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