Thin Lens Formula

The thin lens formula is a key idea in optics that shows how a lens’s focal length is related to the distances of the object and the image from the lens. A thin lens is one where its thickness is so small that it can be ignored compared to the object and image distances. This formula helps us figure out where the image will form and what kind of image it will be—whether it’s real or virtual, bigger or smaller than the object. The formula is derived using the concept of similar triangles and how light bends when passing through curved surfaces. Knowing this formula is important for understanding how both converging and diverging lenses work, and it’s used in many everyday devices like cameras, microscopes, and eyeglasses.

1.0Statement of Thin Lens Formula

• This equation relates the object distance u, image distance v, and focal length f for a spherical lens.

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

• The formula holds true for convex and concave lenses, regardless of whether the image is real or virtual.

2.0Assumption To Derive Thin Lens Formula

• The lens is thin.
• The aperture of the lens is small.
• Both the incident and refracted rays make slight angles with the principal axis.
• The object is a small object placed on the principal axis.

3.0Derivation of Thin Lens Formula

Convex Lens when it forms a real image

• Object AB placed perpendicular to the principal axis of a thin convex lens between its F’and C’.
• A real,inverted and magnified image A’B’ is formed beyond C on the other side of the lens.

$\begin{array}{rl}& △A^{\prime }{B}^{\prime }O\text{ and }△ABO\text{ are similar }\\ & \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{O{B}^{\prime }}{BO}\end{array}$

Also, $△A^{\prime }{B}^{\prime }F\text{ and }△MOF\text{ are similar }$

$\begin{array}{rl}& \frac{A^{\prime }{B}^{\prime }}{MO}=\frac{F{B}^{\prime }}{OF}\Rightarrow \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{F{B}^{\prime }}{OF}(\therefore MO=AB)\\ & \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{F{B}^{\prime }}{OF}\\ & \frac{O{B}^{\prime }}{BO}=\frac{F{B}^{\prime }}{OF}=\frac{O{B}^{\prime }-OF}{OF}\\ & BO=-u,O{B}^{\prime }=+v,OF=+f\end{array}$

$\begin{array}{rl}& \begin{array}{rl}& \frac{v}{-u}=\frac{v-f}{f}\Rightarrow vf=-uv+uf\\ & vf=-uv+uf\end{array}\\ & \text{ Dividing both sides by uvf }\\ & \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\end{array}$

4.0Linear Magnification

• The ratio of the image size to the object size formed by the lens is called magnification, denoted by m.

$m=\frac{h_2}{h_1}=\frac{v}{u}$

Linear Magnification in terms of u and f

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

Multiply both sides by u

$\begin{array}{rl}& \frac{u}{f}=\frac{u}{v}-1\Rightarrow \frac{u}{v}=1+\frac{u}{f}\\ & m=\frac{v}{u}=\frac{f}{f+u}\end{array}$

Linear Magnification in terms of v and f

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

Multiply both sides by v

$\begin{array}{rl}& \frac{v}{f}=1-\frac{v}{u}\Rightarrow m=\frac{v}{u}=1-\frac{v}{f}\\ & m=\frac{v}{u}=\frac{f-v}{f}\\ & m=\frac{v}{u}=\frac{f}{f+u}=\frac{f-v}{f}\end{array}$

5.0Image Position, Size, and Nature in Lenses

(A) Convex lens:

• Suppose a real object is placed at a distance x from a convex lens of focal length $f_0$. The lens formulae may be modified as

$\begin{array}{rl}& \frac{1}{v}-\frac{1}{-x}=\frac{1}{f_o}\\ & v=\frac{x{f}_o}{x-f_o}\end{array}$

Lateral Magnification,

$m=\frac{v}{u}=\frac{v}{-x}=-\frac{f_o}{x-f_o}$

Note:

(1) A convex lens will form a real image for a real object when the object is placed beyond focus $\left(x>f_o\right)$.

(2) When the object comes within focus i.e., x<f_o, then a virtual image is formed for the real object.

(3) The real image formed is always inverted while the virtual image is always erect.

(B) Concave lens

Suppose a real object is placed at a distance x in front of a concave lens of focal length $f_0$. Then the lens formulae can be modified as

$\frac{1}{v}-\frac{1}{-x}=\frac{1}{f_o}\Rightarrow v=\frac{-x{f}_o}{x+f_o}$

Transverse Magnification,

$m=\frac{v}{u}=\frac{v}{-x}=\frac{f_o}{x+f_o}$

Note:

(1) A concave lens always forms a virtual image for a real object.

(2) A concave lens always forms an erect, smaller image compared to the object.

(3) A concave lens can form a real image if the object is virtual.

6.0Displacement Method for Focal Length of Convex Lens

• Place an object of height H and a screen at distance D apart (see figure). Move a converging lens between them. When a sharp image forms on the screen with the lens at distance a from the object, the lens formula gives:

 $\frac{1}{D-a}-\frac{1}{(-a)}=\frac{1}{f}\text{ or }{a}^2-Da+fD=0$

This is a quadratic equation and hence two values of ‘a’ are possible.Thus, a_1 and a_2  are the roots of the equation, and based on the properties of quadratic equations

$\therefore a_1+a_2=D\Rightarrow a_1{a}_2=fD$

Also $\left(a_1-a_2\right)=\sqrt{{\left(a_1+a_2\right)}^2-4a_1{a}_2}=\sqrt{D^2-4fD}=d\text{ (Suppose) }$

d′ denotes the physical separation between the lens's two positions.

