# Chapter 18 Ray Optics Part 2 (Refraction of Light) free study material by TEACHING CARE online tuition and coaching classes

Chapter 18 Ray Optics Part 2 (Refraction of Light) free study material by TEACHING CARE online tuition and coaching classes

The bending of the ray of light passing from one medium to the other medium is called refraction.

### Snell’s law

i.e. sin i = m

#### sin r

(a constant). For two media, Snell’s law can be written as

1 m 2

m 2

m1

= sin i

#### sin r

Þ m1 ´ sin i = m 2 ´ sin r

i.e. m sinq =

#### constant

Also in vector form : ˆi ´ nˆ = μ (rˆ´ nˆ)

# Refractive Index.

#### Refractive index of a medium is that characteristic which decides speed of light in it. It is a scalar, unit less and dimensionless quantity.

• Types : It is of following two types

 Absolute refractive index Relative refractive index (i)    When light travels from air to any transparent medium     (i) When light travels from medium (1) to medium (2) then then R.I. of medium w.r.t. air is called it’s absolute R.I. i.e.                                                                                      R.I. of medium (2) w.r.t. medium (1) is called it’s relativem           = c                                                                                       R.I. i.e.    m = m2 = v1   (where v and v are the speed ofair   medium     v                                                                                                        1   2      m      v                      1             21        2light in medium 1 and 2 respectively).(ii)   Some absolute R.I.                                                           (ii) Some relative R.I.m      = 3 = 1.5 , m         = 4 = 1.33                     (a)    When    light    enters    from    water    to    glass   :a  glass      2            a  water         3                                        m = mg = 3 / 2 = 9w  ga mdiamond = 2.4, a mCs   = 1.62                                                  mw         4 / 3     82                                                                                    (b) When light enters from glass to diamond : a m crown = 1.52, mvacuum = 1 , mair = 1.0003 » 1                                m       2.4    8    m   =    D =          = g D mg 1.5     5

#### Note : @Cauchy’s equation :

m = A +  B

l2

+ C + ……

l4

(lRed

• lviolet so mRed

m violet )

m µ 1

l

#### @    If a light ray travels from medium (1) to medium (2), then

m = m 2 = l1

v1

m µ 1

• Dependence of Refractive index

1   2      m1

l2     v2

v

v µ l

#### (i)   Nature of the media of incidence and refraction.

• Colour of light or wavelength of
• Temperature of the media : Refractive index decreases with the increase in
• Principle of reversibility of light and refraction through several media :

# Refraction Through a Glass Slab and Optical Path

### (1)  Lateral shift

#### The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to

it’s incident direction i.e. the ray undergoes no deviation d = 0. The angle of emergence (e) is equal to the angle of incidence (i)

#### The Lateral shift of the ray is the perpendicular distance between the incident and the emergent ray, and it is given by MN = t sec r sin (i – r)

æ      1 ö

OO‘ = x = ç1 – m ÷ t

 Glass slabm

#### è         ø

O               O

x

t

Or the object appears to be shifted towards the slab by the distance x

### (2)  Optical path :

#### It is defined as distance travelled by light in vacuum in the same time in which it travels a given path length in a medium.

Note : @              Since for all media path length (x).

# Real and Apparent Depth.

m > 1,

#### so optical path length (mx) is always greater than the geometrical

If object and observer are situated in different medium then due to refraction, object appears to be displaced from it’s real position. There are two possible conditions.

 (1) When object is in denser medium and observer is in rarer medium    h¢        m  h                                 O¢ d                            O (1) Object is in rarer medium and observer is in denser medium.O¢  d  h¢O  h (2) m =     Real depth      = hApparent depth       h‘Real depth >Apparent depth that’s why a coin at the bottom of bucket (full of water) appears to be raised) (2) m = h‘hReal depth < Apparent depth that’s why high flying aeroplane appears to be higher than it’s actual height. (3) Shift d = h – h‘ = æ1 – 1 ö hç       m ÷è           ø (3) d = (m – 1)h (4) For water m = 4 Þ d = h3            4For glass m = 3 Þ d = h2            3 (4) Shift for water dw = h3Shift for glass dg = h2

#### Note : @If a beaker contains various immisible liquids as shown then

m1                          d1

m2                          d2

m1

• d2

m 2

• d3

m 3

#### + ….

m3                          d3

m             =  d AC

d1 + d2 + …..

#### (In case of two liquids if d = d

than m =

2m1m2 )

combination       d               d        d

1     2              m + m

App.

