Question

A lens is made with a focal length of -40 cm using a material with index of refraction 1.50. A second lens is made with the SAME GEOMETRY as the first lens, but using a material having refractive index of 2.00. What is the focal length of the second lens

Answer 1

f = - 20 cm

Step-by-step explanation:

This exercise asks us for the focal length, which for a lens in air is

1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

- 1/40 = (n-1) (1 / R₁ -1 / R₂)

(1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

(1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

(1 / R₁ -1 / R₂) = -1/20

Now we change the construction material for one with refractive index

n = 2, keeping the radii,

1 / f = (n-1) (1 / R₁-1 / R₂)

1 / f = (n-1) (-1/20)

let's calculate

1 / f = (2.00-1) (-1/20)

1 / f = -1/20

f = - 20 cm