# Geometric Optics - 5

__ Types of lenses__: Lenses are of following type:

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**Definition**•

__OPTICAL CENTRE__: Optical centre is a point for a given lens through which any ray passes undeviated.

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__PRINCIPAL AXIS__: The line joining the centres of curvature of the two bounding surfaces is called the principal axis.

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__FOCUS (F)__: A parallel beam of light, parallel and close to the principal axis of the lens after refraction passes through a fixed point on the principal axis (in case of convex lens) or appear to diverge from a position principal axis (in case of concave lens). This point is called the principal focus or focus of the lens. It is generally denoted by F.

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**RULES FOR IMAGE FORMATION**• A ray passing through the optical centre of the lens proceeds undeviation through the lens.

• A ray passing parallel to the principal axis after refraction through the lens passes or appear to pass through the focus. (By definition of the focus)

• A ray through the focus or directed towards the focus, after refraction from the lens, becomes parallel to the principal axis. (Principal of reverssibility of light)

__: What do you mean principal ‘reversesibility of light’ ?__

**Dumb Question**Ans: If the light is projected along the emergent ray (in opp. direction) then it will follow the path same as incident ray (in opp. direction)

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**LENS MAKERS FORMULA**If the material of lens has refractive index µ and refractive index of medium in which lens is kept is µ

_{0}. Then focal length of lens is given by:

R

_{1}and R

_{2}radius of the radii of the surfaces from which the light enters the lens and the one from which light leaves out respectively.

Why ??

This equation can be obtainedb applying thin lens formula twice.

Let image due to first surface be formed at v

_{1}from P

………………………………… (i)

for image formula at second surafce, this image will act as virtual object.

………………………………… (ii)

eliminating v

_{1}from (i) & (ii) yeilds.

__: The above proof used lens formula for this case linear magnification is m =__

**Lens Formula**__: How we can determine the focal length ‘f’ of a lens system ?__

**Dumb Question**Ans: ‘f’ can bge determined by using the basic definition of focal length. Assume that object is kept at infinity and for that calculate the final image position for the system. Then in this case f = v. As focus is a point where rays coming from infinity meet after refraction/reflection.

__: Above equation is derived only for a case where the lens has medium of same refractive index across its ends. How to priceed if we have different media.__

**Dumb Question**Ans: In that case start from basic formula and apply this equation with corresponding refractive indices, twice andthe obtain the image.

__: Will immersing in water a lens and mirror, change this focal lengths ?__

**Dumb Question**Ans: Focal length of mirror f

_{M}= and is independent of refractive index of medium hence it will remain unattached, however focal length of lens is depended upon medium, hence it will change.

__: Calculate the focal length of a biconvex lens in air. If the radii of its surfaces are 60 cm and 15 cm.__

**Illustration**Refarctive index of glass = 1.5.

Solution: Consider a light ray going through the lens as shown.

It strikes the convex side of 60 cm radius and concave side of 15 cm radius while coming act.

R

_{1}= + 60 cm R

_{2}= - 15 cm

or

f = 24 cm.

__: If an erquiconvex lens of focal length ‘f’ is cut into two equal halves (a) along its axis (b) perpendicular to its axis. then in which case ‘f’ cahnges. Find its value.__

**Dumb Question**Solution: (a)

neither radius of any of the surface charged nor the media changed, hence focal length will remain same i.e. ‘f’.

(b)

Here the raius of other has changed, hence focal length will change

f = while new f

_{1}=

f

_{1}=

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**POWER OF LENS**The power of lens is defined as P =

where ‘f’ must be expressed in metres. S.I. unit is m

^{-1}also known as Dioptre (D).

__: What is the physical interpretation of power ?__

**Dumb Question**Ans: A lens which bends the light more, is considered more powerful. Smaller is the focal length, more is the power of lens.

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**COMBINATION OF LENSES**If two or more lenses of focal length f

_{1}, f

_{2}and so on one placed in contact, then this combination can be replaced by a lens called ‘equivalent lens’.

The focal length (F) of equivalent lens is given by

i.e. P = P

_{1}+ P

_{2}+ ………….

If two this lenses of focal lengths f

_{1}and f

_{2}are separated by a distance ’d’ then the focal length of the equivalent lens is given by

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**DISPLACEMENT METHOD**Let us consider a real object and a screen fixed at a distance D apart. A convex lens is placed between them.

Let x be the distance of the object from the lens when its real image is obtained on the screen.

we have, u = - x

v = + (D - x)

f = + f

Using lens formula, we have

or

=> x

^{2}- Dx + Df = 0

or x =

Cases:-

(i) If D < 4f, for no position of lens, image formation is possible.

(ii) If D = 4f, then in this case x = = 2f. So only one position of the lens will be possible.

(iii) If D > 4f, in this situation, two positions of lens will give same image position. The positions are x

_{1}= x

_{2}=

x

_{2}- x

_{1}= L =

__: what for is this method used ?__

**Dumb Question**Ans: It is used for determining the focal length of lens experimentally from the abiove relation

f =

by measuring L and D value of focal length can be calculated.