# Optics Homework: Converging Lens and Varying Refractive Index

In summary, a rectangle with a uniformly changing index is equivalent to a more familiar piece of optics equipment.

## Homework Statement

A rectangular slab of length l=20cm and thickness d=4cm is placed in left of a converging lens of focal length f=20cm. A screen is placed in the focal plane plane of lens (right side of lens).Refractive index of the material of slab increases linearly from u0 at the bottom by an amount ##du=2*10-4)## between bottom and top faces.A parallel beam is made to incident on the left side of slab. Prove that only a bright spot forms on the screen? Find its position

## Homework Equations

$$\Delta x=n \lambda$$
$$y=n \lambda /d$$

## The Attempt at a Solution

Here as the refractive index varies, speed of light varies. It will be faster at bottom and slower at top.
$$v=c/ \mu$$
$$\mu = \mu_o + y \delta \mu$$
For any two rays at y1 and y2 once the slowest ray comes out of slab,
$$t=l/v => t = l/ ( \mu_0 + d \delta \mu)$$
Hence $$\Delta x = l(\mu_0 + d \delta \mu) \frac {(y_2-y_1) \delta \mu}{(\mu_0 + y_1 \delta \mu)(\mu_0 c + y_2 \delta \mu)}$$

I'm not sure what you have calculated. What is Δx? The difference in the two path lengths?
I feel it is more useful to describe how the two light beams will be changed by the slab. Will they be bent through the same angle?
(It might help to recognise that a rectangular slab with uniformly changing index is equivalent to a more familiar piece of optics equipment.)

Close the thread. I have solved the problem.

haruspex said:
I'm not sure what you have calculated. What is Δx? The difference in the two path lengths?
I feel it is more useful to describe how the two light beams will be changed by the slab. Will they be bent through the same angle?
(It might help to recognise that a rectangular slab with uniformly changing index is equivalent to a more familiar piece of optics equipment.)
I have solved the problem. The question is of high level. Thank you for the response.
Actually $$\mu_1 = \mu_0 + \delta \mu$$
Find the time taken by slowest ray to come out of slab. Let it be t.
$$t=l \mu_1/c$$
During that time find the distance traveled by fastest ray(at bottom.
Now the situation is like fraunhoffer single slit diffraction.

Last edited:

Thank you for your response. I would like to suggest a few things to improve your solution.

Firstly, it would be helpful to define the variables used in the solution, such as l, d, f, u0, du, etc. This will make it easier for others to follow your solution.

Secondly, it would be beneficial to provide a brief explanation of the physical principles involved in this problem, such as Snell's law and the concept of refractive index. This will provide context for your solution and help others understand the problem better.

Additionally, it is important to use proper units for all the variables and equations. In this case, the units for l, d, and f should be in centimeters, and the units for refractive index should be dimensionless.

Finally, it would be helpful to provide a diagram or figure to illustrate the setup of the problem. This will make it easier for others to visualize the problem and understand your solution.

Overall, your solution is well thought out and demonstrates a good understanding of the problem. By incorporating these suggestions, you can make your solution even clearer and more comprehensive.

## 1. What is a converging lens?

A converging lens is a type of lens that is thicker in the middle and thinner at the edges. It is designed to bend light rays towards a central point, known as the focal point, and is commonly used in devices such as cameras, telescopes, and eyeglasses.

## 2. How does a converging lens work?

A converging lens works by refracting, or bending, light rays that pass through it. As the light rays pass through the lens, they are slowed down and bent towards the center of the lens. This causes the light rays to converge at a point, known as the focal point, on the other side of the lens. The distance between the lens and the focal point is known as the focal length.

## 3. What factors affect the focal length of a converging lens?

The focal length of a converging lens is affected by two main factors: the curvature of the lens and the refractive index of the material it is made of. A lens with a greater curvature or a material with a higher refractive index will have a shorter focal length, meaning that the light rays will converge at a closer point to the lens.

## 4. What is the refractive index of a material?

The refractive index of a material is a measure of how much the speed of light slows down as it passes through that material. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. For example, the refractive index of air is very close to 1, while the refractive index of glass is around 1.5.

## 5. How does varying refractive index affect the behavior of a converging lens?

Varying refractive index can affect the behavior of a converging lens in several ways. A lens made of a material with a higher refractive index will have a shorter focal length, as mentioned earlier. Additionally, if the refractive index varies across the surface of the lens, it can cause the light rays passing through it to bend at different angles, resulting in a distorted or blurred image. This can be corrected by using a lens made of a material with a uniform refractive index, or by using multiple lenses to compensate for the varying refractive index.