# Geometrical Optics

Light can behave in many ways like a wave or a packet of photons which travel in a straight line exhibiting the property of rectilinear propagation. **Geometrical optics** deals with the propagation of light in a straight line and phenomena such as reflection, refraction, polarization, etc. A ray of light gives the direction of propagation of light. In the absence of an obstacle, the rays advance in a straight line without changing direction. When light meets a surface separating two transparent media, reflection and refraction occur and the light rays bend.

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## Reflection at Smooth Surfaces

A light ray is reflected by a smooth surface in accordance with the 2 laws of reflection:

- The angle of incidence is always equal to the angle of reflection.
- The incident ray, the reflected ray, and the normal to the reflecting surface are coplanar. The images formed by a plane mirror are equidistant from the distance between the object and the mirror.

An object is placed in front of the mirror. If the mirror is moved away from the object through a distance of $x,$ by how much distance will the image move?

The answer will be added shortly.

Our final answer is $2x.\ _\square$

A ray of light falls on a plane mirror. Show that if the mirror is tilted through an angle of $\theta,$ then the reflected ray tilts through an angle of 2$\theta.$

The laws of reflection are the same for plane and curved surfaces. A normal can be drawn from any point of the curved surface by first drawing the tangent plane from that point and then drawing the line perpendicular to that plane. Angles of incidence and reflection are defined from this normal. The angle of incidence is equal to angle of reflection.

## Spherical Mirrors

A spherical mirror is a part cut from a hollow sphere. The spherical mirrors are generally constructed from glass. One surface of the glass is silvered. If reflection takes place at the convex surface, it is called a convex mirror and if it takes place on the concave surface, it is called a concave mirror.

**Terms used in spherical mirrors:**

- Center of curvature (C): it is the center of the sphere from which the mirror is made.
- Focus (f): it is the point at which the rays meet (either virtually or physically). Although we have not verified, from further exploration we will understand that $R=2f.$
- Pole (P): it is the surface of the mirror on which light falls.
- Principle axis (PA): it is the imaginary line which is perpendicular to the pole.
- Paraxial rays: These are rays parallel and close to the principal axis.
- $u:$ This is the distance between the object and the pole and is sometimes referred to as the object distance.
- $v:$ This is the distance between the pole and the image and is sometimes referred to as the image distance.

Here $(0,0)$ is the $C,$ and the principal axis is the $x$-axis.

## Sign Conventions

**Cartesian Sign Conventions**

1) The pole is taken to be the origin.

2) The direction in which the light moves is taken as the positive direction.

3) The height above the principal axis is taken as positive while the height below it is taken as negative.

## Mirror Formula

The main formula used for solving problems on mirrors is

$\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f }.$

**Note**: While solving problems, do not forget to apply sign conventions and make sure you always take the object distance, $u,$ as *negative only*. See the examples below which will make you comfortable with solving the problems based on the mirror formula.

If the distance between the object and the mirror is $10\text{ cm}$ and its focal length is $5\text{ cm}$ given it is a convex mirror, find the distance between the mirror and the object.

We are given $u = -10\text{ cm}$ and $f = 5\text{ cm}.$

Applying the mirror formula, we get

$\begin{aligned} \frac { 1 }{ v } +\frac { 1 }{ u } &=\frac { 1 }{ f } \\ \frac { 1 }{ v } +\frac { 1 }{ -10 } &=\frac { 1 }{ 5 } \\ \frac { 1 }{ v } &=\frac { 3 }{ 10 } \\ \Rightarrow v&=3.33\text{ cm}. \ _\square \end{aligned}$

The magnification of a mirror is given by $\frac { { h }_{ i } }{ { h }_{ o } }$, where ${ h }_{ i }$ is height of the image and ${ h }_{ o }$ is height of the object.

Magnification is also given by $-\frac { v }{ u }$.

Again we should remember not to forget applying the sign conventions.

