CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

Question : 10 of 13

Marks: +1, -0

A point object in air is placed symmetrically at a distance of

60 {\cm}

in front of a concave spherical surface of refractive index 1.5 . If the radius of curvature of the surface is

20 {\cm}

, find the position of the image formed.

Solution:

For refraction from concave spherical surface,

\;{µ_2}/{v}-\;{µ_1}/{u}\;=\;{µ_2-µ_1}/{ {\R}}

\;\text " or, "\;\;\;{1.5}/{v}-\;{1}/{(-60)}\;=\;{1.5-1}/{-20}

\;\text " or, "\;\;\;{1.5}/{v}+\;{1}/{60}\;=-\;{0.5}/{20}

\;\text " or, "\;\;\;{1.5}/{v}+\;{1}/{60}\;=-\;{1}/{40}

\;\;\text " or, "\;\;\;{1.5}/{v}\;=-\;{1}/{40}-\;{1}/{60}

\; ∴ \;v\;=-36 {\cm}

Image in formed in the medium where the object lies at a distance

36 {\cm}

from point

P

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