So I am given two clocks A and B moving in $S'$ frame with a velocity $V$ relative to $S$ frame. The two clocks are separated by a distance $L$ and are synchronized in $S'$ frame. The objective is to find the time difference between two clocks as measured in $S$ frame.
My approach:
Using Lorentz transformation,
$t_A = \gamma ({t_A}' + {x_A}'\frac{V}{c^2})$
$t_B = \gamma ({t_B}' + {x_B}'\frac{V}{c^2})$
where $t_A$ and $t_B$ are times in clock $A$ and $B$ as measured in $S$ frame, and ${t_A}'$ and ${t_B}'$ are times in clock $A$ and $B$ as measured in $S'$ frame. But since the two clocks are synchronous in $S'$ frame, ${t_A}'$ and ${t_B}'$ are same.
Hence, subtracting, we get, $t_B - t_A = \gamma ({x_B}'-{x_A}')\frac{V}{c^2} = \frac{\gamma LV}{c^2}$
Since the quantity on the right hand side is positive, this is giving me $t_B > t_A$. The hint on my question says the clock B lags behind. Could someone help me understand what is going on? Also, is this a correct setup of the problem?