Two transparent media having refractive indices 1 . 0 and 1 . 5 are separated by a spherical refracting surface of radius of curvature 30 cm . The centre of curvature of surface is towards denser medi

Question

Two transparent media having refractive indices 1 . 0 and 1 . 5 are separated by a spherical refracting surface of radius of curvature 30 cm . 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 15 cm from the pole of the surface. The distance of image from the pole of the surface is cm .

TestHub · Accepted Answer

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
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} . . . (1)$
Substitute the values of the known parameters into equation (1) and solve to calculate the required image distance.
$\begin{array}{rcl} \frac{1.5}{v} - \frac{1}{- 15} & = & \frac{1.5 - 1.0}{30} \\ \Rightarrow & \frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} \\ = & - \frac{1}{20} \\ \Rightarrow & v = - 30 \end{array}$
Hence, the image is formed at a distance of $30 cm$ from the pole on the same direction as the object is placed.

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