Derivation of Lens Formula - Definition, FAQs

What Is a Lens?

The lens is a transparent object that may be made of glass or plastic, and it bends (or refracts) light rays to create images. These are lenses with curved surfaces that primarily use the principle of refraction. Lenses are found in eyeglasses, cameras, microscopes, and magnifying glasses.

Lenses are based on two basic types:

• Convex lens (converging lens): Thicker in the middle, bending light rays to bring them together, and can create real or virtual images.
• Concave lens (diverging lens): Thinner in the middle, diverging light rays, and always forming a virtual image.

The lens is divided into two types depending on how the light rays behave when they travel through the lens. These are further divided into the following types.

What is the Lens Formula?

The lens equation or lens formula is an equation that links the focal length, image distance, and object distance.

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

where,

v is the distance of the image from the optical centre of the lens, u is the distance of the object from the optical centre of the lens and f is the focal length of the lens.

Lens Formula Derivation

Let AB represent an object at a distance greater than the focal length f of the convex lens that is located at right angles to the primary axis. The image A' B' is generated between O and $F_1$ on the same side as the item, and it is virtual and erect.

• Focul length = $OF_1$ = f
• Object distance = OA =u
• Image distance = $OA_1$ = v

$\triangle O A B$ and $\triangle$ O A' B' are similar

$\because \angle B A O=\angle B^1 A^1 O=90^{\circ}$, vertex is common for both the triangles so $\angle A O B=\angle A^1 O B^1,$
$ \therefore \angle A B O=\angle A^1 B^1 O$

$\frac{A' B'}{A B}=\frac{O A'}{O A}---(1)$

$\triangle O C F_1$ and $\triangle F_1 A' B' $ are similar

$\frac{A' B'}{O C}=\frac{A' F_1}{O F_1}$

But from the ray diagram, we see that OC = AB

$\begin{aligned} & \frac{A' B'}{A B}=\frac{A' F_1}{O F_1}=\frac{O F_1-O A'}{O F_1} \\ & \frac{A' B'}{A B}=\frac{O F_1-O A'}{O F_1}---(2)\end{aligned}$

From equation (1) and equation (2), we get

$\begin{aligned} & \frac{O A'}{O A}=\frac{O F_1-O A'}{O F_1} \\ & \frac{-v}{-u}=\frac{-f--v}{-f} \\ & \frac{v}{u}=\frac{-f+v}{-f} \\ & -v f=-u f+u v\end{aligned}$

Dividing throughout by uvf

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

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

Magnification of the lens

If $h$ is the height of the object and $h^{\prime}$ is the height of its image, then

$m=\frac{h^{\prime}}{h}=\frac{v}{u}$ ;

$m$ is called magnification.

Its magnitude indicates how many times is the size of the image as compared to that of the object and its sign indicates whether it is erect or inverted. Positive sign of $m$ indicates an erect image and negative sign indicates an inverted image.

Applications of Lens Formula

The lens formula helps us understand how and where an image is formed by a lens. Its main applications are:
1. To find the image position: By using the object distance ( $u$ ) and focal length ( $f$ ), we can calculate the image distance ( $v$ ).
2. To know the nature of the image: After finding $v$, we can check whether the image is real or virtual, inverted or erect, magnified or diminished.
3. To determine the focal length: If $u$ and $v$ are known from an experiment, the focal length $f$ can be easily calculate
4. To design optical instruments: Devices like cameras, microscopes, telescopes, and spectacles use the lens formula for proper image formation.
5. To correct vision problems: The formula helps in choosing the correct lens power for myopia and hypermetropia.

Difference Between Lens Formula and Mirror Formula

Solved Examples on Lens Formula

Example 1: A convex lens has a focal length of 10 cm . Where should the object be placed if the virtual image is to be 30 cm from the lens?
Solution :
Here $v=-30 \mathrm{~cm}$

$
\begin{aligned}
& f=+10 \mathrm{~cm} \\
& u=?
\end{aligned}
$

So from lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
We get,

$
\begin{aligned}
& \quad \frac{1}{-30}-\frac{1}{u}=\frac{1}{10} \\
& \text { or } \frac{1}{u}=\frac{-1}{10}-\frac{1}{30} \\
& \Rightarrow \frac{1}{u}=-\frac{4}{30}
\end{aligned}
$

or $u=-7.5 \mathrm{~cm}$
So, the object must be placed in front of the lens at a distance of 7.5 cm from it.

Example 2: A convex lens of focal length $f$ produces a real image, $x$ times the size of an object, then find the distance of the object from the lens.

Solution :
For real image,

$
m=\frac{v}{u}=-x \quad \text { (given) }
$

So, $v=-x u$
From lens formula

$
\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
\Rightarrow & \frac{1}{-x u}-\frac{1}{u}=\frac{1}{f} \\
\Rightarrow & u=-\frac{(x+1)}{x} \cdot f
\end{aligned}
$

So, object should be placed at a distance $\frac{(x+1)}{x} \cdot f$ from the lens.

Example 3: An object of height 12 cm is kept 1.2 m far from a convex lens of focal length 80 cm . Find the distance of image from the lens, its height and nature of the image.
Solution :
Here

$
\begin{aligned}
& u=-1.2 \mathrm{~m}=-120 \mathrm{~cm} \\
& f=+80 \mathrm{~cm} \\
& v=? \\
& h=12 \mathrm{~cm} \\
& h^{\prime}=?
\end{aligned}
$

So from lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We get,

$
\begin{aligned}
& \frac{1}{80}=\frac{1}{v}-\frac{1}{-120} \\
\therefore & \frac{1}{v}=\frac{1}{80}-\frac{1}{120} \\
\therefore & \frac{1}{v}=\frac{3-2}{240} \\
\therefore & v=240 \mathrm{~cm}=2.4 \mathrm{~m} \\
\text { As } & \frac{h^{\prime}}{h}=\frac{v}{u} \\
\therefore & h^{\prime}=\frac{h v}{u} \\
\therefore & h^{\prime}=\frac{12 \times 240}{-120}=-24 \mathrm{~cm}
\end{aligned}
$

As $h^{\prime}$ is negative so the image is real and inverted.
So, the distance of image from the lens is 2.4 m and its height is 24 cm .