# Lenses

What are **lenses**? They are transparent surfaces which refract the light passing through them.

A key property of lenses lies in their curvature. The curvature of lenses allows them to be used for a variety of purposes and to produce different types of images. Lenses can curve in one of two directions, either inward or outward, and thus lenses are broadly categorized into two types: the bi-concave lens (or simply concave lens) and the bi-convex lens (also called a convex lens).

Due to the differences in curvature, image formation is different in concave and convex lenses. These differences give rise to the lens maker's formula, which gives us the relationship between various components of the lens. All types of lenses obey the laws of refraction.

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## Guiding Rays

The following are the laws of refraction:

**1\(^\text{st}\) law:** The angle of incidence, angle of refraction, and the normal at the point of incidence lie in the same plane.

**2\(^\text{nd}\) law:** The sine of the angle of incidence divided by the sine of the angle of refraction is always constant and is equal to the refractive index of the lens. This law is also referred to as the Snell's law.

**Guiding rays:** Guiding rays are guidelines that help determine the behavior of light as it passes through a lens, and assist with the construction of ray diagrams.

**Ray passing through optical center:**

In the above images, we see that the ray of light passes through the \(\ce{O}\) in both types of lenses, and does not deviate. It maintains its rectilinear path, i.e. it does not refract.

**Ray parallel to principal axis:**

If a ray of light is parallel to the principal axis in the case of a convex lens, it refracts and passes through \(\ce{F}_2\).

If a ray of light is parallel to the principal axis in the case of a concave lens, it refracts and the ray appears to pass through \(\ce{F}_1\) when produced backwards.

**Ray passing through focus:**

Any ray which passes through the focus of the lens comes out parallel to the principal axis after refraction. In the case of a concave lens, the light ray appears to pass through the focus.

## Concave Lenses

Concave lenses are curved inward and they diverge the light rays that hit them. Hence, they are also called **diverging or divergent lenses**. There are many uses of this type of lens, which we shall discuss later on.
We saw the image of a concave lens earlier, and we can observe that the lens is curved inward. When the parallel light rays are incident on them, they refract the rays in a diverging manner.

As far as concave lenses are concerned, there can only be two cases in which an object is placed in front of it.

**Object at \(\infty\):**

When the object is placed at \(\infty\), the image formed is not formed by the actual intersection of rays, but imaginary rays, i.e. the image is virtual and erect. The image is formed at \(\ce{F1}\), having the size of a point.

**Object anywhere on the principal axis:**

In this case, the object is placed anywhere on the principal axis, which results in the formation of an image only between \(\ce{F1}\) and \(\ce{O}\). The image formed is also a virtual image, erect, and diminished in size.

**A brief summary table:**

\[\begin{array} {|c|c|c|c|} \hline \text{Position of the Object} & \text{Position of the Image} & \text{Size of the Image} & \text{Nature of the image}\\ \hline \text{at } \infty & \text{at F}_1 & \text{point-sized} & \text{virtual and erect}\\ \hline \text{anywhere on the principal axis} & \text{between F}_1 \ce{and O} & \text{diminished} & \text{virtual and erect}\\ \hline \end{array}\]

**Applications of concave lenses:**

Used in eye glasses to correct myopia

Used in contact lenses

Used in flashlights

Used in peepholes

Used in binoculars and telescopes

Also used in photography

## Convex Lenses

As mentioned earlier, convex lenses are curved outward and are called **converging lenses**. For the concave lens, image formation is in two cases only. However, for convex lenses, the image formation can be sub-divided into six categories.

**Object at \(\infty\):**

When the object is kept at infinity, we assume that the incident rays are parallel. Thus, we observe that the image is formed at \(\ce{F2}\), it is highly diminished, and it is real and inverted.

**Object beyond \(2\ce{F1}\):**

If we place the object at \(2\ce{F1}\), we will see that the image is formed between \(\ce{F2}\) and \(2\ce{F2}\), the image is smaller, and it is real and inverted.

**Object at \(2\ce{F1}\):**

If we move our object to \(2\ce{F1}\), we will observe that the image is formed exactly on \(2\ce{F2}\), it is of the same size as the object, and again, real and inverted.

**Object between \(2\ce{F1}\) and \(\ce{F1}\):**

Next, if we place our object between \(2\ce{F1}\) and \(\ce{F1}\), the image is formed beyond \(2\ce{F2}\), but this time it is magnified, and real and inverted.

**Object at \(\ce{F1}\):**

If we move closer to the focus of the lens, we will observe that the rays emerging are parallel, and this means that the image is formed at \(\infty.\)

**Object between \(\ce{F1}\) and \(\ce{O}\):**

Lastly, if we keep the object between \(\ce{F1}\) and \(\ce{O}\), we will observe that the image is formed on the same side, it is magnified, and virtual and erect.

