A convex lens is in contact with a concave lens. The magnitude of the…

The correct answer is Option (4) → \( f_{\mathrm{convex}}=10\,\mathrm{cm},\ f_{\mathrm{concave}}=-15\,\mathrm{cm} \)

For thin lenses in contact, \( 1/F=1/f_1+1/f_2 \). With the Cartesian sign convention, a convex lens has a positive focal length and a concave lens has a negative one. The combination has \( F=+30\,\mathrm{cm} \) (converging), so the convex lens is the stronger one, with the shorter focal length.

Let the magnitude of the concave lens’s focal length be \( x \). Then:

    \( \displaystyle f_{\mathrm{convex}}=\frac{2x}{3} \)

    \( f_{\mathrm{concave}}=-x \)

Putting these in the formula for lenses in contact:

    \( \begin{aligned} \frac{1}{30\,\mathrm{cm}} &= \frac{3}{2x}-\frac{1}{x} \\ &= \frac{1}{2x} \end{aligned} \)

    \( \Rightarrow\; x=15\,\mathrm{cm} \)

So the focal lengths are:

    \( \begin{aligned} f_{\mathrm{convex}} &= \frac{2}{3}\times15\,\mathrm{cm} \\ &= 10\,\mathrm{cm} \end{aligned} \)

    \( f_{\mathrm{concave}}=-15\,\mathrm{cm} \)