Question

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30 \,cm$ and refraction index of the material for both the lenses is $1.75$. Both the lenses are placed at distance of $40 \,cm$ from each other. Due to the combination, the image of the object is formed at distance $x=$__ $cm$, from concave lens.

Answer 1

$ \frac{1}{ f _1}=(1.75-1)\left(-\frac{1}{30}\right)$
$\Rightarrow f _1=-40 \,cm $
$ \frac{1}{ f _2}=(1.75-1)\left(\frac{1}{30}\right) \Rightarrow f _2=40\, cm$
Image from $L_1$ will be virtual and on the left of $L_1$ at focal length $40 \, cm$. So the object for $L _2$ will be $80 cm$ from $L _2$ which is $2 f$. Final image is formed at $80\, cm$ from $L _2$ on the right.
So $x=120$