ray optics 1 · 2020. 6. 17. · xii- winner masters academy 1 - anasagar circular road, opp....

MASTERS ACADEMY 1 - Anasagar Circular Road, Opp. Chaupati, Ajmer. -0145-2633111XII- Winner

Ray Optics 1

LensA lens is a piece of transparent material with two refracting surfaces such that at least oneis curved and refractive index of its material is different from that of the surroundings.

Types of lenses

R1 R R1R R2 R R2

Bi Equi- Plano- Concavo-

Lens

R1 R1R R RR2 R2

Bi - Equi- Plano- Convexo

ConcaveConvex

Convergent and divergent behaviour of lenses

Terms related to spherical lenses

1. Thin lens & Lens which have less thickness with compare to radius of curvature.

2. Principal axis &Line joining two centre of curvatures of two surfaces is known as principal axis.

3. Optical Centre & If incident ray and immergent ray of lens is parellel to each other, than pointat which refracted ray cut principal axis is called optical centre.

Optical centre of lens divide thickness of lens in two parts, in the ratio of their radius ofcurvatures.

21 1

2 2

CPR=Intensity (aperture)CPR


Ray Optics 2signs of radius are opposite in biconcave and biconvex lenses, thats why optical centre present

in the lens and 1CP and 2CP are of opposite signs.

Incident and Emergent rays are parellel but lateral shift is present. If thickness of lens is veryless then lateral shift is negligible.

So light passes from the optical centre is undeviated for thin lenses.

For plano concave and plano convex lenses 1 ,R

2 2 2

1 10

CP R RCP R

2 0CP

so for these lenses optical centre C present at the pole of curved surface 2P

for convexo concave and concavo convex lenses 1R and 2R are of same signs. So 1CP and 2CP

also of same signs and optical centre present one side of the lens.

If 1 2R R

then1 1

2 2

CP RCP R

1 2CP CP

so optical centre present near the pole of surface for which radius of curvature is less.


Ray Optics 34. Focus points and focal distances

(i) Primary principal focus point and primary focal distance & Primary principal focus point isthe real or virtual point of object on the principal axis whose image is formed at infinity.Distance between optical centre and primary pricinpal focus point is called primary focaldistance.

(ii) Secondary principal focus point and secondary focal distance & Secondary principal focus pointis the real or virtual point of image on the principal axis whose object lies at infinity. Distancebetween optical centre and secondary pricinpal focus point is called secondary focal distance.

If same medium present both side of lens then

2 1f f

If different medium present then

2 1f µf

1 3µ µ

5. Focus plane & It is the perpendicular plane which passes from the focus point.

(6) Aperture : In reference to a lens, aperture means the effective diameter of its lighttransmitting area. So the brightness,i.e., intensity of image formed by a lens whichdepends on the light passing through the lens will depend on the square of aperture,i.e., (aperture)2

(7) Power of a lens : Power of lens is determine the deviation of light ray.

tan hf

hf

(For small)

If h = 1

So, power of lens is equal to the deflection


Ray Optics 4

Pf

Unit of power = Diopter (D).

the refractive index of the medium in which the lens is placed.

Refraction by a LensIn case of image formation by a lens the incident ray is refracted twice, at firstand second surface respectively. The image formed by first surface, 1 acts as anobject for the second and makes final image . And for thin lens thickness oflens t 0

(i) When different mediums present both sides of lens

3 3 22 1

1 2

µ µ µµ µf R R

3 31µ µµv u f

3

1'

µf f

µ

1 3 'f µ f

(ii) When same mediums present both sides of lens

1 2

1 11 1µR Rf

this is known as lens maker formula.

1 1 1v u f

This is known as lens equation

Note :

(1) The Lensmaker’s formula is applicable for thin lenses only. The value of R1 andR2 are to be put in accordance with the cartesian sign convention.

(2) Position of object and image are interchangable. These positions are called – conjugateposition.


Ray Optics 5(3) For same medium, both the side of lens, focal distances are equal.

'f f

Dependance of focal lens on refractive index and radius of curvatures

Lens maker formula

1 2

1 11 1µR Rf

(a) Dependance on radius of curvatures & for biconvex lens 1R ve and 2R ve

1 2

1 11 1µR Rf

for biconcave lens 1R ve and 2R ve

1 2

1 11 1µR Rf

(1) If 1 2R R than position of object and image are interchangeable and focal distances is

unchanged for thin lenses.

(2) for convex lens focal distances is positive (secondary) and for concave lens it is negative.