1 +   2 + ….                                                           1         2

m1    m 2

# Total Internal Reflection.

#### When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90o, this angle of incidence is called critical angle (C).

When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

μ =    1     = cosec C ;     where m ®

sinC

Rerer

mDenser

#### Note : @When a light ray travels from denser to rarer medium, then deviation of the ray is

d = p – 2q Þ d ® max. whenq ® min. = C

i.e. d max

= (p – 2C); C ® critical angle

• Dependence of critical angle
• Colour of light (or wavelength of light) : Critical angle depends upon wavelength as l µ 1 µ sin C

m

• lR

## >  lV

Þ CR

• CV

• Sin C = 1 = m R

lD

vD

#### (for two media)      (c) For TIR from boundary of two media i > sin-1mR

R m D

m D        lR        vR                                                                                                                                                                                                                               m D

#### (ii)   Nature of the pair of media : Greater the refractive index lesser will be the critical angle.

• For (glass- air) pair ® Cglass = 42o
• For (water-air) pair ® Cwater = 49 o

#### (c)  For (diamond-air) pair

® Cdi amond

= 24 o

• Temperature : With temperature rise refractive index of the material decreases therefore critical angle
• Examples of total internal reflection (TIR)

#### (i)

Mirage : An optical illusion in deserts                                           Looming : An optical illusion in cold countries

• Brilliance of diamond : Due to repeated internal reflections diamond

#### (iii)   Optical fibre : Optical fibres consist of many long high quality composite glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core (m1) is higher than that of the cladding (m2).

When the light is incident on one end of the fibre at a small angle, the light passes inside, undergoes repeated total internal reflections along the fibre and finally comes out. The angle of incidence is always larger than the critical angle of the core material with respect to its cladding. Even if the fibre is bent, the light can easily travel through along the fibre

#### A bundle of optical fibres can be used as a ‘light pipe’ in medical and optical examination. It can also be used for optical signal transmission. Optical fibres have also been used for transmitting and receiving electrical signals which are converted to light by suitable transducers.

• Field of vision of fish (or swimmer) : A fish (diver) inside the water can see the whole world through a cone
• Apex angle = 2C = 98 o
• Radius of base r = h tan C = h

#### (c)  Area of base A =

ph2

(m 2 – 1)

Note : @ For water m = 4

so r = 3h

A = 9ph2 .

#### 7

• Porro prism : A right angled isosceles prism, which is used in periscopes or It is used to deviate light rays through 90 o and 180 o and also to erect the image.

B                                                       A¢

B¢

A

Example: 1         A beam of monochromatic blue light of wavelength 4200 Å in air travels in water (m = 4 / 3) . Its wavelength in water will be                                                                                                                                            [MNR 1991]

(a) 2800 Å                               (b) 5600 Å                           (c) 3150 Å                              (d) 4000 Å

Solution: (c)

m µ 1

###### l

Þ   m1

m 2

= l2     Þ

l1

1 =

l2      Þ

4200

l2 = 3150 Å

Example: 2         On a glass plate a light wave is incident at an angle of 60o. If the reflected and the refracted waves are mutually perpendicular, the refractive index of material is                                     [MP PMT 1994; Haryana CEE 1996]

(a)

3                             (b)

2

(c)

3                              (d)  1

2

Solution: (b)        From figure r = 30o

\     m = sin i = sin 60o =

sin r      sin 30o

Example: 3         Velocity of light in glass whose refractive index with respect to air is 1.5 is 2 × 108 m / s and in certain liquid

the velocity of light found to be 2.50 ´ 108 m / s . The refractive index of the liquid with respect to air is

[CPMT 1978; MP PET/PMT 1988]

(a) 0.64                           (b) 0.80                        (c) 1.20                           (d) 1.44

1          m li           vg

ml          2 ´ 108

Solution: (c)

m µ        Þ

v

=

m g        vl

Þ 1.5 = 2.5 ´ 108

Þ ml = 1.2

Example: 4         A ray of light passes through four transparent media with refractive indices m1.m 2 , m 3 , and m 4 as shown in the

figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have                                                                                                             [IIT-JEE (Screening) 2001]

(a) m1 = m 2

(b) m 2 = m 3

(c) m3 = m4

(d) m4 = m1

Solution: (d)        For successive refraction through difference media m sinq = constant.

Here as q is same in the two extreme media. Hence m1 = m4

Example: 5         A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of

water, then the value of m g

• (4/3) sin i
• 1/ sin i

(c) 4/ 3

(d) 1

would be

[IIT-JEE (Screening) 2003]

Solution: (b)        For glass water interface g mw = sin i

sin r

……(i)    and For water-air interface        m = sin r

 w     a

sin 90

…..(ii)

\     g mw ´w ma = sin i

Þ m g = 1

sin i

Example: 6         The ratio of thickness of plates of two transparent mediums A and B is 6 : 4. If light takes equal time in passing through them, then refractive index of B with respect to A will be                                           [UPSEAT 1999]

(a) 1.4                             (b) 1.5                         (c) 1.75                           (d) 1.33

Solution: (b)        By using t = mx

c

Þ   m B

###### m A

= x A   = 6 Þ

x B           4

A m B

= 3 = 1.5

2

Example: 7         A ray of light passes from vacuum into a medium of refractive index m, the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is

(a)

cos 1 (m / 2)

(b)

2 cos 1 (m / 2)

(c)

2 sin1 (m )

(d)

2 sin1 (m / 2)

Solution: (b)        By using

m = sin i

Þ m = sin 2r = 2 sin r cos r

( sin 2q = 2 sinq cosq )

sin r

sin r

sin r

Þ r = cos 1 æ m ö .      So, i = 2r = 2 cos 1 æ m ö .

