and it is shown by symbol `Z_(F)`. Thus,
`Z_(F)=(d^(2))/(lamda)`
When `DltZ_(F)`, we can apply ..Ray Optics...
Note : Dear students, above explanation is just for your better understanding. You don.t have to write it in the examination to solve present example.
Solution : For the situation, given in the statement, required distance between slit and screen is :
`Z_(F)=(d^(2))/(lamda)`
`=((3xx10^(-3))^(2))/((500xx10^(-9)))`
`=(9xx10^(-6))/(5xx10^(-7))`
=18 m
In our daily life, distance between obstacle and wall (i.e. between slit and screen) is less than above distance. Hence, ray optics is quite valid in most of the situations in practice.
For what distance is ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 800 nm ?