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# A mark on the surface of a glass sphere $\mu =1.5$is viewed from a diametrically opposite position. It appears to be at a distance $10cm$ from its actual position. The radius of the sphere is:\begin{align}& \text{A}\text{. }5cm \\ & \text{B}\text{. }10cm \\ & \text{C}\text{. }15cm \\ & \text{D}\text{. }20cm \\ \end{align}

Last updated date: 17th Sep 2024
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Hint: When a light wave propagates from one medium to another having different values of refractive indices, it experiences some bending in its path, this is called refraction. We will apply the lens maker formula for refraction through the glass sphere and determine the radius of curvature of the sphere.

Refraction is described as the change in direction of a wave passing from one medium to another or from a gradual change in the medium. We can say that refraction is the bending of a wave when it passes through one medium to another. The bending of light occurs due to the difference in density between the two mediums.
We are given that when a mark on the surface of a glass sphere$\mu =1.5$is viewed from a diametrically opposite position, it appears to be at a distance $10cm$from its actual position.

Using the lens maker formula for refraction between two surfaces of different refractive indices,
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$
We have,
\begin{align} & {{\mu }_{1}}=1.5 \\ & {{\mu }_{2}}=1 \\ & v=10-2R \\ & u=-2R \\ \end{align}
Therefore,
\begin{align} & \dfrac{1}{10-2R}-\dfrac{1.5}{\left( -2R \right)}=\dfrac{1-1.5}{-R} \\ & \dfrac{1}{10-2R}+\dfrac{3}{4R}=\dfrac{1}{2R} \\ & \dfrac{1}{10-2R}=-\dfrac{1}{4R} \\ & -4R=10-2R \\ & 2R=-10 \\ & R=-5cm \\ \end{align}
The radius of curvature is $5cm$
Hence, the correct option is A.