ç 2 ÷                                              ç 2 ÷

è     ø                                              è     ø

Example: 8 A ray of light falls on the surface of a spherical glass paper weight making an angle a with the normal and is refracted in the medium at an angle b . The angle of deviation of the emergent ray from the direction of the incident ray is                                                                                                                      [NCERT 1982]

(a) (ab )

(b)

2 (ab )

(c) (ab )/ 2

(d) (a + b )

Solution: (b)        From figure it is clear that DOBC is an isosceles triangle,

Hence

ÐOCB = b

and emergent angle is a

Also sum of two in terior angles = exterior angle

\ d = (ab ) + (ab ) = 2(ab )

Example: 9 A rectangular slab of refractive index m is placed over another slab of refractive index 3, both slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of m will the coin appear to be placed at the interface between the slabs when viewed from the top

(a) 1.8                             (b) 2                            (c) 1.5                            (d) 2.5

Solution: (c)         Apparent depth of coin as seen from top = x + x = x

m1    m 2

x

Þ     1 + 1 = 1

Þ 1 + 1 = 1

Þ m = 1.5

 m1 = 3

m1    m 2                 3    m                                                                                                 x

Example: 10 A coin is kept at bottom of an empty beaker. A travelling microscope is focussed on the coin from top, now water is poured in beaker up to a height of 10 cm. By what distance and in which direction should the microscope be moved to bring the coin again in focus

(a) 10 cm up ward             (b) 10 cm down ward   (c) 2.5 cm up wards       (d) 2.5 cm down wards

Solution: (c)         When water is poured in the beaker. Coin appears to shift by a distance d = h = 10 = 2.5cm

4     4

Hence to bring the coil again in focus, the microscope should be moved by 2.5 cm in upward direction.

 ç

Example: 11       Consider the situation shown in figure. Water æ mw

è

= 4 ö

 3 ÷

ø

is filled in a breaker upto a height of 10 cm. A plane

mirror fixed at a height of 5 cm from the surface of water. Distance of image from the mirror after reflection from it of an object O at the bottom of the beaker is

(a) 15 cm                                 (b) 12.5 cm                         (c) 7.5 cm                               (d) 10 cm

Solution: (b)        From figure it is clear that object appears to be raised by 10 cm (2.5 cm)

4

Hence distance between mirror and O‘ = 5 + 7.5 = 12.5 cm

So final image will be formed at 12.5 cm behind the plane mirror.

10 cm

4

O‘                  10 cm

O

Example: 12       The wavelength of light in two liquids ‘x‘ and ‘y‘ is 3500 Å and 7000 Å, then the critical angle of x relative to y will be (a) 60o                                       (b) 45o                         (c) 30o                            (d) 15o

Solution: (c)

sin C = m 2 = l1

= 3500 = 1 Þ C = 30 o

m1     l2

7000     2

Example: 13  A light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive index m = 1.5) making an incidence angle of 60o on the lateral surface, so that it undergoes a total internal reflection. How much time would it take to traverse the straight fiber of length 1 km                                                                                                       [Orissa JEE 2002]

• 33 m sec
• 67 m sec
• 77 m sec
• 85 m sec

Solution: (d)        When total internal reflection just takes place from lateral surface then i = C i.e. C = 60o

From

m    1

sin C

Þ m     1    = 2 sin 60

2 ´ (1 ´ 103 )

Hence time taken by light traverse some distance in medium t = mx

C

Þ t =    3                = 3.85 m sec.

3 ´ 108

Example: 14       A glass prism of refractive index 1.5 is immersed in water (m = 4 / 3) . A light beam incident normally on the face AB is totally reflected to reach the face BC if                                                                        [CPMT 1981; IIT-JEE 1981]

(a) sinq > 8 / 9

(b) 2 / 3 < sinq < 8 / 9

• sinq £ 2 / 3
• cosq ³ 8 / 9

Solution: (a)        From figure it is clear that

Total internal reflection takes place at AC, only if q > C

Þ sinq > sin C

Þ sinq >    1

w m g

Þ sinq >   1

9 / 8

Þ sinq > 8

9

Example: 15       When light is incident on a medium at angle i and refracted into a second medium at an angle r, the graph of sin i vs sin r is as shown in the graph. From this, one can conclude that

• Velocity of light in the second medium is 73 times the velocity of light in the I medium
• Velocity of light in the I medium is 73 times the velocity in the II medium

• The critical angle for the two media is given by sin ic
• sin ic = 1

2

1

Solution: (b, c)     From graph tan 30o = sin r =  1  Þ           m   =

Þ m 2 = v1

= 1.73

Þ v   = 1.75 v

sin i

1 m 2             1   2

m1      v2                               1                 2

Also from

m =      Þ

sin C

sin C =         1

Rarer mDenser

Þ sin C =    =   1 .

