Multiple catch is necessory when the try when the try block 
hase statements that raise different types of exceptions.

The single catch block will not be able to handle the 
multiple error condition without complex coding. For 
example close the database connection and closing of a file
For a particular exception closing of database connection 
can be written in one catch block, for another kind of 
exception we can close the file in another catch block. 
That means if a file is opened and a database connection is 
established, we want to release only one resourse at a time 
for a particular exception.

When an exception is thrown it traverses through the catch 
block until the matching catch block is found.

I hope my understanding is correct......

Handle e\any es\exception at the layer that can handle  it 
in an efficient manner, else propagate it upwards!

I under stood ur problem, I will expalin with an example

public class Test
{
 public static void main(String args[])
{
 try
 {
 int n =2/args.length;
 int x[]={1,2};
 x[30]=4;
 }
 catch(ArithmeticException e)
 {
  System.out.println("Divede by Zero");
 }
 catch(IndexOutOfBoundsException e)
 {
  System.out.println("ArrayIndex Out of 

Bounds");
 }
}
}

if u run with out any commandline argument then u got 
divide by zero exception. otherwise u get indexbounds 
exception

You will need to have this for a basic reason that the
exceptions need to be caught and needed to be displayed if
your try block throws an exception say Array out of bound
exception which is a checked exception and you have only one
catch block which has the object of Exception class then
that Exception will be caught but you will not be able to
handle the particular array out of bound exception. so all
you need is to put all the exception in diff catch statement
which your try block may throw followed by the superclasses
of the exceptiontype subclasses .thus this is how you will
be able to handle each exception by its own way.