Bukhovtsev-et-al-Problems-in-Elementary-Physics

PHYSICAL OPTICS
429
~---:c -------3~ Fig. 558
Since the angle a is small, h ~ xa.
Therefore, the distance between the interference bands on the wedge is
~x=_A_ .
. 2an
According to the lormula lor the magnlficatton of a lens, ~:=1-, where
a is the distance from the screen to the lens and b from the lens to the wedge.
Since b=d-a, then according to the formula of a lens, -!..+-d I =.!-,. Upon
a -a
cancelling a and b from these expressions, we can find the sought value of
the angle a.
A d T Vd
a=---........;.----:--
2-4/d
2nl11 d ± Y d 2-4fd
The solution of .thls problem is not a single one, because a sharp image
can be obtained on the screen with fixed d and f when the lens is in one of
two positions.
6-2. Diffraction of Light
785. The radius of the first Fresnel zone can be found from triangles ADE
and DEB (Fig. 559):
r:=al-(a-X)I=(b+;r_(b+X)1
Since the wavelength is small, X=:2(ab~b)'
Therefore, r~ = 2ax- Xi.
Neglecting the small value of Xl,
we finalIy obtain
.. 1""""iiii'A
'1 == JI a+b