1 m 2

Example: 16 A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive indices of the material of the prism for the above red, green and blue wavelength are 1.39, 1.44 and 1.47 respectively. The prism will                                                                                                       [IIT-JEE 1989]

• Separate part of red colour from the green and the blue colours
• Separate part of the blue colour from the red and green colours
• Separate all the colours from one another
• Not separate even partially any colour from the other two colours

Solution: (a)        At face AB, i = 0 so r = 0, i.e., no refraction will take place. So light will be incident on face AC at an angle of

incidence of 45o. The face AC will not transmit the light for which i > q C , i.e.,                 sin i > sinq C

or     sin 45 o > (1 / m )

i.e.,   m >

(= 1.41)

Now as

m R < m

while

m G and

m B > m,

so red will be transmitted through the

face AC while green and blue will be reflected. So the prism will separate red

colour from green and blue.                                                                                                                C

Example: 17       An air bubble in a glass slab (m = 1.5) is 6 cm deep when viewed from one face and 4 cm deep when viewed from the opposite face. The thickness of the glass plate is

• 10 cm (b) 67 cm                         (c) 15 cm                                 (d) None of these

Solution: (c)         Let thickness of slab be t and distance of air bubble from one side is x

When viewed from side (1) :

When viewed from side (2) :

1.5 = x Þ x = 9cm

6

1.5 = (t  x) Þ 1.5 = (t  9) Þ t = 15cm

Side 1

Side 2

4                    4

Tricky example: 1

One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first face, forms an image 12 cm behind the silvered face. The refractive index of the glass is                                     [CPMT 1999]

(a) 0.4                             (b) 0.8                                (c) 1.2                          (d) 1.6

Solution : (c)      From figure thickness of glass plate t = 6 cm.

Let x be the apparent position of the silvered surface.

According to property of plane mirror

x + 8 = 12 + 6 – x             Þ     x = 5 cm

x

Object                             Image

Also

Tricky example: 2

m = t

x

Þ      m = 6 = 1.2

5

8 cm

12 cm

12 +(6–x)

t

A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence so that the ray which enters the sphere doesn’t come out of the sphere

(a)

tan-1 æ 2 ö

 ç   ÷

3

sin-1 æ 2 ö

 ç   ÷

3

• 90o (d)

cos -1 æ 1 ö

 ç   ÷

3

è   ø                                             è   ø                                                                                                 è ø

Solution : (c)     Ray doesn’t come out from the sphere means TIR takes place.                            A                                 B

Hence from figure

ÐABO = ÐOAB = C

i                   C        C

O

\     m =      Þ sin C = 1 = 2

sin C                          m

Applying Snell’s Law at A

3

sin i3

Þ sin i = 3 sin C = 3 ´ 2 = 1

Þ i = 90 o

Tricky example: 3

sin C       2

2            2    3

The image of point P when viewed from top of the slabs will be

(a) 2.0 cm above P               (b) 1.5 cm above P                   (c) 2.0 cm below P            (d) 1 cm above P

Solution: (d)        The two slabs will shift the image a distance

 ç       m ÷         ç       1.5 ÷

d = 2æ1 – 1 ö t = 2æ1 – 1 ö (1.5) = 1cm

è           ø          è              ø

1.5 cm

1.5 cm

1.5 cm

2.0 cm

P

# Refraction From Curved Surface.

#### m1 = Refractive index of the medium from which light rays are coming (from object).

m 2 = Refractive index of the medium in which light rays are entering.

#### u = Distance of object, v = Distance of image, R = Radius of curvature

Refraction formula :

m 2 – m1 = m 2 – m1

#### (use sign convention while solving the problem)

R               v         u

#### Note : @Real image forms on the side of a refracting surface that is opposite to the object, and virtual image forms on the same side as the object.

@    Lateral (Transverse) magnification m = I =

##### O

m1v .

m 2u

In a thin spherical fish bowl of radius 10 cm filled with water of refractive index 4/3 there is a small fish at a distance of 4

cm from the centre C as shown in figure. Where will the image of fish appears, if seen from E

(a) 5.2 cm                                       (b) 7.2 cm                                        (c) 4.2 cm                               (d) 3.2 cm

Solution : (a) By using

where

On putting values

v = ?

# Lens.

Lens is a transparent medium bounded by two refracting surfaces, such that at least one surface is spherical.

• Type of lenses

 Double convex Plano convex Concavo convex Double concave Plane concave Convexo concave Thick at middle Thin at middle It forms real and virtual images both                                     It forms only virtual images

• Some definitions

C1, C2 – Centre of curvature,

R1, R2 – Radii of curvature

• Optical centre (O) : A point for a given lens through which light ray passes undeviated (Light ray passes undeviated through optical centre).

