Thank you for reporting, we will resolve it shortly
Q.
A beam of parallel rays is brought to focus by a planoconvex lens. A thin concave lens of the same focal length is joined to the first lens. The effect of this is
The combined focal length of plano-convex lens
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$
Hence, $f_1=\infty$ (for plane surface), $f_2=f$ (say)
$ \therefore \frac{1}{F}=\frac{1}{\infty}+\frac{1}{f} $
$ \Rightarrow F=f$
Now, when concave lens of same focal length is joined to first lens, then combined focal length
$\frac{1}{F^{\prime}} =\frac{1}{F_1}+\frac{1}{F_2}$
$=\frac{1}{f}-\frac{1}{f} \left(\because F_1=f, F_2=-f\right) $
$ =0$
$F^{\prime} =\infty$
Thus, the image can be focussed on infinity $(\infty)$ or focus shifts to infinity.