The radii of curvature of two surfaces of a convex lens are 0.2m and 0.22m. Find the focal length of the lens If refractive index of the material of lens is 1.5. also find the change in focal length, if it is immersed in water of refractive index 1.33.
Answers
focal length of lens is 21cm (approx)
and when lens is immersed in water focal length of lens is increased by 61cm
radii of curvature of two surfaces of a convex lens are 0.2 m and 0.22 m.
i.e., R1 = 0.2 m, R2 = -0.22m
using formula, 1/f = (μ - 1)[1/R1 - 1/R2]
where μ is refractive index of lens with respect to medium.
in air medium, μ = 1.5/1 = 1.5
so, 1/f = (1.5 - 1) [1/0.2 - 1/-0.22 ]
= 0.5[5 + 4.54 ]
= 0.5 × 9.54
= 4.77
or, f = 1/4.77 = 0.2096 m ≈ 21cm
when lens is immersed in water.
μ' = μ_l/μ_w = 1.5/1.33 = 1.1278
now, 1/f' = (μ' - 1)[1/R1 - 1/R2]
= (1.1278 - 1) [1/0.2 -1/-0.22]
= 0.1278 [5 + 4.54]
= 1.219
or, f' = 1/1.219 = 0.82 m = 82 cm
hence, when lens is immersed in water focal length of lens is increased by (82 - 21) = 61cm .
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radii of curvature of two surfaces of a convex lens are 0.2 m and 0.22 m.
i.e., R1 = 0.2 m, R2 = -0.22m
using formula, 1/f = (μ - 1)[1/R1 - 1/R2]
where μ is refractive index of lens with respect to medium.
in air medium, μ = 1.5/1 = 1.5
so, 1/f = (1.5 - 1) [1/0.2 - 1/-0.22 ]
= 0.5[5 + 4.54 ]
= 0.5 × 9.54
= 4.77
or, f = 1/4.77 = 0.2096 m ≈ 21cm
when lens is immersed in water.
μ' = μ_l/μ_w = 1.5/1.33 = 1.1278
now, 1/f' = (μ' - 1)[1/R1 - 1/R2]
= (1.1278 - 1) [1/0.2 -1/-0.22]
= 0.1278 [5 + 4.54]
= 1.219
or, f' = 1/1.219 = 0.82 m = 82 cm
hence, when lens is immersed in water focal length of lens is increased by (82 - 21) = 61cm .