### Principle focus

#### Note : @Second principle focus is the principle focus of the lens.

@    When medium on two sides of lens is same then | F1 |=| F2 |.

f1 = m1

f2       m 2

#### (iii)   Focal length (f) : Distance of second principle focus from optical centre is called focal length

fconvex ® positive,

fconcave ® negative,

fplane ® ¥

#### (iv)   Aperture : Effective diameter of light transmitting area is called aperture. Intensity of image µ (Aperture)2

• Power of lens (P) : Means the ability of a lens to converge the light Unit of power is Diopter (D).

#### P =   1   = 100 ;

Pconvex ® positive, Pconcave ® negative, Pplane ® zero .

f(m)    f(cm)

#### Note : @                Thick lens          Thin lens

P ­ f ¯ R ¯               P ¯ f ­ R ­

### (3)  Image formation by lens

 Lens                       Location of       Location of the                           Nature of imagethe object               image                                                                                   Magnification       Real                Erect At infinityi.e. u = ¥ Away from 2f i.e. (u > 2 f)At 2f or (u = 2 f) Between f and 2f2if.e. f < u < 2 fAt focus i.e. u = f  Between optical centre and focus, u < f  At infinityi.e. u = ¥Anywhere between infinity and optical centre At focus i.e. v = f   Between f and 2f i.e. f < v < 2 f At 2f i.e. (v = 2 f) Away from 2f i.e.(v > 2 f)At infinity i.e. v = ¥ At a distance greater than that of object v > uAt focus i.e. v = f Between optical centre and focus virtual inverted Convex m < 1 Real Inverted diminished m < 1 Real Inverted diminished m = 1 Real Inverted same size m > 1 Real Inverted 2f f f magnifiedm = ¥ Real Inverted magnified m > 1 Virtual Erect magnified Concave m < 1 Virtual Erect diminished m < 1 Virtual Erect diminished

#### Note : @Minimum distance between an object and it’s real image formed by a convex lens is 4f.

@    Maximum image distance for concave lens is it’s focal length.

### (4)  Lens maker’s formula

#### 1           æ 1     1 ö

The relation between f, m, R1 and R2 is known as lens maker’s formula and it is

#### = (m – 1)ç        –       ÷

##### f                      R         R

è   1         2 ø

### (5)  Lens in a liquid

#### Focal length of a lens in a liquid ( fl ) can be determined by the following formula

fl (a μg 1)

fa       (l μg 1)

#### (Lens is supposed to be made of glass).

Note : @Focal length of a glass lens (m = 1.5) is f in air then inside the water it’s focal length is 4f.

@    In liquids focal length of lens increases (­) and it’s power decreases (¯).

### (6)  Opposite behaviour of a lens

In general refractive index of lens (m L ) > refractive index of medium surrounding it (m M ).

### (7)  Lens formula and magnification of lens

#### (i)  Lens formula :

11 – 1

; (use sign convention)

f       v     u

#### (ii)   Magnification : The ratio of the size of the image to the size of object is called magnification.

• Transverse magnification : m = I= v =  f         = f  v

#### (use sign convention while solving the problem)

##### O       u       f + u           f

I       v2 – v1

dv      æ v ö 2    æ    f     ö 2

æ f v ö 2

#### (b)   Longitudinal magnification : m =         =

. For very small object m =

#### = ç   ÷   = ç        ÷   = ç        ÷

O      u2 – u1

du     è u ø       è f + u ø

è    f     ø

A                    æ    f     ö 2

#### (c)  Areal magnification : ms

=    i = m2 = ç        ÷ ,   (Ai = Area of image, Ao = Area of object)

Ao                             è f + u ø

### (8)  Relation between object and image speed

#### If an object move with constant speed

(Vo ) towards a convex lens from infinity to focus, the image will move

æ    f     ö 2

. Vo

è         ø

### (9)  Focal length of convex lens by displacement method

#### (i)  For two different positions of lens two images (I1 and I 2 ) of an object is formed at the same location.

• Focal length of the lens

f D 2x 2

=         x

#### where m1

I1

and m2

4 D

I 2

##### O

m1 – m2

• Size of object O =

### (10)  Cutting of lens

#### (i)   A symmetric lens is cut along optical axis in two equal parts. Intensity of image formed by each part will be same as that of complete lens.

• A symmetric lens is cut along principle axis in two equal Intensity of image formed by each part will

be less compared as that of complete lens.(aperture of each part is             times that of complete lens)

### (11)  Combination of lens

#### (i)  For a system of lenses, the net power, net focal length and magnification given as follows :

P = P1 + P2 + P3………… ,

#### 1 = 1 + 1 + 1

+ ……….. ,

m = m1 ´ m2 ´ m3 ´…………

F       f1       f2       f3

#### (ii)   In case when two thin lens are in contact : Combination will behave as a lens, which have more power or lesser focal length.

