Object is placed at 40 cm from a spherical surface whose radius of curvature is 20cm. Refractive index of medium is µ =1.54. Find height of image formed.

Based on the provided diagram and question text, we can extract the following parameters by applying the standard Cartesian sign convention (taking the direction of incident light from left to right as positive):

1. Refractive index of the first medium (air):
${\mu }_1=1$
2. Refractive index of the second medium (spherical surface):
${\mu }_2=1.54$
3. Object distance from the pole of the surface:
$u=-40\text{cm}$
4. Radius of curvature of the spherical surface (since the center of curvature $C$ lies in the first medium to the left of the pole):
$R=-20\text{cm}$
5. Height of the object:
$h_o=+2\text{cm}$

To find the position of the image ($v$), we use the formula for refraction at a spherical surface:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Substitute the known values into the equation:

$\frac{1.54}{v}-\frac{1}{-40}=\frac{1.54-1}{-20}$

Simplifying the terms:

$\frac{1.54}{v}+\frac{1}{40}=\frac{0.54}{-20}$

$\frac{1.54}{v}+0.025=-0.027$

Subtract $0.025$ from both sides:

$\frac{1.54}{v}=-0.027-0.025$

$\frac{1.54}{v}=-0.052$

Solving for $v$:

$v=-\frac{1.54}{0.052}\approx -29.62\text{cm}$

Next, the lateral magnification ($m$) for a spherical refracting surface is given by the formula:

$m=\frac{h_i}{h_o}=\frac{{\mu }_{1}v}{{\mu }_{2}u}$

Therefore, the height of the image ($h_i$) is:

$h_i=h_o\times \left(\frac{{\mu }_{1}v}{{\mu }_{2}u}\right)$

Substituting the calculated value of $v$ and the other parameters:

$h_i=2\times \left(\frac{1\times \left(-29.62\right)}{1.54\times \left(-40\right)}\right)$

$h_i=2\times \left(\frac{29.62}{61.6}\right)$

$h_i\approx 2\times 0.4808=0.9616\text{cm}$

Rounding to two decimal places, we get:
$h_i\approx 0.96\text{cm}$

Thus, the height of the image formed is 0.96cm.