CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

Question : 10 of 13

Marks: +1, -0

The power of a thin lens is

+5 {\D}

. When it is immersed in a liquid, it behaves like a concave lens of focal length

100 {\cm}

. Calculate the refractive index of the liquid. Given refractive index of glass = 1.5

Solution:

Power of the lens

=+5 D

Focal length of the lens

=f=\;{1}/{D}

Or,

\;f=\;{1}/{5} {\cm}

∴ \;f=20 {\cm}

Using lens makers' formula,

\;{1}/{f_{\;\text "air "\;}}\;= (\;{µ_{\;\text "glass "\;}-µ_{\;\text "air "\;}}/{µ_{\;\text "air "\;}}) (\;{1}/{R_1}-\;{1}/{R_2})

Or,

\;{1}/{20}\;= (\;{1.5-1}/{1}) (\;{1}/{R_1}-\;{1}/{R_2})

Or,

\;{1}/{20}\;=0.5 (\;{1}/{R_1}-\;{1}/{R_2})

∴ \;\; (\;{1}/{R_1}-\;{1}/{R_2}) =\;{1}/{10}

When immersed in liquid,

\;\;{1}/{f_{\;\text "liquid "\;}}= (\;{µ_{\;\text "glass "\;}-µ_{\;\text "liquid "\;}}/{µ_{\;\text "liquid "\;}})

(\;{1}/{R_1}-\;{1}/{R_2})

\;\text " Or, "\;\;-\;{1}/{100}= (\;{1.5-µ_{\;\text "liquid "\;}}/{µ_{\;\text "liquid "\;}})

(\;{1}/{R_1}-\;{1}/{R_2})

\;\text " Or, "\; \;\; -\;{1}/{100}= (\;{1.5-µ_{\;\text "liquid "\;}}/{µ_{\;\text "liquid "\;}})

(\;{1}/{10})

[Putting

({1}/{R_1}-{1}/{R_2})={1}/{10}

]

\;\text " Or, "\;\;\;{1}/{10}= (\;{1.5-µ_{\;\text "liquid "\;}}/{µ_{\;\text "liquid "\;}})

\;\text " Or, "\;\;9 µ_{\;\text "liquid "\;}=15

∴ \;µ_{\;\text "liquid "\;}=\;{15}/{9}=\;{5}/{3}

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