1 = 1 + 1 Þ      F =    f1 f2

and

P = P1 + P2

F       f1       f2                          f1 + f2

#### (iii)   If two lens of equal focal length but of opposite nature are in contact then combination will behave as a

plane glass plate and

Fcombination = ¥

1 = 1 + 1 d

#### and

P = P1 + P2 – dP1 P2

F       f1       f2       ff2

### (12)  Silvering of lens

1 = 2 + 1

#### where fl = focal length of lens from which refraction takes place (twice)

F     fl        fm

#### fm = focal length of mirror from which reflection takes place.

• Plano convex is silvered

 2

fm = R , fl

=     R     so

(m – 1)

F  R

2m

fm = ¥, fl

=     R     so

(m – 1)

F =         R

2(m – 1)

#### (ii)   Double convex lens is silvered

Since f =        R       , f      = R

l       2(m – 1) m         2

##### R

So F = 2(2m – 1)

Note : @ Similar results can be obtained for concave lenses.

### (13)  Defects in lens

#### (i)   Chromatic aberration : Image of a white object is coloured and blurred because m (hence f) of lens is different for different colours. This defect is called chromatic aberration.

mV > m R so fR > fV

#### Mathematically chromatic aberration = f R– fV

w = Dispersion power of lens.

= ωf y

#### fy = Focal length for mean colour =

Removal : To remove this defect i.e. for Achromatism we use two or more lenses in contact in place of single lens.

#### Mathematically condition of Achromatism is :

w1w 2

= 0 or w1 f2 = –w 2 f1

f1        f2

#### Note : @              Component lenses of an achromatic doublet cemented by canada blasam because it is transparent and has a refractive index almost equal to the refractive of the glass.

• Spherical aberration : Inability of a lens to form the point image of a point object on the axis is called Spherical

#### In this defect all the rays passing through a lens are not focussed at a single point and the image of a point object on the axis is blurred.

Removal : A simple method to reduce spherical aberration is to use a stop before and infront of the lens. (but this method reduces the intensity of the image as most of the light is cut off). Also by using plano-convex lens, using two lenses separated by distance d = F F ‘, using crossed lens.

#### Note : @              Marginal rays : The rays farthest from the principal axis.

Paraxial rays : The rays close to the principal axis.

#### @          Spherical aberration can be reduced by either stopping paraxial rays or marginal rays, which can be done by using a circular annular mask over the lens.

@          Parabolic mirrors are free from spherical aberration.

#### (iii)    Coma : When the point object is placed away from the principle axis and the image is received on a screen perpendicular to the axis, the shape of the image is like a comet. This defect is called Coma.

It refers to spreading of a point object in a plane ^ to principle axis.

#### Removal : It can be reduced by properly designing radii of curvature of the lens surfaces. It can also be reduced by appropriate stops placed at appropriate distances from the lens.

• Curvature : For a point object placed off the axis, the image is spread both along and perpendicular to the principal The best image is, in general, obtained not on a plane but on a curved surface. This defect is known as Curvature.

#### Removal : Astigmatism or the curvature may be reduced by using proper stops placed at proper locations along the axis.

• Distortion : When extended objects are imaged, different portions of the object are in general at different distances from the axis. The magnification is not the same for all portions of the extended object. As a result a line object is not imaged into a line but into a

#### Astigmatism : The spreading of image (of a point object placed away from the principal axis) along the principal axis is called Astigmatism.

Example: 18       A thin lens focal length

f1 and its aperture has diameter d. It forms an image of intensity I. Now the central

part of the aperture upto diameter d/2 is blocked by an opaque paper. The focal length and image intensity will change to                                                                                          [CPMT 1989; MP PET 1997; KCET 1998]

(a)

f and I                            (b) f and I                      (c)   3 f and I                         (d)    f and 3I

2         2                                     4                         4           2

d                                     1

4

 ç
 ÷

æ         pd 2 ö

Solution: (d)        Centre part of the aperture up to diameter

2

is blocked i.e.      th

area is blocked

ç A =

è

÷ . Hence

4 ø

remaining area

A¢ = 3 A . Also, we know that intensity µ Area Þ

I ¢ = A¢ = 3

Þ    I ¢ = 3 I .

4

Focal length doesn’t depend upon aperture.

I        A       4                 4

Example: 19       The power of a thin convex lens (a mg = 1.5) is + 5.0 D. When it is placed in a liquid of refractive index a m l ,

then it behaves as a concave lens of local length 100 cm. The refractive index of the liquid a m l

will be

(a) 5 / 3                           (b) 4 / 3                        (c)                                   (d) 5 / 4

Solution: (a)        By using

fl  = a m g  1 ; where

m  m g  1.5

and f

= 1 = 1 m = 20 cm

fa            l m g – 1

l  g           ml             ml

a       P       5

Þ      -1001.5 1 Þ

m l = 5 / 3

20       1.5 – 1

###### m l

Example: 20       A double convex lens made of a material of refractive index 1.5 and having a focal length of 10 cm is immersed in liquid of refractive index 3.0. The lens will behave as                                                       [NCERT 1973]

• Diverging lens of focal length 10 cm (b) Diverging lens of focal length 10 / 3 cm

(c) Converging lens of focal length 10 / 3 cm                       (d) Converging lens of focal length 30 cm

Solution: (a)        By using

fl  = a m g 1 Þ        fl

1.5 1 Þ

f   = -10 cm

(i.e. diverging lens)

fa            l m g – 1

+ 10

1.5 – 1          l

3

Example: 21 Figure given below shows a beam of light converging at point P. When a concave lens of focal length 16 cm is introduced in the path of the beam at a place O shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value x will be equal to                                              [AMU (Med.) 2002]

• 12 cm
• 24 cm
• 36 cm
• 48 cm

Solution: (d)        From the figure shown it is clear that

For lens : u = 12 cm and v = x = ?

