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  4. A convex refracting surface of radius of curvature 20 cm separates two media of refractive indices 4/3 and 1.60. An object is placed in the first medium (μ = 4/3) at a distance of 200 cm from the refracting surface. The position of the image formed is

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Q. A convex refracting surface of radius of curvature $20 \,cm$ separates two media of refractive indices $4/3$ and $1.60$. An object is placed in the first medium $(\mu = 4/3)$ at a distance of $200 \,cm$ from the refracting surface. The position of the image formed is

Ray Optics and Optical Instruments

Solution:

Using, $-\frac{\mu_{1}}{u}+\frac{\mu_{2}}{v} = \frac{\mu_{2}-\mu_{1}}{R} $
Here, $R = 20\, cm$,
$ \mu_{1}=\frac{4}{3}, \mu_{2} = 1.60, u = -200\, cm $
$ \therefore - \frac{4/3}{-200} +\frac{1.60}{v} $
$= \frac{1.60-(4/3)}{20}$
$\Rightarrow v = 240\,m$