Refer the figure given below. μ_1 and μ_2 are refractive indices of...

Taking the pole $P$ as the origin and the direction of incident light as positive:
$\mu_1 = 1$ (air), $\mu_2 = 1.5$ (lens)
$u = -40\text{ cm}$
$R = -20\text{ cm}$ (centre of curvature lies to the left of $P$)
$h_o = +2\text{ cm}$

Using the formula for refraction at a single spherical surface:
$\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$

$\dfrac{1.5}{v} - \dfrac{1}{-40} = \dfrac{1.5 - 1}{-20}$

$\dfrac{1.5}{v} + \dfrac{1}{40} = -\dfrac{1}{40}$

$\dfrac{1.5}{v} = -\dfrac{2}{40} = -\dfrac{1}{20}$

$v = -30\text{ cm}$

The negative sign indicates a virtual image formed $30\text{ cm}$ to the left of $P$.

The transverse magnification for refraction at a spherical surface is:
$m = \dfrac{\mu_1 \cdot v}{\mu_2 \cdot u} = \dfrac{1 \cdot (-30)}{1.5 \cdot (-40)} = \dfrac{-30}{-60} = 0.5$

Therefore:
$h_i = m \cdot h_o = 0.5 \times 2 = 1\text{ cm}$

Hence, the correct option is $(1)\ 1$.