*There are two adjacent tissues in an image with a T*_{2} of 30 and 50 ms.
You wish to produce a spin-echo image of the tissues such that the contrast between the tissues is maximized.
What TE should be used?
C = S_{1} - S_{2}

where:

C = contrast

S_{1} = signal from tissue 1

S_{1} = signal from tissue 2

We need to find the maximum C, or dC/dTE = 0.
dC/dTE = d(S_{1} - S_{2})= 0

d(e^{-TE/T2(2)})/dTE - d(e^{-TE/T2(1)})/dTE = 0

(e^{-TE/T2(2)})/T_{2(2)} = (e^{-TE/T2(1)})/T_{2(1)}

T_{2(2)}/T_{2(1)} = (e^{-TE/T2(2)})/(e^{-TE/T2(1)})

T_{2(2)}/T_{2(1)} = exp(-TE/T2(2)+ TE/T2(1))

ln(T_{2(2)}/T_{2(1)}) = TE/T2(1) + TE/T2(2)

ln(T_{2(2)}/T_{2(1)}) = TE (T_{2(2)} - T_{2(1)})/ (T_{2(2)} T_{2(1)})

TE = T_{2(2)} T_{2(1)} ln[T_{2(2)}/T_{2(1)}]/(T_{2(2)} - T_{2(1)})

TE = 38.3 ms