Sure. Suppose we have a light source and two detectors that are each a distance ##L## from the light source, one in the ##+x## direction and the other in the ##-x## direction. The light source flashes at ##t=0##, so the equations of motion for the light pulses and the detectors can be written as follows: $$x_{+light}(t)=ct$$$$x_{-light}(t)=-ct$$$$x_{+detector}(t)=L$$$$x_{-detector}(t)=-L$$ We can easily see that $$|\dot x_{+light}|=|\dot x_{-light}|=c$$ and we can also easily see that the separation between the detectors is $$x_{+detector}(t)-x_{-detector}(t)=2L$$ If ##t_+## is the arrival of the light at the ##+## detector and ##t_-## is the arrival of the light at the ##-## detector then $$x_{+detector}(t_+)=x_{+light}(t_+)$$$$t_+=\frac{L}{c}$$$$x_{-detector}(t_-)=x_{-light}(t_-)$$$$t_-=\frac{L}{c}$$ The light arrives simultaneously at both detectors.
Now, consider a primed reference frame moving at ##v## with respect to the unprimed frame. The equations of motion in the primed frame are obtained from the unprimed frame using the
Lorentz transform. With that we can find $$x'_{+light}(t')=ct'$$$$x'_{-light}(t')=-ct'$$$$x'_{+detector}(t')=\frac{L}{\gamma}-vt'$$$$x'_{-detector}(t')=-\frac{L}{\gamma}-vt'$$ Again, we can easily see that$$|\dot x'_{+light}|=|\dot x'_{-light}|=c$$, so the speed of light is the same in both frames. We can also easily see that the separation between the detectors is $$x'_{+detector}(t')-x'_{-detector}(t')=\frac{2L}{\gamma}$$ meaning that lengths are contracted. Again, if ##t'_+## is the arrival of the light at the ##+## detector and ##t'_-## is the arrival of the light at the ##-## detector then $$x'_{+detector}(t'_+)=x'_{+light}(t'_+)$$$$t'_+=\frac{L}{(c+v)\gamma}$$$$x'_{-detector}(t'_-)=x'_{-light}(t'_-)$$$$t'_-=\frac{L}{(c-v)\gamma}$$ The light arrives at each detector at a different time.