Question

Light from a point source in air falls on a convex spherical glass surface ( refractive index = 1.5, radius of curvature = 20 cm). The distance of light source from the glass surface is 100 cm. At what position is the image formed ?

Answer 1

Here, the refraction in a spherical surface takes place. Light is moving from a rarer medium to a denser medium, the ray moves towards the normal. Keeping this in mind, we can use the required formula and deduce the answer.

Formulas used:
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$
Where ${{\mu }_{1,}}{{\mu }_{2}}$ are refractive indices of both media, v is the image distance, u is the object distance and R is the radius of curvature of the spherical medium.

Complete step by step answer:
We are given that the refractive index of the convex spherical glass surface is 1.5.Its radius of curvature is 20cm and the object distance is 100 cm.We have to find the distance of the image from the spherical surface.
We know that
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$
${{\mu }_{1}}$ is 1 since the refractive index of air is 1, and ${{\mu }_{2}}$ is 1.5. The other values to be substituted are given above.
Substituting the other values, we get
$\begin{aligned} & \dfrac{1.5}{v}+\dfrac{1}{100}=\dfrac{1.5-1}{20} \\\ & \Rightarrow \dfrac{1.5}{v}=\dfrac{0.5}{20}-\dfrac{1}{100} \\\ & \Rightarrow \dfrac{1.5}{v}=\dfrac{3}{200} \\\ & \therefore v=\dfrac{200}{3}\times 1.5=100cm \\\ \end{aligned}$ (By convention, u is taken as negative and v is positive)

Hence, the distance at which the image is formed is 100cm.

Note: The calculation formula that is used here must satisfy the condition that the lenses are very thin, that is the radius of the lens should be neglected. This formula is used in the derivation of the lens makers formula.