Question 170 of 295

Two transparent media having refractive indices $1.0$ and $1.5$ are separated by a spherical refracting surface of radius of curvature $30cm$. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of $15cm$ from the pole of the surface. The distance of image from the pole of the surface is $cm$.

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Answer: 30

Let's consider the following diagram-
The relation between the object distance, the image distance and the radius of curvature of the spherical surface can be written as
$\dfrac{{\mu }_{2}}{v}-\dfrac{{\mu }_{1}}{u}=\dfrac{{\mu }_{2}-{\mu }_{1}}{R}...\left(1\right)$
Substitute the values of the known parameters into equation (1) and solve to calculate the required image distance.
$\begin{matrix}\dfrac{1.5}{v}-\dfrac{1}{-15} & = & \dfrac{1.5-1.0}{30} \\ & \Rightarrow & \dfrac{1.5}{v}=\dfrac{1}{60}-\dfrac{1}{15} \\ & = & -\dfrac{1}{20} \\ & \Rightarrow & v=-30\end{matrix}$
Hence, the image is formed at a distance of $30cm$ from the pole on the same direction as the object is placed.