Как принять данные на PHP с Ajax?

formData.append('avatar', blob, 'avatar.jpg');
            $.ajax('upload_photo.php', {
              method: 'POST',
              data: formData,
              processData: false,
              contentType: false,

<?php

$file_to_upload = $_FILES['croppedImage']['tmp_name'];
$file_name = 'avatar.jpg';
move_uploaded_file($file_to_upload, $file_name);

?>