# Optics - Two lenses touching each other with oil inbetween

1. Sep 23, 2011

### Xyius

1. The problem statement, all variables and given/known data
Two identical, thin, plano-convex lenses with radii of curvature of 15 cm are situated with their curved surfaces in contact at their centers. The intervening space is filled with oil of refractive index 1.65. The index of the glass is 1.50. Determine the focal length of the combination. (Hint: Think f the oil layer as an intermediate thin lens.)

2. Relevant equations
The Lens Maker Equation
$$\frac{1}{f}=\frac{n_2-n_1}{n_1}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Where $R_1$ and $R_2$ are the radii of curvature for each surface on the lens, and the n's are the respective refractive indexes of the lens and the outside medium.

3. The attempt at a solution

So the first lens is plano-convex lens. The first surface it goes through is the plane, then it travels to the spherical side. The radius of curvature for the plane is infinity and the radius of curvature for the curved side is -15cm. (Negative because from left to right it is a concave lens.) Plugging it into the equation above (And plugging in 1.5 for n2 and 1 for n1.) I get f=30cm.

For the second "lens" it is the layer of oil which acts as a bi-concave lens. Using the lens makers equation again with R1=R2=15cm, and R1<0 because it is concave from left to right and R2>0 because it is convex from left to right. Plugging all this in (and plugging 1.65 for n2 and 1.5 for n1) I get f=-75cm.

For the last lens, it is again a plano-convex lens. The first surface it goes through is the curved side with R1=15cm (Positive because it is convex from left to right), and the second surface is a plane right R2=infinity. Plugging this in (and plugging n1=1.5 and n2=1.65) I get f=-45cm.

So I am guessing I need to sum all of these up, I get f=-90cm. The answer in the back of the book says -60 cm! What am I doing wrong? :(??

2. Sep 24, 2011

### Xyius

Can anyone help me on this one??

3. Sep 25, 2011

### mehStevens

I have the same problem and have reached the same solution, except my book says -50cm in the back : /

4. Oct 17, 2011

### mrwdq

what is this right answer 50 or 60
i have this question also

5. Oct 17, 2011

### Xyius

It is 50. The trick is to treat each lens separately as it is in air. And the equivalent focal distance is

$$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$$

Sorry I forgot to post the solution when I figured it out!

6. Oct 18, 2011

### mrwdq

i get the solution if you still want i can send it

7. Oct 18, 2011

### mrwdq

consider the three media as a squence of three thin lenses. each has a focal lenght given by the lensmaker's equation, and the equivalent focal lenth is

1/feq=1/f1+1/f2+1/f3

then
1/f1=(1.5-1)(1/∞-1/(-15))

f1=30 cm

1/f2=(1.65-1)(1/(-15)-1/15

f2=-150/3

f1=f3=30 cm

then

1/feq= 1/30 + (-13)/150 +1/ 30

so feq=-50 cm