MongoDB 4 でドキュメントを並べ替えて 1 つのフィールドだけを表示するにはどうすればよいですか?

> db.demo611.insertOne({"Name":"Chris"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987110f6b89257f5584d83")
}
> db.demo611.insertOne({"Name":"Adam"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987115f6b89257f5584d84")
}
> db.demo611.insertOne({"Name":"John"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987118f6b89257f5584d85")
}
> db.demo611.insertOne({"Name":"Bob"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e98711bf6b89257f5584d86")
}

> db.demo611.find();
This will produce the following output:
{ "_id" : ObjectId("5e987110f6b89257f5584d83"), "Name" : "Chris" }
{ "_id" : ObjectId("5e987115f6b89257f5584d84"), "Name" : "Adam" }
{ "_id" : ObjectId("5e987118f6b89257f5584d85"), "Name" : "John" }
{ "_id" : ObjectId("5e98711bf6b89257f5584d86"), "Name" : "Bob" }

> db.demo611.find().sort({Name:1});

{ "_id" : ObjectId("5e987115f6b89257f5584d84"), "Name" : "Adam" }
{ "_id" : ObjectId("5e98711bf6b89257f5584d86"), "Name" : "Bob" }
{ "_id" : ObjectId("5e987110f6b89257f5584d83"), "Name" : "Chris" }
{ "_id" : ObjectId("5e987118f6b89257f5584d85"), "Name" : "John" }