9.38: Figure 9.37 shows a biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

                                        Equiconvex lens

Hint: Use lens makers formula.

Step 1: Find the length of the liquid.
The focal length of the convex lens, f1=30 cm
The focal length of the liquid = f2

Focal length of the system (convex lens+liquid), f = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length:

1f=1f1+1f2
1f2=1f-1f1=145-130=-190
f2=-90 cm 

Step 2: Find the radius of curvature of the lens.
Let the refractive index of the lens be μ1 and the radius of curvature be R.

Using the lens maker's formula:

1f1=μ1-11R-1-R 130=1.5-1 2R R=300.5×2=30 cm

Step 3: Find the refractive index of the liquid.
Let μ2 be the refractive index of the liquid.

Again using the lens maker's formula:

1f2=μ2-1-1R-1
-190=μ2-1-130-0
μ2-1=13
μ2=43=1.33

Hence, the refractive index of the liquid is 1.33.