If magnification is $-ve,$ it means the formed image is inverted.

If the image formed by a spherical mirror is half the size of the object and it is virtual, then find the distance between the mirror and the image given that the object's distance is $16\text{ cm}.$

We are given $m=\frac { { h }_{ i } }{ { h }_{ o } } =\frac { 1 }{ 2 } =\frac { -v }{ u }.$

So, $u=-2v=-16 \text{ cm} \Rightarrow v=8\text{ cm}. \ _\square$

## Refraction

When light passes from one medium to another and undergoes a change in speed and direction, it is called refraction. The light does reflect but continues on its path with a change in its direction.

When light moves from a rarer to denser medium, i.e. moves from a light to heavy medium, the light tends to bend towards the normal while the opposite happens when light travels from a denser to rarer medium.

For the concept of refraction, one important thing to know is to learn about Snell's law.

This law tells us that

${ \mu } _{ 1 } \sin i = { \mu }_{ 2 } \sin r,$

where ${ \mu }_{ 1 }$ is the medium's refractive index from which light travels and ${ \mu }_{ 2 }$ is the medium's refractive index to which light goes. The angle of incidence is $i$ and the angle of refraction is $r.$

Generally, we deal with a situation where light goes from air to another medium and since the refractive index of air is 1, we get the equation

$\sin i = { \mu } \sin r.$

The lens formula is given by

$\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }.$

Again it is important to use the sign conventions, or the answers are going to be wrong.

**Most importantly**, the sign conventions are going to remain the same as that of the mirrors, but here we have two focal lengths rather than one in the case of mirrors. So the focal length of concave is $-\text{ve}$ by convention and that of convex is $+\text{ve}.$

If the image formed by a convex lens is $20\text{ cm}$ from the lens and the radius of curvature is $30\text{ cm},$ then find the distance of the object from the lens.

We have $R=30\text{ cm} \implies f=15\text{ cm}$. Using the lens formula, we have $\frac { 1 }{ +20 } -\frac { 1 }{ u } =\frac { 1 }{ 15 }$. Remember not to put the negative sign of $u$ before solving it:

$\begin{aligned} \frac { 1 }{ 20 } -\frac { 1 }{ 15 } &=\frac { 1 }{ u } \\ \frac { 3-4 }{ 60 } &=\frac { 1 }{ u } \\ \frac { -1 }{ 60 } &=\frac { 1 }{ u } \\ u &= -60\text{ cm}. \ _\square \end{aligned}$

The lens formula we used above was a result of using a biconvex or biconcave lens. These lenses have equal radii of curvature and equal foci. However, the same cannot be used for a lens having different radii of curvature. So a new method of finding the focus, $u,$ and $v$ was developed, called the lens maker formula.

$\frac { 1 }{ F } =\left(\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } -1\right)\left(\frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right).$

Here $F$ is the resultant focal length, ${ \mu }_{ 2 }$ is the refractive index of the object in which light enters, and ${ \mu }_{ 1 }$ is the refractive index from which light travels to reach the object. Normally ${ \mu }_{ 1 }$ is ignored when light travels from air because it has a refractive index of $1,$ so ${ \mu }_{ 1 }$ becomes $1.$ ${ R }_{ 1 }$ is the radius of curvature of the $1^\text{st}$ lens, and ${ R }_{ 2 }$ is the radius of curvature of the ${2}^\text{nd}$ lens.

## Atmospheric Refraction

We know that the earth is surrounded by a layer of air called the atmosphere. The density of air is not the same everywhere. The refractive index of air depends on its density, i.e. the higher the density of air, the greater its refractive index. The changes in refractive index can give rise to many phenomena.

For example, a rising current of hot air makes the objects viewed through it flicker and sway. You might have seen that on a summer day; hot air seems to be rising above the surface of the roads. It also happens when you see above a fire.

