Refer the figure given below. \(\mu_1\) and \(\mu_2\) are refractive indices of air and lens material. The height of image will be _______ cm.

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 \((1)\ 1\).