A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm

A Thin Equiconvex Lens Of Refractive Index 32 And Radius Physics QuestionA Thin Equiconvex Lens Of Refractive Index 32 And Radius Physics Question

A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm is put in water (refractive index = 4/3) , its focal length is

Options

(a) 0.15 m
(b) 0.30 m
(c) 0.45 m
(d) 1.20 m

Correct Answer:

1.20 m

Explanation:

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  • According to lens maker's formula

    1f=(μ−1)(1R1−1R2)

    where μ=μLμM

    Here, μL=32, μM=43, R1=+30cm, R2=−30cm

    ∴1f=(3243−1)(130−1−30)

    =(18)(230)

    1f=14×30=1120

    or f=120cm=1.2m

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