Question

If radius of curvature of both convex surfaces is 20 cm, then focal length of the object placed in air in the given arrangement is given by

Answer 1

case 1 : for first surface

μ1 = 1, μ2 = 1.5 , R = 20cm

using refraction formula for curve surface.

μ2/v' - μ1/u = (μ2 - μ1)/R

⇒1.5/v' - 1/u = (1.5 - 1)/20......(1)

case 2 : for second surface,

μ1 = 1.5, μ2 = 4/3 , u = v' and R = -20cm

so, (4/3)/v - 1.5/v' = (4/3 - 1.5)/-20 .....(2)

adding both equations we get,

-1/u + (4/3)/v = (0.5 + 1.5 - 4/3)/20

⇒-1/u + (4/3)/v = 1/30

for getting value of focus, taking u = ∞ then, v = f

so, (4/3)/f = 1/30

⇒1/f = 1/30(4/3) = 1/40

⇒f = 40cm

hence, focal length is 40cm