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PhysicsClass 12

Light from a point source in air falls on a spherical glass surface\((n=1{\cdot}5\) and radius of curvature \(=20\;\text{cm})\). The distance of the light source from the glass surface is \(100\;\text{cm}\). At what position the image is formed?

Answer

Given data,
Refractive index of air \(n_1=1\)
Refractive index of glass \(n_2=1.5\)
Radius of curvature \(R=20\;\text{cm}\)
Distance between light source and glass surface \(u=-100\;\text{cm}\)
As object is in rarer medium, so refraction takes place from rarer to denser medium.
We know, for refraction from rarer to denser medium by spherical surface is given by,
\(\frac{n_2} v-\frac{n_1} u=\frac{n_2-n_1} R\)\({\cdots}(i)\) \(\{\)where \(v\)  is distance of image from glass surface\(\}\)
Putting all values in equation \((i)\)
\(\frac{1{.}5} v-\frac 1{-100}=\frac{1{.}5-1}{20}\)  
\(\Rightarrow v=100\;\text{cm}\)
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