Focal length of converging meniscus lens filled with a liquid

[Edge Of Pain]

Homework Statement

A converging meniscus lens is made of glass with index of refraction n = 1.55,
and its sides have radii of curvature of 4.5 cm and 9 cm. The concave
surface is placed upward and filled with carbon tetrachloride which has index
of refraction n' = 1.46. Determine the focal length of the combination of glass and carbon tetrachloride.

Homework Equations

I think these equations are relevent.

$1/f = 1/f_1 +1/f_2$ (equation 1, effective focal length)

$1/f = (n/n' -1)(1/R_1 - 1/R_2)$ (equation 2, lensmaker's equation)

Alternatively if the thin lens is surrounded by air, then

$1/f = (n -1)(1/R_1 - 1/R_2)$

The Attempt at a Solution

Substituting all known values into the second equation then taking the reciprocal of it to get the value of f, I get -146cm. I think the sign is correct because these are refracting interfaces, so drawing a principle ray diagram gives the focal point on the opposite side of the outgoing light, so the focal length should be negative. But I'm not confident in the magnitude of my answer, 146cm. It doesn't seem right considering that this question is worth 8 marks.

I also tried to treat the meniscus lens as two concave lenses which had an one effective focal length, and the carbon liquid stuff as another concave lens which had a radius of curvature of 9cm (equal to the R.O.C of the inner part of the meniscus lens) and its own effective focal length. Then, the effective focal length of the entire combination would be

$1/f = 1/f_lens + 1/f_carbon$

which is

$1/f = 1/(1/2.25 + 1/4.5) + 1/4.5$

$∴ 1/f = 1/1.5 + 1/4.5 => f = 1.125cm$

The magnitude of 1.125cm seems far more reasonable but I think the sign is incorrect based on the side of the outgoing light. Here's a sketch of how I imagine this to be:

(am I imagining it correctly? Is one of my answers correct?)

EDIT: I sketched the R1 incorrectly! I drew the line a little too far down. The line for R1 should stop at the first part of the lens.

 

[Edge Of Pain]

I believe it's been more than 24 hours. Therefore I am bumping this.

 

[Vibhor]

You need to use both the formulas , one at a time .

Stage 1:Using the lensmaker's formula (equation 2) find the focal length of the individual lenses .

Stage 2:Then using equation 1 find the equivalent focal length .

 

[Edge Of Pain]

You need to use both the formulas , one at a time .

Stage 1:Using the lensmaker's formula (equation 2) find the focal length of the individual lenses .

Stage 2:Then using equation 1 find the equivalent focal length .

OK, thank you, but how does equation 2 apply to the liquid, since it has only 1 radius of curvature?

EDIT: Wait, we can treat the edge of the liquid as a plane refracting surface so its other radius of curvature is equal to infinity and we have 1/R - 0. Correct? Using this I get around 17cm effective focal length.

 

[Vibhor]

OK, thank you, but how does equation 2 apply to the liquid, since it has only 1 radius of curvature?

EDIT: Wait, we can treat the edge of the liquid as a plane refracting surface so its other radius of curvature is equal to infinity and we have 1/R - 0. Correct?

Yes.

Using this I get around 17cm effective focal length.

This doesn't look correct . Please show your working .

 

[Edge Of Pain]

Yes.



This doesn't look correct . Please show your working .

OK, I've done it once more and got a different answer. This time I'll show my working.

Firstly, image will be formed on opposite side of outgoing like so we predict a negative effective focal length. Correct?

Then, we apply the lensmaker's equation to both lenses. Call $n_1$ index of refraction of air, $n_2$ index of refraction of carbon tetrafloride, and $n_3$ index of refraction of glass.

For the meniscus lens we have

$1/f_1 = (n_2/n_3 -1)(1/R_1 - 1/R_2) = (1.46/1.55 -1)(1/4.5 -1/9) = -1/155$ (f1 is around -0.006cm)

and for the carbon liquid we have

$1/f_2 = (n_1/n_2 -1)(1/9 - 0) = -457/13140$ (f2 is around -29cm)

so for our effective focal length

$1/f = 1/f_1 +1/f_2$

so

$1/f = -1/155 - 457/13140$

so

$f = -24.3cm$ (3 s.f.)

 

[Vibhor]

Suppose there is only liquid lens (i.e ignore glass lens for a while ) .Think of liquid lens with air medium on both the sides.

What is the focal length ? The one you have calculated above i.e -29cm is incorrect .Focal length of a converging lens is positive .

 

[Edge Of Pain]

Suppose there is only liquid lens (i.e ignore glass lens for a while ) .Think of liquid lens with air medium on both the sides.

What is the focal length ? The one you have calculated above i.e -29cm is incorrect .Focal length of a converging lens is positive .

OK.

For the focal length of the liquid lens surrounded by air we have

$1/f_2 = (n-1)(1/R_1 - 1/R_2)$.

But in this case R1 is 9cm and R2 is infinity.

So

$1/f_2 = (1.46 - 1)(1/9 - 0) = 0.05111... ∴ f_2 ≈ 19.6cm$.

Now, we do the same thing for the glass lens and then find the effective focal length of them together?

If I am correct then

$1/f_1 = (1.55 -1)(1/4.5 - 1/9) = 0.6111... ∴ f_1 ≈ 16.4cm$.

So the effective focal length of the whole thing is

$1/f = 1/f_1 +1/f_2 ∴ f ≈ 8.9cm$. Correct?

 

[Vibhor]

OK.

For the focal length of the liquid lens surrounded by air we have

$1/f_2 = (n-1)(1/R_1 - 1/R_2)$.

But in this case R1 is 9cm and R2 is infinity.

So

$1/f_2 = (1.46 - 1)(1/9 - 0) = 0.05111... ∴ f_2 ≈ 19.6cm$.

Now, we do the same thing for the glass lens and then find the effective focal length of them together?

If I am correct then

$1/f_1 = (1.55 -1)(1/4.5 - 1/9) = 0.6111... ∴ f_1 ≈ 16.4cm$.

So the effective focal length of the whole thing is

$1/f = 1/f_1 +1/f_2 ∴ f ≈ 8.9cm$. Correct?

Looks good

 

[Edge Of Pain]

Looks good

Thanks! I was thinking about this question for several days now :)

I already thanked one of your posts in this thread so it won't allow me to thank you again.