# Lenses

Power of lens:

The power of a lens is its ability to bend light – the greater the power the greater the refraction of light. Power of a lens is measured in Dioptres (D).

Uses of a lens:

The lens are used in optical instrument for various purposes and in spectacles to correct deflects in vision, etc.

Angle of deviation is independent with angle of incidence in case of small angled prism: Figure; Deviation produced by a small angle prism

Consider a prism ABC of small angle A about 10 – 12 degree as shown in the figure

When a ray of light PQ making angle of incidence I incident on face AB, it is refraction in first face, r2 is angle of incidence on RS. Here e is angle of emergence, r1 is angle of refraction in first face, r2 is angle of incidence on second face and δ is the angle of deviation. The angle of deviation of the ray of PQ is given by

δ = (I - r1) + (e - r2)

= (I + e) – (r1 + r2)……….1

At face AB,

µ = $\frac{{\sin {\rm{i}}}}{{\sin {\rm{r}}1}}$….2

Since angle of incident is small then,

Equation 2 become

µ = $\frac{{\rm{i}}}{{{\rm{r}}1}}$

Or, I = μ r1

Similarly at face AC,

e=μ r2

Substituting value of I and e in 1 we get

$= \left( {\mu - 1} \right)\left( {{\rm{r}}1 + {\rm{r}}2} \right)$

= (μ -1) A, where r1 + r2 + A

δ = (μ -1) A

That is angle of deviation of a ray is independent with angle of incidence in a small angled prism.

The lens formula for a convex lens; Figure;  convex lens forming a real image

Δ’s ABC and A’B’C’ are similar so their corresponding sides are proportional.

∴$\frac{{{\rm{AB}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CA}}}}{{{\rm{CA'}}}} \ldots \ldots \ldots \ldots \ldots \ldots ..1$

Similarly,

Δ’s CDF and A’B’F’ are similar

∴${\rm{\: }}\frac{{{\rm{CD}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CF}}}}{{{\rm{FA'}}}}$

But CD = AB

∴$\frac{{{\rm{AB}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CF}}}}{{{\rm{FA'}}}}$…………….2

From equation 1 and 2

$\frac{{{\rm{CA}}}}{{{\rm{CA'}}}} = \frac{{{\rm{CF}}}}{{{\rm{FA'}}}}$

$\frac{{{\rm{CA}}}}{{{\rm{CA'}}}} = \frac{{{\rm{CF}}}}{{{\rm{CA'}} - {\rm{CF}}}}$

Or,$\frac{{\rm{u}}}{{\rm{v}}} = \frac{{\rm{f}}}{{{\rm{v}} - {\rm{f'}}}}$

Where CA=u is object distance

CA’=V is image distance

CF = f is focal length

Dividing both sides we get the formula

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

This is lens formula.

Lens formula for a concave lens; Figure: concave lens forming a virtual image

Δ’s ABC and A’B’C’ are similar so their corresponding sides are proportional.

∴$\frac{{{\rm{AB}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CA}}}}{{{\rm{CA'}}}} \ldots \ldots \ldots \ldots \ldots \ldots ..1$

Similarly,

Δ’s CDF and A’B’F’ are similar

∴ $\frac{{{\rm{CD}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CF}}}}{{{\rm{FA'}}}}$

But CD = AB

∴$\frac{{{\rm{AB}}}}{{{\rm{A'B'}}}} = \frac{{{\rm{CF}}}}{{{\rm{FA'}}}}$…………….2

From equation 1 and 2

$\frac{{{\rm{CA}}}}{{{\rm{CA'\: }}}} = \frac{{{\rm{CF}}}}{{{\rm{A'F}}}}{\rm{\: }}$

$\frac{{{\rm{CA}}}}{{{\rm{CA'}}}} = \frac{{{\rm{CF}}}}{{{\rm{CA'}} - {\rm{CA'}}}}$

Or,$\frac{{\rm{u}}}{{ - {\rm{v}}}} = \frac{{ - {\rm{f}}}}{{ - {\rm{f}} + {\rm{v\: '}}}}$

Where CA=u is object distance

CA’=-V is image distance

CF = f is focal length

Dividing both sides by uv, we get the formula

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

This is lens formula.

The formula for the focal length of two thin lenses in contact: Figure: focal length of two thin lenses in contact

From lens formula,

$\frac{1}{{\rm{f}}} - \frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}$……………1

When lens L2 is placed in contact, the image at I’ acts virtual object for lens L2 since the converging beam is incident on L2 as shown in the figure

For lens L2

Object distance = -v’

Image distance = v

So,
$\frac{1}{{{{\rm{f}}_2}}} - \frac{1}{{ - {\rm{v'}}}} + \frac{1}{{\rm{v}}}$

Or,  ${\rm{\: }}\frac{1}{{{{\rm{f}}_2}}} - \frac{1}{{\rm{v}}} + \frac{1}{{{\rm{v'}}}}{\rm{\: \: \: \: \: \: }}$…………..2

$\frac{1}{{\rm{f}}} - \frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}{\rm{\: }}$+ $\frac{1}{{{{\rm{f}}_2}}} - \frac{1}{{\rm{v}}} + \frac{1}{{{\rm{v'}}}}{\rm{\: \: \: \: \: \: }}$

$\frac{1}{{{{\rm{f}}_1}}} + \frac{1}{{{{\rm{F}}_2}}}$= $\frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}$………….3

If F is the combined focal length of two thin lenses placed in contact having object distance u and image distance v then,

$\frac{1}{{{{\rm{F}}_2}}}$= $\frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}$……….4

From 3 and 4

$\frac{1}{{{{\rm{f}}_1}}} + \frac{1}{{{{\rm{F}}_2}}}$= $\frac{1}{{\rm{F}}}$

This formula for the focal length of two thin lenses in contact

An object moved from infinity to the convex lens; Figure: an object moved from infinity to the convex lens

When the object moves in the infinity, the image formed at the focus as shown in the figure

m≈0

Condition does a concave lens form a real image;

When converging beam of light is incident on a concave lens, the rays of light are diverged and meet at a point on the principle axis. In this condition a concave lens form a real image.

Virtual image formed by a lens:

virtual image is formed at the position where the paths of the principal rays cross when projected backward from their paths beyond the lens

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