Explanation

(a) The combined focal length of plano-convex lens
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
Given, $f_{1} = \infty(for\ plane\ surface),\ f_{2} = f(say)$
$\therefore\frac{1}{F} = \frac{1}{\infty} + \frac{1}{f}$
$\Longrightarrow F = f$
Now when concave lens of same focal length is joined to first lens , then combined focal length
$\frac{1}{F'} = \frac{1}{F_{1}} + \frac{1}{F_{2}}$
$= \frac{1}{f} - \frac{1}{f}\ \ \ \ \ \ \ \ (\because F_{1} = f,\ F_{2} = - f)$
$= 0$
$F' = \ \infty$
Thus, the image can be focused on infinity ($\infty$) or focus shifts to infinity.