Radii of curvature of a convex lens are 20cm and 10 cm. Find focal length and power of lens in air and in water, if refractive of lens is 3/2 and of water is 4/3.
Answers
Given, radii of curvature of double convex lens are 20cm and 10cm.
refractive index of lens in air is 3/2 and
refractive index of lens in water is 4/3.
To find focal length and power of lens in air :
let n as refractive index of lens in air medium.
and R1 is the first radius of curvature
R2 is the second radius of curvature
so,
1/ focal length = (refractive index -1 ) (1/ first radius of curvature - 1/ second radius of curvature)
1/f = (n-1)(1/R1 - 1/R2)
1/f = [(3/2) -1] (1/20- (-1/10))
1/f = (1/2) (1/20+1/10)
1/f = (1/2) (3/20)
1/f = 3/40 cm
f = 40/3 cm OR
f = 13.33 cm
now,
power of lens = 100/ focal length (in cm)
P = 100/f (in cm)
P = 100/ 40÷3
P = 300/40
P = 30/4 D OR
P = 7.5 D
To find focal length and power of lens in water :
let n as refractive index of lens in water medium.
and R1 is the first radius of curvature
R2 is the second radius of curvature
we have to find n first in water medium,
so, n of water = n of lens/n of surrounding air
n of water = 3÷2/4÷3
n of water = 3×3 / 2×4
n of water = 9/8 OR
n of water = 1.125
so,
1/ focal length = (refractive index -1 ) (1/ first radius of curvature - 1/ second radius of curvature)
1/ focal length = (refractive index -1 ) (1/ first radius of curvature - 1/ second radius of curvature)
1/ focal length = (refractive index -1 ) (1/ first radius of curvature - 1/ second radius of curvature) 1/f = (n-1)(1/R1 - 1/R2)
1/ focal length = (refractive index -1 ) (1/ first radius of curvature - 1/ second radius of curvature) 1/f = (n-1)(1/R1 - 1/R2) 1/f = [(9/8) -1] (1/20- (-1/10))
1/f = (1/8) (1/20+1/10)
1/f = (1/8) (3/20)
1/f = 3/160 cm
f = 160/3 cm OR
f = 53.33 cm
now,
now,power of lens = 100/ focal length (in cm)
now,power of lens = 100/ focal length (in cm) P = 100/f (in cm)
now,power of lens = 100/ focal length (in cm) P = 100/f (in cm) P = 100/ 160÷3
P = 300/160
P = 30/16 D OR
P = 1.875 D