By using

1 = 1 – 1                                                                                              P

f      v     u

Þ     1

+ 16

= 1 –

x

1

+ 12

Þ x  = 48 cm.

Example: 22       A convex lens of focal length 40 cm is an contact with a concave lens of focal length 25 cm. The power of combination is                                                                            [IIT-JEE 1982; AFMC 1997; CBSE PMT 2000]

(a) – 1.5 D                               (b) – 6.5 D                           (c) + 6.5 D                             (d) + 6.67 D

Solution: (a)        By using

1 = 1 + 1

F       f1      f2

Þ   1 =

F

1    +

+ 40

1

– 25

Þ     F = – 200 cm , hence

3

P = 100 =

f(cm)

100

– 200 / 3

= -1.5 D

Example: 23       A combination of two thin lenses with focal lengths

f1 and

f2 respectively forms an image of distant object at

distance 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination

when two lenses are separated by 10 cm. The corresponding values of

f1 and

f2 are                  [AIIMS 1995]

(a) 30 cm, – 60 cm                (b) 20 cm, – 30 cm            (c) 15 cm, – 20 cm                (d) 12 cm, – 15 cm

Solution: (b)        Initially F = 60 cm (Focal length of combination)

Hence by using

1 = 1 + 1

F       f1      f2

Þ 1 + 1 = 1

f1      f2       60

Þ     f1 f2 f1 + f2

……(i)

Finally by using

1 = 1 + 1

F ¢       f1      f2

•  ff2

where  F ¢ = 30 cm  and d = 10 cm         Þ

1 = 1 + 1

30     f1       f2

– 10

ff2

……(ii)

From equations (i) and (ii)

f1 f2 = – 600.

From equation (i)

f1 + f2 = -10

…..(iii)

Also, difference of focal lengths can written as

f1 – f2 =

Þ   f1 – f2  = 50

…..(iv)

From (iii) ´ (iv)

f1 = 20

and

f2 = -30

Example: 24       A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with refractive index 1.65. Its focal length is nearly                                                                                                              [MP PMT 1997]

(a) 20 cm                                 (b) 31 cm                             (c) 35 cm                                 (d) 50 cm

 2 m – 1

Solution: (b)        By using f =   ( R     ) Þ

40

 2 1.65 – 1)
 =

f         (

= 30.7 cm » 31 cm.

Example: 25 A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ . The distance PO is equal to

[MP PMT 1994; Haryana CEE 1996]

(a) 5 R                                     (b) 3 R                                 (c) 2 R                                      (d) 1.5 R

Solution: (a)        By using

m 2 – m1 = m 2 – m1

v         u              R

Where

m1 = 1,

m 2 = 1.5,

u = – OP,         v = OQ

Hence

1.5 –

OQ

1

• OP

= 1.5 – 1 Þ

(+ R)

1.5 +   1

OP      OP

= 0.5

R

Þ OP = 5 R

Example: 26       The distance between an object and the screen is 100 cm. A lens produces an image on the screen when placed at either of the positions 40 cm apart. The power of the lens is                                                  [SCRA 1994]

(a) 3 D                                      (b) 5 D                                 (c) 7 D                                     (d) 9 D

Solution: (b)        By using

f D 2x 2   Þ

 =

4 D

f = 1002 – 402

4 ´ 100

= 21 cm

Hence power

100

P     F(cm)

= 100 » + 5D 21

Example: 27  Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length +20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is                                                                                                                          [NSEP 1994; DPMT 2000]

(a) + 80 cm

• – 80 cm
• – 24 cm
• – 100 cm

Solution: (d)        Here

1 = (1.6 – 1) æ 1 – 1 ö =  3

…….(i)

 ¥
 ø

ç                ÷

f1                      è          20      100

1 = (1.5 – 1) æ 1 – 1 ö =

…….(ii)

 è
 ø
 2

f                        ç 20     – 20 ÷       20

1 = (1.6 – 1) æ 1 – 1 ö =  3

…….(iii)

 è
 ¥
 3
 ø

f                        ç – 20        ÷     100

F                                f1         f2         f3

By using

1 = 1 + 1 + 1 Þ

F       f1      f2       f3

1 = -3 +

F       100

1 –   3

20    100

Þ F = – 100 cm

Example: 28       A concave lens of focal length 20 cm placed in contact with a plane mirror acts as a                                  [SCRA 1998]

