In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and transparent material (µ=4/3). A point object O is placed in a iron the principal axis of the surface 20 cm from P. The distance of the image of O from P will be ______.

Fill in the blank with the correct answer from the options given below.

The correct option is: 16 cm left of P in air

1. Identify the given parameters from the problem statement and the provided figure:
From the given figure, we can observe that the curved boundary APB separates air (on the left) and a transparent material with refractive index $\mu =\frac{4}{3}$ (on the right).
The curved surface curves such that its center of curvature lies to the left of the vertex P (in air). Therefore, using the standard Cartesian sign convention (where the direction of incident light from left to right is taken as positive):
- Refractive index of the first medium (air), ${\mu }_1=1$
- Refractive index of the second medium (transparent material), ${\mu }_2=\frac{4}{3}$
- Object distance, $u=-20\text{cm}$ (since the point object O is placed 20 cm to the left of P)
- Radius of curvature, $R=-10\text{cm}$ (since the center of curvature is to the left of P)

2. Apply the formula for refraction at a spherical surface:
The relation between the object distance ($u$), image distance ($v$), radius of curvature ($R$), and the refractive indices of the media is given by:

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

3. Substitute the values into the formula:

$$\frac{4/3}{v}-\frac{1}{-20}=\frac{\frac{4}{3}���1}{-10}$$

Simplify the equation step-by-step:

$$\frac{4}{3v}+\frac{1}{20}=\frac{1/3}{-10}$$

$$\frac{4}{3v}+\frac{1}{20}=-\frac{1}{30}$$

Rearrange to solve for $v$:

$$\frac{4}{3v}=-\frac{1}{30}-\frac{1}{20}$$

Find a common denominator (60) for the terms on the right side:

$$\frac{4}{3v}=\frac{-2-3}{60}$$

$$\frac{4}{3v}=-\frac{5}{60}$$

$$\frac{4}{3v}=-\frac{1}{12}$$

Now, solve for $v$:

$$3v=4\times (-12)$$

$$3v=-48$$

$$v=-16\text{cm}$$

4. Conclusion:
The negative sign indicates that the virtual image is formed to the left of the refracting surface vertex P (in the same medium as the object, which is air).
Thus, the image is formed at a distance of 16 cm left of P in air.