Charge over 2 layer dielectric, image method

If I have a charge $Q$ located over a 2 layer dielectric as represented:

According to the image method: the charge $Q’1$ will be located at a distance $h_1$ under the first interface and the $Q’2$ will be located at a distance $h_2$ under the first interface.

With:

$$
Q’1 = frac{epsilon_0-epsilon_1}{epsilon_0+epsilon_1} = frac{1-5}{1+5} Q
$$

My problem is determining the value of $Q’2$, is it:
$$
Q’2 = frac{epsilon_1-epsilon_2}{epsilon_1+epsilon_2} = frac{5-100}{5+100} Q
$$

or

$$
Q’2 = frac{epsilon_0-epsilon_2}{epsilon_0+epsilon_2}= frac{1-100}{1+100} Q
$$