Refraction through a lens

Without the lens the converging rays would meet at a point 15 − 5 = 10 cm behind the lens position. This point acts as a virtual object for the lens.
For a virtual object on the far side of the lens, u = +10 cm.
With the lens the rays actually meet 15 cm from the lens on the opposite side, so v = +15 cm.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}$
f = −30 cm (negative, as expected for a diverging lens)

The refractive index is not given; taking glass with $\mu = 1.5$.
Lens maker's formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.5\times\frac{1}{10} = \frac{1}{20}$
f = 20 cm
Lens formula with u = −30 cm: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$
v = +60 cm; v is positive, so the image is real and on the other side of the lens.
Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$; the negative sign means the image is inverted.
Image height = $|m|\times h_o = 2\times2 = 4$ cm
So the image is real, inverted and 4 cm high.

The focal length depends only on the refractive index and the radii of curvature of the lens, not on its aperture; so it remains f.
The intensity of the image is proportional to the area of the lens that transmits light, i.e., $I \propto d^2$.
Original transmitting area $\propto d^2$; area covered $\propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4}$
Remaining area $\propto d^2 - \frac{d^2}{4} = \frac{3d^2}{4}$
New intensity $= I - \frac{I}{4} = \frac{3I}{4}$
So the focal length is f and the intensity is $\frac{3I}{4}$.