[snip]
If that was true then why does this work:

img1off = new Image();   img1off.src = "<?php echo $base_url;
?>Graphics/home_off.gif";

I certainly didn't generate it with PHP and it was included as an
external file.
[/snip]

If this is true then your problem is solved. if not....

echo "img1off = new Image();   img1off.src = . "$base_url .
"Graphics/home_off.gif";

I think I have the syntax correct

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