For combination of prisms
net deviation = δ1 + δ2 = (ny -1)A +(n’y-1)A
where δ1 and δ2 are deviations produced by individual prisms.
Net dispersion = (nv - nR) A + (n’v –n’R) A
According to problem net deviation is zero
hence (ny -1)A = -(n’y-1)A’
A’ = - [(ny -1)A /(n’y-1)]
Negative sign implies that second prism is inverted relative to first
ny = (nv + nR)/2, n’y = (n’v –n’R)/2
A’=-[(ny -1)A /(n’y-1)]
ny = (1.51 + 1.49)/2
n’y = (1.77 + 1.73)/2 , A = 6**** on substituting numerical values of ny, n’y, A
we get A’ = -4****
Net dispersion = (nv - nR)A – (n’v - n’R)A’ = 0.04 ans.