The Big Bang Polyglot
Foo, Python, JavaScript, Retina, >&lt;>: 166 148 bytes

1//1;print('Sheldon Lee Cooper');"""
alert`Howard Joel Wolowitz`

// Rajesh Ramayan Koothrapali
//!??

//\"Leonard Leakey"&gt;o&lt;'Penny'" Hofstadter"""



The trailing newline is significant.

Foo outputs Leonard Leakey Hofstadter

Python outputs Sheldon Lee Cooper

JavaScript outputs Howard Joel Wolowitz

Retina outputs Rajesh Ramayan Koothrapali

>&lt;> outputs Penny Hofstadter

Explanations

Foo

In Foo, any text written inside double quotes is sent to STDOUT. So it outputs Leonard Leakey and Hofstadter. It also prints it with a leading newline. Honestly I'm really not sure about how it parses strings, but it works as-is. It also tries to do some other operations (/ is divide, parentheses form loops, etc.), but all of that just causes errors and prints to STDERR without affecting the main output in any way.

Python

1//1 is 1 integer divided by 1. In a statement by itself, that does nothing relevant. Then it prints Sheldon Lee Cooper, and the rest is in a triple quoted string, which is basically a block comment.

JavaScript

JavaScript sees most lines as either empty or commented (//). The first statement is 1 by itself which, like in the Python part, does nothing. Then it alerts Howard Joel Wolowitz.

Retina

1//1;print('Sheldon Lee Cooper');"""
alert`Howard Joel Wolowitz`


First all matches of 1//1;print('Sheldon Lee Cooper');""" in the input are replaced with alert`Howard Joel Wolowitz`. Seeing as the input is empty, nothing happens.


// Rajesh Ramayan Koothrapali

Then the empty input is replaced with // Rajesh Ramayan Koothrapali.

//!??



Then all matches of // optionally followed by a ! are replaced with nothing. This deletes the initial //, leaving one leading space (which is allowed).

//\"Leonard Leakey"&gt;o&lt;'Penny'" Hofstadter"""



Then all matches of the regex above are replaced with nothing. This regex doesn't match anything in the working string, so nothing happens.

The working string is then output implicitly.

>&lt;>

The IP starts in the top-left, initially moving right. It first pushes 1, then bounces around on mirrors (/ and \) for a while (looping around when it goes out of bounds). The 1 will be popped when it hits ?.

Eventually it will be on the right side of the second last line moving left. On this line, it sees an empty string "", pushing nothing. Then it encounters two strings, " Hofstadter" and 'Penny', pushing the characters from each as it sees them. When it hits &gt;o&lt;, it repeatedly pops and prints the top of the stack as a character, until there's nothing left to pop. It exits with an error.