A converging beam of light travelling in air converges at a point $P$ as shown in the figure. When a glass sphere of refractive index 1·5 is introduced in between the path of the beam, calculate the new position of the image. Also, draw the ray diagram for the image formed.

For the first refraction, we have $u=20cm$ and $R=5cm$,     $n_{2}=1.5,$ $n_{1}=1$

$\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}$

$\frac{1.5}{v}-\frac{1}{20}=\frac{(1.5-1)}{5}$

$\frac{3}{2v}=\frac{1}{10}+\frac{1}{20}=\frac{3}{20}$

$v=10cm$

For the 2nd refraction

$u=0$

then,

$\frac{1}{v}-\frac{1.5}{0}=\frac{1-1.5}{-5}$

$v=0$

So, the final image will be a pole of the surface 2.

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