Answer
$46.2cm$ from the lens
Work Step by Step
We know that
$\frac{\frac{1}{f_r}}{\frac{1}{f_v}}=\frac{n_r-1}{n_v-1}$
$\frac{\frac{1}{d_{\circ}}+\frac{1}{d_r}}{\frac{1}{d_{\circ}}+\frac{1}{d_v}}=\frac{n_r-1}{n_v-1}$
This simplifies to:
$d_v=[(\frac{1}{d_{\circ}}+\frac{1}{d_r})(\frac{n_v-1}{n_r-1})-\frac{1}{d_{\circ}}]^{-1}$
We plug in the known values to obtain:
$d_v=[(\frac{1}{0.240m}+\frac{1}{0.550m})(\frac{1.605-1}{1.572-1})-\frac{1}{0.240m}]^{-1}$
$d_v=46.2cm$ from the lens