**Early sunrise and late sunset**

Consider an oblique ray from a heavenly body such as the sun or a star. While travelling through the atmosphere, it continuously moves into regions of higher refractive index. So, it continuously bends towards the normal, resulting in a path similar to the figure given below.

Since we see the object, that is the sun now, in the direction of the ray incident on the eye, the heavenly body appears higher than its actual position.

Now, consider a situation when the sun is just below the horizon. Rays of light coming from it get bent such that they seem to be coming from above the horizon (to the observer). Thus, when the sun's position is just below the horizon, the sun is visible to us. So, at sunrise, we see the sun before it actually comes to the horizon. At sunset, we see it even after it has dipped below the horizon. The fact is that it increases daylight by 4 minutes everyday (2 minutes at sunrise and 2 minutes at sunset).

You might also have noticed that the sun looks oval/flattened at sunrise and sunset. The rays from the lower regions of the sun travel a bit greater distance than those rays which are from the middle and upper regions. So, they bend more. As a result, the image of the lower region gets shifted upwards more than that of the upper region. This makes the sun appear like a flattened circle or an oval.

Similarly, *twinkling of stars* is also a result of atmospheric refraction.

Now, we can define atmospheric refraction as follows:

Atmospheric refractionis a phenomenon, where light/any electromagnetic wave deviates from its rectilinear path because of passing throughatmospherewhere there are variations in the air density and refractive index.

## Total Internal Reflection

When light travels from a denser to rarer medium with an angle greater than the critical angle, the ray of light does not deviate in its path or does not refract, but it undergoes a reflection known as total internal reflection.

The critical angle differs from medium to medium.

It is given by the formula

$\mu =\frac { 1 }{ \sin{ \theta }_{ c } }.$

This is very useful as it is used in fiber glasses where total internal reflection helps in fast movement of wavelengths.

Optical fibers are devices used for guiding light in many applications, most notably for fast communication. A fiber consists of a glass cylinder surrounded by a wall covered in a special coating.

The fibers work on a principle called *total internal reflection*: light enters the fiber at an angle such that it does not get transmitted through the wall of the fiber when it hits the inside of the wall. Therefore, the refraction index of the glass part of the fiber has to be higher than that of its coating.

What is the maximum entering angle **in degrees** a light ray can pass from the air to the glass fiber for the total internal reflection to occur?

**Details and Assumptions:**

- Measure the entering angle from the axis of the fiber.
- Use the following refraction indexes: $n_{\text{air}} = 1.00$, $n_{\text{glass}} = 1.50,$ and $n_{\text{coating}} = 1.46$.

## Shifting of Image and Object

When viewing an object from different media the refraction of light often creates a shifting of the image. Because of this, for example, when viewing a pencil through a thick glass slab there is a shift of the image.

This shift of image is given by

$\mu =\frac {(\text{real height})}{(\text{apparent height})}.$

This is useful when viewing an object from water where the light bends only once.

However, when viewing an object from a glass slab, the light rays bend twice: once when going to the glass and again when coming out from the glass. So here we use

$\triangle x=t\left( 1-\frac { 1 }{ \mu } \right),$

where $\triangle x$ is the shift in the image and $t$ is the thickness of the glass slab.

## Additional Problems

**Q1:** Prove that only two real images can be formed when the distance between the screen and the image is $4f$ in the case of a convex lens.

- Details and Assumptions: The image should only form on the screen and the lens can be placed anywhere in between the object and the screen, so that no virtual image should form.
- Hint: Use the information $v+u=4f$ to form a quadratic equation. A quadratic equation has 2 roots, which implies it will have two such positions of $u$ and $v.$

**Q2:** The speed of an object travelling towards a concave mirror is $4 \text{ m/s}$ and the image distance is $10\text{ cm},$ while the object distance is $20\text{ cm}$. Find the speed of the image.

- Speed =$1\text{ m/s}$.
- Hint: Use differentiation for this question.

**Cite as:**Geometrical Optics.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/geometrical-optics/