Can someone check my answers (Closed and open tubes)

[viciousp]

Homework Statement

A tube, open only at one end, is cut into two shorter (non-equal)
lengths. The piece open at both ends has a fundamental frequency
of 425 Hz, while the piece open only at one end has a fundamental
frequency of 675 Hz. What is the fundamental frequency of the
original tube ?

also
Which of the following describes the relative temperatures in the refraction of sound. N is the normal line.

http://img511.imageshack.us/my.php?image=refractionfe5.jpg

Homework Equations

L1+L2=L
V/2L=frequency of open tube
V/4L=Frequency of closed tube

The Attempt at a Solution

For the first question I said the V/2(L1)=425 and V/4(L2)=675. so V=850L1 and V=2700L2. so I solved for L1 and got L1=3.176L2 and using the above equation L=4.176L2. So 2700L2/4(4.176L2)=161.63Hz, it seems resonable but I just want to see if I am right.

As for the other question I said that it moves from cold air to warm air since the angle gets larger and then from warm air to cold air since the angle gets smaller.

 

[krausr79]

Couldn't see the image, but I think this work is good.

 

[viciousp]

Here is the image again

http://img511.imageshack.us/img511/4403/refractionfe5.jpg

 

[viciousp]

Anyone?