Locate the image formed by the refraction in the situation shown in figure

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Solution

$μ_{1} = 1, μ_{2} = 1.5, R = 20 c m$ Radius of curvature), $u = - 25 c m$
$∴ \frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \Rightarrow \frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{25} = \frac{1}{40} - \frac{1}{25} = \frac{- 3}{200}$
$\Rightarrow v = - 200 \times 0.5 = - 100 c m$
So, the image is $100 c m$ from (P) the surface on the side of S.

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Locate the image formed by the refraction in the situation shown in figure

μ1=1,μ2=1.5,R=20cm Radius of curvature), u=−25cm∴μ2v−μ1u=μ2−μ1R⇒1.5v=0.520−125=140−125=−3200⇒v=−200×0.5=−100cmSo, the image is 100cm from (P) the surface on the side of S.