(3) For equi convex lens focal distance 2 1Rfµ

(4) f(focal length) for convex (Convergent) lens is positive and for concave (Divergent) lens isnegative.

(5) As every part of a lens forms complete image, if a portion (say lower half) isobstructed (say covered with black paper) full image will be formed but brightnessi.e. intensity will be reduced (to half). Also if a lens is painted with black stripsand a donkey is seen through it, the donkey will not appear as a zebra but willremain a donkey with reduced intensity.

(6) If an equi-convex lens of focal length f is cut into two equal parts by a horizontalplane AB, then as none of µ, R1 and R2 will change, the focal length of eachpart be equal to that of initial lens, i.e.

R R A B

f

f

f ' f '

C

D

1f = (µ – 1)

1 1R R

=

2 1( ) R

However in this situation as light transmitting area of each part becomes half ofinitial, so intensity will be reduced to half and aperture to (1/ 2 ) times of itsinitial value


Ray Optics 6

[as I (Aperture)2].

However, if the same lens is cut into two equal parts by a vertical plane CD,the focal length of each part will become.

1f ' = (µ – 1)

1 1R

=

( ) 1R

= 12f

f' = 2f

Focal length of each part will be double of initial value. In this situation as thelight transmitting area of each part of lens remains equal to initial, intensity andaperture will not change.

f

F = f F = f

and

f

and

F = f/2 F =

(b) Dependance on refractive index µ of lens -

11µ

f

for violet colour focal length is maximum and for red colour it is minimum. becuase µ ismaximum for violet colour and minimum for red colour

focal lens of lens in air

1 2

1 11 1a gair

µR Rf

.....(1)

focal lens of lens in liquid

1 2

1 11 1l gliquid

µR Rf

.....(2)


Ray Optics 7Equation 1 @ 2

1 11 1

liquid a g g

gair l g

l

f µ µµf µµ

1a g g aµ µ µ

Three conditions are possible :

(i) if 1 ,gµ µ then 1g

l

µµ

]

liquid airf f

so focal length is increase but nature of lens is unchanged.

(ii) If ,l gµ µ then 1

1 1liquid g

air

f µf

infinity

liquidf infinity ¼P 0 ½

so lens behave like a plane glass plate and invisible.

(iii) If ,l gµ µ in this case sign of focal length is change so nature is also change.

Nature of varios lenses depending upon their surroundings :

If 1 is the R.I of surrounding and

2 is that of lens then

Shape of lens Nature of lens

For 1 <

2For

1 >

2

µ1 µ1µ2Converging Diverging

µ1 µ1

µ2

Diverging Converging


Ray Optics 8

µ1 µ1µ2 Converging Diverging

µ1 µ1

µ2

Diverging Converging

µ1 µ1µ2

Depends on the radius of curvature of the first

and second surface

• If a lens is made of number of layers of different refractive index for a given 1

2

1

2

1

wavelength

then no. of images is equal to number of refractive index, as 1

1f ( )

In figure number of images = 2

• If half portion of lens is covered by black paper then intensity of

B

A

A

B

image will be reduced but complete image will be formed.

Ray diagramsGraphically we can locate the position of image for a given object by drawing anytwo of the following three rays.

(A) A ray, initially parallel to the principal axis will pass (or appear to pass) throughfocus.

(B) A ray which initially passes (or appear to pass) through focus will emerge fromthe, lens parallel to the principal axis.

(C) A ray passing through the optical centre of the lens goes underviated as it passesthrough the lens region.


Ray Optics 9 Special Points(A) On Convex lens

(1) A convex lens will form a real image for a real object when the object is placedbeyond focus (x > fo)

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

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

(4) Anything (object or image) which is farther from the lens is always larger.

(B) On Concave lens

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

(2) The image formed by a concave lens is always erect and diminished in size realobject.

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

Position, Size and Nature of the image formed by a lens

(a) For Convergent or Convex Lens

S.No. Position of Position of Ray – Diagram Nature (size)Object image

1. At infinity At focus Real, inverted ,[ Highly

Diminished ] (m<<–1)

2. Between Between Real, inverted,

and 2F F & 2F [Diminished (m < –1)]

3. At 2F At 2F Real, inverted,

[Equal (m = –1)]

4. Between Between Real, inverted,

2F and F 2F & [Enlarged (m > –1)]


Ray Optics 10

5. At F At Real, inverted, [Highly

Enlarged (m >> – 1)]

6. Between Between Virtual, erect

F and P [Enlarged (m > +1)]

on same side

(b) For Divergent or Concave Lens

S.No. Position of Positin of Ray – Diagram Nature of Size

Object image image

1. At infinity At F Virtual, Highly

erect diminished

(m << +1)

2. In front of Between Virtual, Diminished

lens F & optical erect (m < +1)

centre

Magnification A. Linear Magnification

Linear magnification is the ratio of the size of image to the size of object.

