Problem 7 - Depth of the well
On a relatively hot day, 30 C; a student measures two successive resonances above a well at 52Hz and 60Hz. Determine (a) the order of the
resonances and (b) the depth of the well.
(a) The wave lengths and frequencies for resonances above a partially
…lled column of water satisfy
2m+1
=4
L
and f2m+1 =
2m + 1
vs
= (2m + 1)
2m+1
vs
4L
This means that the di¤erence between these two frequencies satis…es
f2m+1
f2m
1
= (2m + 1
(2m
1))
vs
vs
=
= 8Hz:
4L
2L
The corresponding m0 s are
2m + 1
vs
=
8 = 60
4L
2
1 = 6:
f2m+1 = (2m + 1)
m = 7; m
(b) The speed of sound at 30 C is
r
r
T
303
vs = 343
= 343
= 349m:
293
293
Hence the depth of the well is given by
L=
348:8
vs
=
= 21:8m:
16
16
Problem 8 - Triangular Prism
An isosceles triangular glass prism with apex angle
has an index of
refraction n. Show that the condition on the incident
angle 1 so that a light
p
ray can emerge from the other side is sin 1 > n2 1 sin
cos :
1
De…ne 1 as the incident angle, 2 as the refracted angle, and 3 as the
incident angle on the opposite face. To avoid total internal re‡ection 3 must
satisfy
sin 3 < 1=n ! 3 < sin 1 (1=n) :
From the triangle above the light ray we have
2
This means that
2
2
+
+
2
3
!
=
2
=
3:
must satisfy
2
>
sin
1
(1=n) :
From Snell’s law we …nd
sin
sin
1
sin
1
sin
1
sin
1
1
= n sin 2 ! sin 1 > n sin
sin 1 (1=n)
> n sin cos sin 1 (1=n)
(1=n) cos
q
> n sin
1 sin2 sin 1 (1=n)
(1=n) cos
q
> n sin
1 (1=n)2 (1=n) cos
p
>
n2 1 sin
cos
This is the desired condition.
2