CBSE Class 12 Physics 2019 Delhi Set 1 Paper

Question : 7 of 27

Marks: +1, -0

Calculate the radius of curvature of a equiconcave lens of refractive index 1.5 , when it is kept in a medium of refractive index 1.4, to have a power of

-5 {\D}

?
OR
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of the minimum deviation of the prism, when kept in a medium of refractive index

\;{4 {√{2}}}/{5}

Solution:

Calculation of focal length
Lens maker's formula
Calculation of radius of curvature

\;f=\;{1}/{P}=\;{1}/{-5} {\m}=-\;{100}/{5} {\cm}=-20 {\cm}

\;\;{1}/{f}= (\;{µ_2}/{µ_1}-1) (\;{1}/{R_1}-\;{1}/{R_2})

\;µ_2=1.5, µ_1=1.4, R_1=-R, R_2=R

\;\;{1}/{-20}= (\;{1.5}/{1.4}-1) (-\;{1}/{R}-\;{1}/{R})

\;\;{1}/{-20}= (\;{0.1}/{1.4}) (-\;{2}/{R})

\;R=\;{20}/{7} {\cm}(=2.86 {\cm})

OR
Formula
Substitution and calculation

\;µ=\;{ sin \;{ (A+δ_m) }/{2}}/{ sin \;\;{A}/{2}}

\;µ=\;{µ_1}/{µ_2}=\;{1.6}/{{4}/{5} {√{2}}}=\;{8}/{4 {√{2}}}={√{2}}

µ\;=\;{µ_1}/{µ_2}=\;{1.6}/{{4}/{5} {√{2}}}=\;{8}/{4 {√{2}}}={√{2}}

{√{2}}\;=\;{ sin \; ({60^{∘}+δ_m}/{2}) }/{ sin \;\;{60^{∘}}/{2}}

=\;{ sin \; ({60^{∘}+δ_m}/{2}) }/{ sin \;30^{∘}}

\; ∴ \; sin \; (\;{60+δ_m}/{2}) \;={√{2}} ⋅ \;{1}/{2}

=\;{1}/{{√{2}}}= sin \;45^{∘}

∴ \;\;{60+δ_m}/{2}\;=45^{∘}

\; ∴ \;δ_m\;=30^{∘}

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