1. Transverse Linear Magnification 2. Longitudenal Linear Magnification

1. Transverse Linear Magnification - Object present perpendicular to the principal axis.

I vmO u

Transverse Linear Magnification

I v f v fmO u f f u

2. Longitudenal Linear Magnification - Object present along to the principal axis.

Longitudenal Linear Magnification2 22

2

f v fI dv v mf f uO du u


Ray Optics 11(B) Arial or Superficial Magnification

(i) If a 2-D object is placed with its plane perpendicular to the principal axis than itsmagnification = m2

(ii) If one side of object is parellel to the principal axis than its magnification = m × m2 = m3

(C) Volume Magnification = m × m × m2 = m4

Note :- m is the transverse linear magnification.

When object present perpendicular to the principal axis than I vO u , if v u so I O and if

v u so I O A

Real Image Virtual Image

Image is on other side of the object Towards the object

Inverted Erect

Magnification = – ve Magnification + ve

u = –ve, v = +ve u = –ve, v = –ve

Smaller, equal and larger depend on the Always magnified

position of object can be obtained on screen. Can not be obtained on screen

For Real Extended Objects

If the image formed by a single lens is erect (m is positive) it is always virtual.

In this situtation :

If the image is enlarged : then it is for convex lens (convering lens) with object between focusand optical centre

If the image is diminished : then it is for concave lens (diverging) with image between focus andoptical centre.

F

light

realobject

virtualimage

vu

diminished and virtual image

v = –veu = –ve ,F

v = –velight u = –ve ,

realobject

virtualimage

uv

virtual and enlarged image

Check Point 32: Find the position, size and nature of image,

for the situation shown in figure. Draw ray diagram .

Solution For refraction near point A, u = – 30 ; R = – 20; n1 = 2 ; n

2 = 1.

Applying refraction formula

vn2

un1 =

Rnn 12

v1 – 30

2

= 2021


Ray Optics 12v = – 60 cm

m = 1

2

hh

= unvn

2

1 = )30(1

)60(2

= 4 h2 = 4 mm.

Check Point 33: Find the behaviour of a concave lens placed in a rarer medium.

Solution the focal length of the lens, which depends on the product of these factors ,is negative and hence the lens will behave as diverging lens.

Check Point 34:Show that the factor

21 R1

R1

(and therefore focal length) does not depend on

which surface of the lens light strike first.

Solution Consider a convex lens of radii of curvature p and q as shown.

CASE 1: Suppose light is incident from left side and strikes the surface with radius ofcurvature p, first.

Then R1 = +p ; R

2 = -q and

21 R1

R1

=

q

1p1

CASE 2: Suppose light is incident from right side and strikes the surface with radius ofcurvature q, first.

Then R1 = +q ; R

2 = -p and

21 R1

R1

=

p

1q1

Though we have shown the result for biconvex lens , it is true for every lens.

Check Point 35: Find the focal length of the lens

shown in the figure.

Solution : f1 = (n

rel – 1)

21 R1

R1

f1 = (3/2 – 1)

)10(

1101

f1 =

21 × 10

2 f = + 10 cm.

Check Point 36:Find the focal length of the

lens shown in figure

Solution :f1 = (n

rel – 1)

21 R1

R1

=

1

23

10

1101

f = – 10 cm


Ray Optics 13Check Point 37: Find the focal length of the

lens shown in figureROC = 60 cm

ROC = 20 cm

(a) If the light is incident from left side.

(b) If the light is incident from right side.

Solution : (a) f1 = (n

rel – 1)

21 R1

R1

=

1

23

201

601

f = 60 cm

(b) f1 = (n

rel – 1)

21 R1

R1

=

1

23

601

201

f = 60 cm

Check Point 38: Point object is placed on the principal axis of a thin lens with parallel curvedboundaries i.e., having same radii of curvature. Discuss about the position of theimage formed .

Solution1f = (nrel 1)

21 R1

R1

= 0 [ R1 = R

2]

1v

1u = 0 or v = u i.e. rays pass without appreciable bending.

Check Point 39: Focal length of a thin lens in air, is 10 cm. Now medium

on one side of the lens is replaced by a medium of refractive index=2. The radius of curvature of surface of lens, in contact withthe medium, is 20 cm. Find the new focal length.