 Focal length of the lens as a function of D and d.

$\begin{array}{rl}& a_1-a_2=\sqrt{{\left(a_1+a_2\right)}^2-4a_1{a}_2}\\ & \sqrt{D^2-4fD}=d\Rightarrow f=\frac{D^2-d^2}{4D}\end{array}$

For d=0, the two positions overlap.

$f=\frac{D^2}{4D}\Rightarrow D=4f$

Roots of the equation $a^2-Da+fD=0$, become imaginary if

$\begin{array}{rl}& b^2-4ac<0=D^2-4fD<0\\ & =D(D-4f)<0\end{array}$

To ensure that a has real values in the equation, $a^2-Da+fD=0$

$\begin{array}{rl}& b^2-4ac\ge 0=D^2-4fD\ge 0\\ & D\ge 4f\Rightarrow D_{\min }=4f\end{array}$

Lateral Magnification in Displacement Method:

Let $m_1$ and $m_2$ represent the two magnifications corresponding to the two positions in the displacement method.

Assume $m_1$ and $m_2$ are the magnifications obtained in the two configurations of the displacement method.

$m_1=\frac{v_1}{u_1}=\frac{\left(D-a_1\right)}{-a_1}\text{ and }{m}_2=\frac{v_2}{u_2}=\frac{a_1}{-\left(D-a_1\right)}$

So $m_1{m}_2=\frac{\left(D-a_1\right)}{-a_1}\times \frac{a_1}{-\left(D-a_1\right)}=1$

If the image lengths in the two cases are  $h_1$ and $h_2$​ respectively.

Then $m_1=-\frac{h_1}{H}{m}_2=-\frac{h_2}{H}\Rightarrow m_1{m}_2=1$

$\therefore \frac{h_1{h}_2}{H^2}=1\Rightarrow h_1{h}_2=H^2\Rightarrow H=\sqrt{h_1{h}_2}$

7.0Solved Examples on Thin Lens Formula

Q-1.The figure shows a point object placed in front of a converging lens. Determine the position and nature of the final image formed.

Solution:

$\begin{array}{rl}& \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ & \frac{1}{v}-\frac{1}{-15}=\frac{1}{10}\\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}\Rightarrow v=+30\mathrm{cm}\end{array}$

Q-2.The figure shows two converging lenses with incident rays parallel to the principal axis. What should be the value of d so that the emergent rays are also parallel to the principal axis?"

Solution:For the final rays to be parallel, the focal point of lens $L_1$ must coincide with the focal point on the object side of lens $L_2$.

$d=10+20=30\mathrm{cm}$

Here the diameter of the ray beam becomes wider.

Q-3.Determine the position of the final image formed.

Solution: For Lens, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{-15}=\frac{1}{10}\Rightarrow v=+30\mathrm{cm}$

Therefore, it serves as the object for the mirror.  $u=-15\mathrm{cm}$

$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}\Rightarrow v=-30\mathrm{cm}$

For the second time, it passes through the lens again $u=-15\mathrm{cm}$

$\frac{1}{v}-\frac{1}{-15}=\frac{1}{10}\Rightarrow v=+30\mathrm{cm}$

Therefore, the final image is formed 30 cm to the left of the lens."

Q-4.What should be the value of d such that the image is formed at the same position as the object?

Solution: For Lens , $\frac{1}{v}-\frac{1}{-15}=\frac{1}{10}\Rightarrow v=+30\mathrm{cm}$

Case 1 : If  d=30, the object for the mirror will be at the pole and its image will be formed there itself.

Case 2 : If the rays strike the mirror perpendicularly (i.e., along the normal), they will retrace their path, resulting in the image being formed at the location of the object itself.

$\therefore d=30-20=10\mathrm{cm}$

Q-5.An extended real object with a height of 2 cm is positioned perpendicular to the principal axis of a converging lens (convex lens) with a focal length of 20 cm. The object is placed 30 cm in front of the lens along the principal axis.

1. Calculate the lateral magnification produced by the lens.
2. Determine the height of the image formed.
3. If the object is moved 1 mm closer to the lens along the principal axis, calculate the resulting change in lateral magnification.

Solution:

1. Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ and $m=\frac{v}{u}$, we get $m=\frac{f}{f+u}$ …………(A)

$m=\frac{+20}{+20+(-30)}=\frac{+20}{-10}=-2$

–ve sign indicates that the image is inverted.

2. $\frac{h_2}{h_1}=m\Rightarrow h_2=m{h}_1=(-2)(2)=-4cm$
3. Differentiating (A) we get

$dm=-\frac{f}{(f+u{)}^2}du=\frac{-(20)}{(-10{)}^2}(0.1)=\frac{-2}{100}=-0.2$