a convex lens of focal length 20 cm and made of glass , is immersed in water of refrating index of 1.33 .Calculate change in focal length of lens?
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We know ,
1/f = [n(g) - n(m) ] [ 1/R(1) = 1/R(2)]
f = focal length
n(g) = refractive index of glass - 1.5
n(m) - refractive index of medium
R1 and R2 are radius of curvature
Refractive index of air is 1
Hence focal length of air
1/f(a) = (1.5 - 1 ) ( 1 / R1 - 1/R2)___1
refractive index of water = 1.33
Hence focal length of water
1/f(w) = (1.5 - 1.33) ( 1/R1 - 1/R2)__2
From 1 and 2
we got
f(w) / f(a) = (1.5-1)/ (1.5-1.33)
= 0.5/0.17 = 2.94
GIven f(a) = 20 cm
f(w) = 20 x 2 . 94 = 58.80 cm
1/f = [n(g) - n(m) ] [ 1/R(1) = 1/R(2)]
f = focal length
n(g) = refractive index of glass - 1.5
n(m) - refractive index of medium
R1 and R2 are radius of curvature
Refractive index of air is 1
Hence focal length of air
1/f(a) = (1.5 - 1 ) ( 1 / R1 - 1/R2)___1
refractive index of water = 1.33
Hence focal length of water
1/f(w) = (1.5 - 1.33) ( 1/R1 - 1/R2)__2
From 1 and 2
we got
f(w) / f(a) = (1.5-1)/ (1.5-1.33)
= 0.5/0.17 = 2.94
GIven f(a) = 20 cm
f(w) = 20 x 2 . 94 = 58.80 cm
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