Solution : Let radius of surface be R1 and refractive index of lens be . Let parallel

rays be incident on the lens. Applying refraction formula at first surface

1–v1

= 1R1

...(1)

At surfacev2 –

1v = 20

2

...(2)

Adding (1) and (2)

1–v1

+ v2 –

1v =

1R1 + 20

2

= ( – 1)

201

R1

1 – 20

1 – 20

2 =

f1 (in air) + 20

1 – 20

2

v = 40 cm f = 40 cm

Check Point 40: Figure shows a point object and a converging lens.

Find the final image formed.


Ray Optics 14

Solution :v1 – u

1 =

f1

v1 – 15

1

= 101

v1 = 10

1 – 15

1= 30

1

v = + 30 cm

Check Point 41:See the figure

Find the position of final image formed.

Solution For converging lens

u = –15 cm, f = 10 cm v = uffu

= 30 cm

For diverging lens

u = 5 cm

f = –10 cm v = uffu

= 10 cm

Check Point 42:Figure shows two converging lenses. Incident

rays are parallel to principal axis. What should be the

value of d so that final rays are also parallel.

Solution Final rays should be parallel. For this the focus of L1

must coincide with focus of L2.

d = 10 + 20

= 30 cm

Here the diameter of ray beam becomes wider.

Check Point 43: See the figureFind the position of final image formed.

Solution : For lens,

v1 – u

1 =

f1

v1 – 15

1

= 101

v = + 30 cm

Hence it is object for mirror

u = – 15 cm


Ray Optics 15

v1 + 15

1 = 10

1 v = – 30 cm

Now for second time it again passes through lens

u = – 15 cm

v = ? ; f = 10 cm

v1 – 15

1

= 101

v = + 30

Hence final image will form at a distance 30 cm from the lens towards left.

Check Point 44: What should be the value of d so that image

is formed on the object itself.

Solution For lens :

v1 – 15

1 = 10

1 v = + 30 cm

Case I :

If d = 30, the object for mirror will

be at pole and its image will be

formed there itself.

Case II :

If the rays strike the mirror normally, they willretrace

and the image will be formed on the object itself

d = 30 – 20 = 10 cm

Check Point 45:An extended real object of size 2 cm is placed perpendicular to the principalaxis of a converging lens of focal length 20 cm. The distance between the objectand the lens is 30 cm.

(i) Find the lateral magnification produced by the lens.

(ii) Find the height of the image.

(iii) Find the change in lateral magnification, if the object is brought closer to thelens by 1 mm along the principal axis.

Solutionv1 – u

1 =

f1

m = uv

m = uff

.........(A)


Ray Optics 16

m = )30(2020

= 10

20

= – 2

–ve sign implies that the image is inverted.

(ii)1

2

hh

= m

h2 = mh

1 = (–2) (2) = – 4 cm

(iii) Differentiating (A) we get

dm = 2)uf(f

du = 2)10(

)20(

(0.1) = 100

2 = –.02

Note that the method of differential is valid only when changes are small.

Alternate method : u (after displacing the object) = –(30 + 0.1) = – 29.9 cm

Applying the formula

m = uff

m = )9.29(2020 = – 2.02

change in ‘m’ = – 0.02.

Since in this method differential is not used, this method can be used for any

changes, small or large.

Newton's Formula for Lens

Distance of real object from first focus point F' is 'x and distance of real image from second

focus point F is x than

' 'xx ff

when same medium present both the sides of lens ' f f

2' xx f

In this formula x' is negative and x is positive.

Note : 1. Lens equation 1 1 1

f v u is used only for thin lenses. Newton's formula use for thick lenses

also.

2. For unequale radius of curvatures optical center of lens is not exectly at mid point of lens.So u and v is not measured exectly.

3. This furmula is also used for concave mirrors when image is real. ¼u f ½

Limitations of Newton's Formula

This formula is not applied when image is virtual. So this cannot use for

(i) Concave lens

(ii) Convex lens, when object present between focus point and optical centre.

(iii) Concave mirror, when object present between focus point and pole.


Ray Optics 17 Combination of lenses

When lenses are in contact with each other

When several lenses are kept co-axially , the image formation is considered oneafter another in steps. The image formed by the lens facing the object serves asan object for the next lens, the image formed by the second acts as an objectthe third and so on .