**A brief tabular summary:**

\[\begin{array} {|c|c|c|c|} \hline \text{Position of the Object} & \text{Position of the Image} & \text{Size of the Image} & \text{Nature of the image}\\ \hline \text{At infinity} & \text{At Focus} & \text{Highly Diminished} & \text{Real and Inverted}\\ \hline \text{Beyon 2F}_1 & \text{Between }\ce{2F2}\text{ and }\ce{F2} & \text{Diminished} & \text{Real and Inverted}\\ \hline \text{At 2F}_1 & \text{At 2F}_2 & \text{Same size} & \text{Real and Inverted}\\ \hline \text{Between }\ce{2F1}\text{ and }\ce{F1} & \text{Beyond 2F}_2 & \text{Magnified} & \text{Real and Inverted}\\ \hline \text{At F}_1 & \text{At infinity} & \text{Highly magnified} & \text{Real and Inverted}\\ \hline \text{Between }\ce{F1}\text{ and }\ce{O} & \text{On the same side as the object} & \text{Magnified} & \text{Virtual and Erect}\\ \hline \end{array}\]

**Applications of concave lenses:**

Camera lenses are convex.

They are used in contact lenses.

Magnifying glasses are made of convex lenses.

They are used in binoculars and telescopes.

## Sign Convention

The **sign convention** gives us specific rules and regulations that allow for proper construction of the image distance, object distance, focal length, and heights. The sign convention says:

Heights above the principal axis are positive and those below the principal axis are negative.

Distances to the right of the optical center are positive and those to the left of the optical center are negative.

The focal length of a concave lens is negative and that of a convex lens is positive.

The object is always placed to the left of the lens.

The image below summarizes all the sign conventions in one place:

## Thin Lens Formula

The **thin lens formula** says that: \[\dfrac 1f = \dfrac 1v + \dfrac 1u.\]

We can prove the above result using simple trigonometry.

By referring to the convex lens diagram above, we can note the following:

\[\text{OA} = u,\quad \text{OF}= f,\quad \text{OA'}= v,\quad \text{AB}=h_o,\quad \text{AB}=h_i.\]

The distance between two points is always positive, so sign is not needed.

In \(\Delta\text{OAB}\) we can see that

\[\tan\theta = \dfrac{\text{AB}}{\text{OA}}=\dfrac{h_o}{u}.\]

where \(\theta\) represents the angle AOB. From \(\Delta\text{OA'B'}\) we can see that

\[\tan\theta = \dfrac{\text{A'B'}}{\text{OA'}}=\dfrac{h_i}{v}.\]

AOB and A'OB' are vertically opposite angles, so:

\[\dfrac{h_o}{-u}=\dfrac{h_i}{v}\implies \dfrac{h_o}{h_i}=\dfrac{u}{v}.\]

Now, we see that

\[\begin{array} &\tan\alpha = \dfrac{h_o}{f} &\text{and} &\tan\alpha &= \dfrac{h_i}{v-f}.\end{array}\]

where \(\alpha\) represents the angle CF

2O CF2O and B'F_2A' are vertically opposite angles, so:\[\dfrac{h_o}{f} = \dfrac{h_i}{v-f}.\]

which leads to

\[\dfrac{h_o}{h_i} = \dfrac{f}{v-f}.\]

Equating both expressions, and with some algebraic manipulation. \[\dfrac{u}{v} = \dfrac{f}{v-f}.\]

\[f = \dfrac{uv}{u+v}.\]

\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}.\]

Note: This derivation ignores the sign convention, so when this equation is used it will only accept distances as input.

If the sign convention is considered, then:\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}.\]

**Magnification of a lens:**

The image formed by a lens can be larger or smaller than the object, or it can also be of the same size. If \(h_o\) and \(h_i\) denote object-height and image-height respectively, the ratio \(\dfrac{h_i}{h_o}\) is defined as the magnification of the lens. This ratio is also equal to the ratio \(\dfrac{v}{u}\):

\[\text{magnification}=\dfrac vu=\dfrac {h_i}{h_o}.\]

If \(m\) is positive, then \(h_e\) is also positive. This denotes that the image and the object are on the same side of the principal axis (erect image). If \(m\) is negative, they are on the opposite sides of the principal axis (inverted image).

For a convex lens, the image can be erect or inverted, depending upon the position of the object on the principal axis. When the image is erect (virtual), \(m\) is positive, and when the image is inverted (real), \(m\) is negative.

For a *concave lens*, \(m\) is always positive.

A \(\ce{2 cm}\) long pin is placed perpendicular to the principal axis of a

convexlens of focal length \(\ce{12 cm}\). The distance of the pin from the lens is \(\ce{15 cm}\). Find the size of the image.

