The Problem

We have a plano-convex lens with a focal length of f and a refractive index of 1.5. The lens is placed over a glass plate with its curved surface touching the plate. The gap between the lens and the plate is filled with a liquid, which results in the effective focal length of the combination becoming 2f. We need to find the refractive index of the liquid.

Understanding the Situation

To solve this problem, we need to understand the concept of effective focal length and the behavior of light at the interface between different mediums.

The effective focal length of the combination is the focal length when the lens and the glass plate with the liquid act as a single optical system. The effective focal length is given by the lens-maker's formula:

1/f = (n-1)(1/R1 - 1/R2)

Where:
- f is the focal length of the lens
- n is the refractive index of the lens material
- R1 is the radius of curvature of the curved surface of the lens
- R2 is the radius of curvature of the flat surface of the lens

Applying the Given Information

In our case, the lens is plano-convex, which means one surface is flat (R2 = ∞) and the other surface is curved. Since the curved surface is in contact with a glass plate, its radius of curvature is equal to the radius of curvature of the glass plate. Let's denote this radius of curvature as R.

Given that the effective focal length of the combination is 2f, we can rewrite the lens-maker's formula as:

1/(2f) = (1.5 - 1)(1/R - 1/∞)

Simplifying this equation, we get:

1/(2f) = 0.5/R

Finding the Refractive Index of the Liquid

Now, let's consider the behavior of light at the interface between the liquid and the lens material. When light travels from one medium to another, its direction changes due to the change in refractive index. We can apply Snell's law to relate the refractive indices of the liquid and the lens material:

n1 * sin(θ1) = n2 * sin(θ2)

Where:
- n1 is the refractive index of the liquid
- θ1 is the angle of incidence (measured from the normal)
- n2 is the refractive index of the lens material (given as 1.5)
- θ2 is the angle of refraction (measured from the normal)

In our setup, the light is incident from the liquid onto the lens material, so θ1 is the angle between the normal and the incident ray. Since the lens is in contact with the glass plate, the normal to the lens surface is perpendicular to the glass plate.

Let's assume the angle between the normal and the incident ray is θ. Then the angle of refraction within the lens material is also θ, as the light travels parallel to the normal within the lens.

Applying Snell's law, we have:

n1 * sin(θ) = 1.5 * sin(θ)

Simplifying this equation, we find:

n1 = 1.5

Therefore, the refractive index of the liquid is 1.