Question

derive the lens formula with the help of a ray diagram​

Answer 1

Explanation:

In optics, the relationship between the distance of an image (v), the distance of an object (u) and the focal length (f) of the lens is given by the formula known as Lens formula. Lens formula is applicable for convex as well as concave lenses. These lenses have negligible thickness. The formula is as follows:

1v−1u=1f

Lens Formula Derivation

Consider a convex lens with object AB kept on the principal axis. Two rays are considered such that one ray is parallel to the principal axis and after reflection, it passes through the focus. The second ray is towards the optical center such that it passes undeviated.

A’ is the point where the two rays intersect and also the image formed by point A. Point B image is obtained on the principal axis as the point B is on the principal axis.

As the object is perpendicular to the principal axis, even the object is perpendicular to the principal axis. To get the location of the image formed by point B, we need to draw a perpendicular from point A’ to the principal axis. Following are the things obtained after drawing the figure:

ABA′B′=BOOB′ (from similar ΔABO and ΔA’B’O) (equ. 1)

POOF=A′B′FB′ (from similar ΔPOF and ΔFB’A’)

∴ABOF=A′B′FB′ (from figure PO = AB)

ABA′B′=OFFB′ (equ. 2)

BOOB′=OFFB′ (from equ. 1 and equ. 2)

∴BOOB′=OFOB′−OF −u+v=+fv−f (substituting optical distance values)

∴−uv+uf=fv −1f+1v=1u (dividing by u,v and f on both the sides)

1v=1u+1f 1f=1v−1u

Therefore, this is known as Lens formula

Answer 2

Lens formula:

It describes the relationship between

• The distance of an image, which is represented as v,
• The distance of an object, which is represented as u, and
• The focal length, which is represented as f of the lens

The lens formula works for both convex and concave lenses. The thickness of these lenses is minimal. The following is the formula:  

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

• Let us take a convex lens, which has a focal length of 'f' with an optical centre of O.
• F represent the primary focus.
• f represents the focal length.

The figure represents, ΔABO is similar to  ΔA'B'O  

$\frac{A'B'}{AB}$ = $\frac{OB'}{OB}$      eq 1

Similarly, ΔA'B'F is similar to  ΔOCF  

$\frac{A'B'}{OC}$ = $\frac{FB'}{OF}$

OC = AB

Therefore, $\frac{A'B'}{AB}$ = $\frac{FB'}{OF}$   eq 2

from eq 1 and 2,$\frac{ OB'}{ OB}$ = $\frac{FB'}{OF}$ = $\frac{OB'-OF}{OF }$

By substituting sign convention,

OB = -u

OB' = v

OF = f

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

uv = uf-vf

vf = -uv+uf

By uvf, divide both sides,

$\frac{uv}{uvf} = \frac{uf}{uvf} - \frac{vf}{uvf}$

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

It is known as the lens formula.