Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 293

of this element is ensured by the difference in the

corresponding pressures:

fp (r Ar) — p (0] AS = p Ar AS co^2 r.

Therefore, for the variation of pressure, we obtai r

the following equation:

d

dp

=

Since the relation pp = pR T is satisfied for au

ideal gas (R is the universal gas constant), we

obtain

dp R0)2

r.

dr P^ RT^

By hypothesis, for r < rbeam, we have p (r) —
Po < p 0 , and hence

2
P Po (14

Igo

T

r2

2R

Accordingly, for the gas density at r < rbeam,
we obtain

(r) Po (1 (^2) 11R6)T 2
r2 ) ,^ PO = PO RT
and for the refractive index, we get
n no + kr (^2) , no 1+ apo, k= ceP0 2 \^2
RT I •
Let us now find the angle of refraction of a ray
passing through the vessel at a distance r from the
axis. The optical path length in the vessel is
n (r) 1.
The optical path difference 6opt between two
close rays emerging from the vessel must be equal