1f =

1

1f +

1

2f

P = P1 + P2

1 2

1 2

1 2

1 1 1F f f

f fFf f

Case (i) When both lenses are convex than 1f and 2f positive. So F is also positive. Combination of

lenses behave like a convex lens.

Case (ii) When both lenses are concave than 1f and 2f negative. So F is also negative. Combination

of lenses behave like a concave lens.

Case (iii) If first lens is convex and second lens is concave than f1 is positive and f

2 is negative.

2 1 2 1

1 2 1 2 1 2

1 1 1 f f f fF f f f f f f

1 2

2 1

f fF

f f

(a) If 2 1f f ¼ 2 1p p ½ ] Then F is negative, so combination behaves like concave lens.

(b) If 2 1f f ¼ 2 1p p ½ ] then F and 0p ] so combination behaves like plane plate.

(c) If 2 1f f ¼ 2 1p p ½] then F positive, so combination behaves like convex lens.

Memorise:-

1. For more than two lenses

1 2 3

1 1 1 1 ......F f f f

1 2 3p p p p

2. If distance between two lenses is d

1 2 1 2

1 1 1 dF f f f f

1 2

1 2

f fF

f f d


Ray Optics 18

If 1 2f f f and 2d f

than 2ffF

f d

so F is negative and combination of two convex lenses behave like a concave lens.

Power 1 2 1 2p p p dp p

Check Point 46:

Find the lateral magnification produced by the

combination of lenses shown in the figure.

Sol:f1 =

1f1 +

2f1 = 10

1 – 20

1 = 20

1 f = + 20

v1 – 10

1

= 201

v1 = 20

1 – 10

1= 20

1= – 20 cm

m = 1020

= 2

Check Point 47: Find the focal length of equivalent system.

Sol:1f1 =

1

23

101

101

= 21 × 10

2 = 10

1

2f1=

1

56

201

101

= 51 ×

201030

= 1003

3f1=

1

58

201

201

= 503

f1 =

1f1 +

2f1 +

3f1

= 101 + 100

3 + 50

3f = 13

100

Focal Distance of convex lens by displacment method


Ray Optics 19When Distance between two pins is greater than 4f than for two positions of lens image ofone pin is formed on another pin.

Focal Distance of lens 2 2

4a df

a

2a du

and 2

a dv

By equation of focus distance 2 24af a d

2 2 4d a af

20 4 0d a af

2 4a af

4a f

1 2O I I

2 21

2

I v a dI u a d

Special Point -

In displacement method

2 2

1 2v u v um mu v uv

11 1v u v u dv uuv fv u

1 2

dfm m

Lens with one silvered surface

Power of Lens PL =

1fL

Convex lens PL = +ve Concave lens P

L = –ve

Power For mirror PL =

1fm

Convex mirror PM = –ve Concave mirror P

M = +ve

If the back surface of a lens is silvered and an object is placed in front of it then.

(1) First, light will pass through the lens and it will form the image I1.


Ray Optics 20

(2) The image I1 will act as an object for silvered surface which acts as curved mirror and

forms an image I2 of object I

1.

(3) The light after reflection from silvered surface will again pass through the lens and lens willform final image I

3 of object I

2.

This all is shown in figure. In such situation power of the silvered lens will be

L M LP P P P

with1

LL

Pf

where 1 2

1 111

L R Rf

and1

MM

Pf

where 2

2MRf

So the system will behave as a curved mirror of focal length F given by.

1FP

To make this all clear we now consider the case of a silvered plano-convex lens underfollowing circumstances.

(A) When the plane surface is silvered and the object is in fron of curved surface: In this situation

11 1 11Lf RR

fl

+

F fm

and2MF

So 11

LL

Pf R

and1 1 0MM

Pf

And hence power of system

2L M L L MP P P P P P

i.e, 21 1

2 0PR R

......(1)

So 1

2 1RF

P

......(2)


Ray Optics 21

i.e, the lens will behave as a concave mirror of focal length / 2 1R .

(B) When the curved surface is silvered and the object is in front of plane surface:

In this situation

11 1 11Lf RR

and

2MRf

So 11

LL

Pf R

and

1 2M

MP

f R

F fl fm

+

And hence power of system

2L M L L MP P P P P P

i.e, 2 1 2 2P

R R R

.....(3)

So1

2RF

P

......(4)

i.e, the lens will be equivalent to a converging mirror of focal length / 2R .

Double convex lens is silvered

fl fm

+

F

2

,)1(2

RfRf ml

)12(2

RF

ray optics 1 · 2020. 6. 17. · xii- winner masters academy 1 - anasagar circular road, opp....

Documents