Problem: A biconvex lens of refractive index 1.5 and focal length 0.2m acts as a divergent lens of power 1D when immersed in a liquid. Find the refractive index of the liquid.

Solution:
To solve the problem, we can use the lens formula:
1/f = (μ - 1)(1/R1 - 1/R2)
where f is the focal length of the lens, μ is the refractive index of the lens material, R1 and R2 are the radii of curvature of the two surfaces of the lens.

We can assume that the lens is symmetric, so R1 = R2 = R. Then the lens formula becomes:
1/f = (μ - 1)2/R

We are given that the focal length of the lens when it is immersed in a liquid is 1D or 1 m^-1. Therefore, we have:
1/1 = (μ - 1)2/R
or, μ - 1 = 2R

We are also given that the focal length of the lens in air is 0.2 m. Using the lens formula again, we have:
1/0.2 = (1.5 - 1)2/R
or, R = 0.15 m

Substituting this value of R in the expression for μ, we get:
μ = 2R + 1 = 2(0.15) + 1 = 1.3

Therefore, the refractive index of the liquid is 1.3.

Explanation:
The biconvex lens has a refractive index of 1.5 and a focal length of 0.2 m in air. When the lens is immersed in a liquid, it acts as a divergent lens of power 1D or 1 m^-1. We need to find the refractive index of the liquid.

To solve the problem, we use the lens formula and the given information about the focal length of the lens in air and in the liquid. We assume that the lens is symmetric and has equal radii of curvature on both surfaces. We also use the fact that a divergent lens has a negative focal length.

By solving the lens formula equations, we get the value of the radius of curvature of the lens surface. Using this value, we can find the refractive index of the liquid.

Finally, we can conclude that the refractive index of the liquid is 1.3.