# Describing Image of a Bi-Concave Lens with n=1.5

• fluidistic
In summary: I can't see how to calculate the magnification or anything else. The matrix method calculates the position and magnification of an object after it has been refracted by a lens.
fluidistic
Gold Member

## Homework Statement

A bi-concave lens (n=1.5) has radii of 20 cm and 10cm and a thickness of 5 cm. Describe the image of an object of 2.5 cm of hight and situated at 8 cm from the first side of the lens.

Attempt:
$$\frac{1}{f}=(n-1) \left [ \frac{1}{R_1}-\frac{1}{R_2} + \frac{(n-1)d}{nR_1R_2} \right ]$$. After some arithmetics, $$f=-\frac{4}{3}$$.
So I know that the image is on the same side of the oject and according to my sketch is it smaller. I'm having a hart time figuring out how to calculate its distance from the lens.
I've been searching on the Internet but there are very few information for thick lens or I'm just blind that I don't find anything that could help me.
Once I have the place of the image, I can easily calculate the magnification and hence say how hight is the image. But I'm stuck as how to calculate the distance between the image and the lens.
Thanks for any help.

Find the image distance formed by the first surface only by using the formula
1/do + μ/di = (μ - 1)/R1 ...(1)
In this problem, this image is virtual. This image becomes a real object inside the lens for the second surface. Its distance from the second surface is (di + 5). Using the same above equation you can write
μ/(di + 5) + 1/di' = ( μ - 1 )/R2 ...(2).
Solve for image distance formed by the second surface. Use the proper sign conventions.

rl.bhat said:
Find the image distance formed by the first surface only by using the formula
1/do + μ/di = (μ - 1)/R1 ...(1)
In this problem, this image is virtual. This image becomes a real object inside the lens for the second surface. Its distance from the second surface is (di + 5). Using the same above equation you can write
μ/(di + 5) + 1/di' = ( μ - 1 )/R2 ...(2).
Solve for image distance formed by the second surface. Use the proper sign conventions.
Thanks for the reply. Could you please tell me what does mu represent?
I know that for thick lenses there is an equivalent formula to $$\frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i}$$ though I don't know the exact one, which seems to be $$\frac{1}{S_o}+\frac{\mu}{S_i}$$ if I'm not misunderstanding you. Mu would be the refractive index of the lens?
Can I use the ray transfer matrix method? In any case I think I should locate the principal planes of the lens, but I don't know how to do so.

As you have guessed, mu is the refractive index of the lens.
The equations which I have used are for the refraction through the spherical surfaces. In the first case the object is in the air and the image is in the glass. This image acts as the object for the second surface with in the lens. The final image in in the air.

rl.bhat said:
As you have guessed, mu is the refractive index of the lens.
The equations which I have used are for the refraction through the spherical surfaces. In the first case the object is in the air and the image is in the glass. This image acts as the object for the second surface with in the lens. The final image in in the air.

Ok thank you, I understand now.
Just a little remark: I just looked in Hecht's book (page 273 if I remember well) and for thick lenses, he says that we can use the formula I posted in my first post or yours, but that $$S_o$$ is not the distance from the object to the vertex of the lens. Rather, it's the distance between the object and the first principal plane. The equation to find each principal planes are $$h_1=-\frac{f(n_l-1)d_l}{R_2 n_l}$$ and $$h_2=-\frac{f(n_l -1)d_l}{R_1n_l}$$, measured from the first and rear vertices respectively.
So we must first calculate the focus, which can be done with the formula I provided in the first post.
Using $$h_1$$ and $$h_2$$ we can calculate the position of the image with respect to the optical axis. Then the magnification... I'm not really sure. Something like $$-\frac{S_i}{S_o}$$ if my memory works well.

Now, maybe this method is an equivalent to yours. Tomorrow I'm having a test. Last course the "helper" of the class made a fool of me because I said that for thick lenses, $$S_o$$ is the distance between the object and the first vertex of the lens. Rather than helping me, he got pissed off by my ignorance. Not a good sign, he might be the one who corrects my test.

I just understood the matrix method. However I don't really see how it can calculate all the cardinal points of an optical system, like Hecht says. Rather I see how I can predict how an entering ray will leave the optical system, knowing its initial conditions and knowing what the system is made of. It doesn't seem really helpful, at least not as helpful as Hecht tends to say. Unless I'm misunderstanding what the method is all about.

## 1. How does a bi-concave lens differ from a bi-convex lens?

A bi-concave lens has two concave surfaces, while a bi-convex lens has two convex surfaces. This means that a bi-concave lens causes light to diverge, while a bi-convex lens causes light to converge.

## 2. What does the value of n=1.5 represent in the description of the bi-concave lens?

The value of n=1.5 represents the refractive index of the lens material. This indicates how much the speed of light changes when passing through the lens material.

## 3. How does the refractive index affect the behavior of light passing through the bi-concave lens?

The refractive index determines the degree of bending of light as it passes through the lens. In a bi-concave lens with a refractive index of 1.5, the light will be bent more than in a lens with a lower refractive index.

## 4. Can you describe the image formed by a bi-concave lens with n=1.5?

The image formed by a bi-concave lens with n=1.5 will be virtual, upright, and smaller than the object. It will also appear to be located closer to the lens than the object.

## 5. What are some real-life applications of bi-concave lenses with a refractive index of 1.5?

Bi-concave lenses are commonly used in eyeglasses to correct nearsightedness. They are also used in microscopes and telescopes to magnify images. In addition, they are used in optical devices such as cameras and projectors.

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