There is no variable interpolation in my $regexp = 'some$thing';. The content of $regex is now a literal string some$thing, since you used single quotes.

The literal string some$thing used as a regex is not equivalent to /some\$thing/, because $ is a interpreted as a regex metacharater denoting EOL. It is equivalent to /some$thing/ though, I hope the difference is obvious now.

If you wanted to interpolate $thing into $regexp, you would need to use double quotes not single quotes. 
my $thing = 'THING';
my $regexp = "some$thing";  # now $regexp is 'someTHING'

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Maybe the following code will illustrate it all a little better. 
#!/usr/bin/perl -W
use strict;

my $thing1 = 'THING';
my $thing2 = '$thing';

sub test {
    my $regex = shift;
    local $_ = 'some thing someTHING some$thing';
    printf " %15s =~ %s", $regex, $_; 
    print " == TRUE , matched [", $&, "]" if /$regex/;
    print "\n";
}


test ( 'some$thing1' );
test ( "some$thing1" );
test ( 'some$thing2' );
test ( "some$thing2" );

test ( quotemeta 'some$thing1' );
test ( quotemeta "some$thing1" );
test ( quotemeta 'some$thing2' );
test ( quotemeta "some$thing2" );

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In reply to Re: variable interpolation in regexps by Firefly258
in thread variable interpolation in regexps by Locutus

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