(d) Length of tube = 10 cm
$f_{0} + f_{e} = 10\ cm$
Magnification
$m = \frac{f_{0}}{f_{e}} = 4$
$f_{0} = 4f_{e}$
Putting in Eq. (I)
$5f_{e} = 10\ cm$
$or\ f_{e} = 2\ cm$
$and\ \ \ f_{0} = 8\ cm$
$f_{0} = 8cm,\ f_{e} = 2cm$
Hence, $L_{4}\ $and $L_{1}$ will be used.