We have \(u=-15\text{ cm}\) and \(f=+12\text{ cm} \ \ \ \text{(by sign convention)}\). Then \[\begin{align} \dfrac{1}{v}-\dfrac{1}{u}&=\dfrac{1}{f}\\ \dfrac{1}{f}+\dfrac{1}{u}&=\dfrac{1}{v}\\ \dfrac{1}{12\text{ cm}}-\dfrac{1}{15\text{ cm}}&=\dfrac{1}{v}\\ \implies v&=60\text{ cm}. \end{align}\]

Since we get \(v=+60cm\), the image is to the right of the lens. Now, by the magnification formula, \(m=\dfrac{v}{u}=\dfrac{60\text{ cm}}{-15\text{ cm}}=-4,\) which is equivalent to \[\begin{align} \dfrac{h_i}{h_0}&=-4\\ {h}_i&=-4{h}_o\\ &=-4\times2\text{ cm}\\ &=-8\text{ cm}. \end{align}\]

The image of the pin is \(8\text{ cm}\) high. The minus sign signifies that the image is formed below the principal axis, which also means that the image is inverted. Hence, the image is also real. \(_\square\)

The focal length of a convex lens is \(0.4\text{ m}\). At what distance should an object be placed so that the image formed is three times the size of the object?

We know that the magnification of a lens is given by \(m=\dfrac{h_o}{h_i}=\dfrac{-v}u\), and here it is equal to \(3\). Thus

\[\dfrac{-v}u=3 \implies v = -3u.\]

Then, substituting \(v=-3u\) into the lens formula, we get

\[\begin{align} \dfrac 1f &= \dfrac{-1}{3u} - \dfrac{1}{u}\\ \dfrac 1{0.4} &= \dfrac{-1}u - \dfrac{3}{3u}\\ \dfrac 52 &= \dfrac{-4}{3u}\\ \Rightarrow u&=\dfrac{-8}{15}\\&\approx -0.5334. \end{align}\]

Therefore, if we place our object \(0.5334\text{ m}\) before the lens, the image will be magnified three times. \(_\square\)

A concave lens has a focal length of \(0.5\text{ m}\). If the object is placed \(0.75\text{ m}\) from the lens, find where the image will be formed.

We are given \(f=0.5\text{ m}\) and \(u=\text{ m}\). Substituting these values in the lens formula, we arrive at

\[\begin{align} \dfrac 1f &= \dfrac 1v - \dfrac 1u\\ \dfrac 1{0.5} &= \dfrac 1v - \dfrac 1{0.75}\\ \dfrac 1{0.5}+1{0.75} &= \dfrac 1v. \end{align}\]

Thus, we get \(v=\dfrac{3}{10}=0.3\text{ m}\). \(_\square\)

## Power of a Lens

The

powerof a lens is defined to be the reciprocal of itsfocal length:\[\text{Power}=\dfrac{1}{f}.\ _\square\]

The focal length of a lens is measured in meters, and the power of that lens will be \(\text{meter}^{-1}\), and this unit \(\text{meter}^{-1}\) is also called as **dioptre** and is represented as \(\text{D}\).

The focal length of a lens is \(25\text{ cm}\). Find the power of the lens.

Converting the focal length to meters gives \(f=0.25 \text{ m}\). Then

\[\begin{align} P&=\dfrac{1}{0.25}\text{ m}^{-1}\\ &=4 \text{ m}^{-1}\\ &=4\text{D}. \ _\square \end{align}\]

From the example above, we can decipher that the lens with \(f=0.25\text{ m}\) is a *convex lens* since we get the power of the lens as **positive**. If the same had been **negative**, the lens would be a *concave* lens.

Hence, we can conclude that the focal length of a *convex lens* is always taken to be **positive**, whereas the focal length of a *concave lens* is always taken to be **negative**, which also means that the power of a *convex lens* is always **positive** and the power of a *concave lens* is always **negative**.

**Lenses in contact:**

When two or more lenses are kept in contact with each other such that they have the *same* principal axis, the combination can be treated as a single lens.
Let's take a combination of many lenses in contact with focal lengths \(f_1\), \(f_2\), \(f_3\), and so on, respectively.
Now the focal length \(F\) of the equivalent single lens would be given by the formula

\[\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_1}+\dfrac{1}{f_3}+\cdots.\]

Now for two lenses in contact, the equation becomes \(F=\dfrac{f_1f_2}{f_1+f_2}.\).

Hence, we can now easily arrive at a formula for the combination of power of the collective lenses in contact. The formula for calculating power of lenses in contact is

\[P=P_1+P_2+P_3+\cdots.\]

## References

[1] Image from https://en.m.wikipedia.org/wiki/Lens*(optics)#/media/File%3AConcave*lens.jpg under the creative commons license for reuse and modification.

[2] Image from https://en.m.wikipedia.org/wiki/Lens*(optics)#/media/File%3ALarge*convex_lens.jpg under the creative commons license for reuse and modification.

[3] Light reflection and refraction http://www.ncert.nic.in/ncerts/l/jesc110.pdf. retrieved 17:59, March 25, 2016.