• Convex mirror of focal length 10 cm (b) Concave mirror of focal length 40 cm

(c) Concave mirror of focal length 60 cm                       (d) Concave mirror of focal length 10 cm

Solution: (a)        By using

1 = 2 + 1

F       fl           fm

fl       20                                                            Þ                +

Since

fm = ¥ Þ

F =       =

2

2 = 10 cm

(After silvering concave lens behave as convex mirror)

F                                Fe                  Fm

Example: 29       A candle placed 25 cm from a lens, forms an image on a screen placed 75 cm on the other end of the lens. The focal length and type of the lens should be                                                                           [KCET (Med.) 2000]

(a) + 18.75 cm and convex lens                              (b) – 18.75 cm and concave lens

(c) + 20.25 cm and convex lens                              (d) – 20.25 cm and concave lens

Solution: (a)        In concave lens, image is always formed on the same side of the object. Hence the given lens is a convex lens for which u = – 25 cm,                        v = 75 cm.

By using

11 – 1

Þ 1 = ( 1

 – 25

) (      )

Þ   f = + 18.75 cm.

 f
 v
 u
 f
 + 75

Example: 30       A convex lens forms a real image of an object for its two different positions on a screen. If height of the image in both the cases be 8 cm and 2 cm, then height of the object is                                             [KCET (Engg./Med.) 2000, 2001]

• 16 cm (b) 8 cm                                 (c) 4 cm                                     (d) 2 cm

Solution: (c)         By using

O =                         Þ  O =

= 4 cm

Example: 31       A convex lens produces a real image m times the size of the object. What will be the distance of the object from the lens                                                                                                                        [JIPMER 2002]

(a)

æ m + 1 ö f

(m – 1)f

æ m – 1 ö f

m + 1

 m

ç            ÷                                                                                     ç            ÷

è            ø                                                                                     è   m    ø                                          f

f                                     (+ f )

• f + u u

æ m + 1 ö

Solution: (a)        By using m =

f + u

here – m = (+ f ) + u Þ        – m =       f

= 1 + f

Þ u = – ç

è

÷ . f

 m

ø

Example: 32       An air bubble in a glass sphere having 4 cm diameter appears 1 cm from surface nearest to eye when looked along diameter. If a m g = 1.5 , the distance of bubble from refracting surface is                                        [CPMT 2002]

(a) 1.2 cm                                (b) 3.2 cm                           (c) 2.8 cm                                (d) 1.6 cm

Solution: (a)        By using

m 2 – m1

v         u

m 2m1

R

where u = ? , v = 1 cm,

m1 = 1.5 ,

m 2 = 1 , R = – 2 cm.

1 – 1.51 1.5

Þ u = – 6 = -1.2 cm.

• 1 u

(- 2)                     5

Example: 33       The sun’s diameter is 1.4 ´ 109 m and its distance from the earth is 1011 m . The diameter of its image, formed by a convex lens of focal length 2m will be                                                                                                   [MP PET 2000]

• 7 cm (b) 1.4 cm                           (c) 2.8 cm                               (d) Zero (i.e. point image)

Solution: (c)         From figure

D = 1011

d          2

Þ   d =

2 ´ 1.4 ´ 109

1011

= 2.8 cm.

Sun (D)

Example: 34 Two point light sources are 24 cm apart. Where should a convex lens of focal length 9 cm be put in between them from one source so that the images of both the sources are formed at the same place

(a) 6 cm                                   (b) 9 cm                               (c) 12 cm                                 (d) 15 cm

Solution: (a)    The given condition will be satisfied only if one source (S1) placed on one side such that u < f   (i.e. it lies under the focus). The other source (S2) is placed on the other side of the lens such that u > f (i.e. it lies beyond the focus).

If S1 is the object for lens then

1  1 –   1   Þ 11 – 1

……..(i)

f       y      x           y       x      f

If S2 is the object for lens then 1 = 1 –             1      Þ

11      1

……..(ii)

From equation (i) and (ii)

f       + y

– (24 – x)

y       f     (24 – x)

1 – 1

x      f

= 1 –

f

1      Þ

(24 – x)

1 +        1      =

x      (24 – x)

2 = 2

f       9

Þ x 2 – 24 x + 108 = 0

On solving the equation x = 18 cm , 6 cm                                                                                y

Example: 35       There is an equiconvex glass lens with radius of each face as R and a m g

water in object space and air in image space, then the focal length is

= 3 / 2 and a mw = 4 / 3 . If there is

(a) 2R                                       (b) R                                     (c) 3 R/2                          (d) R 2

Solution: (c)         Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium

æ 3 ö – æ 4 ö        4     3

m   m       m      m

ç 2 ÷     ç 3 ÷

2        1 =    1 +    2 Þ

è   ø     è   ø = 3 + 2

Þ v1 = 9R

R             u      v1

R               ¥      v1

I1

Now consider the refraction at the second surface of the lens i.e.

refraction from denser medium to rarer medium

1 – 3         3

2 = – 2 + 1 Þ v

= æ 3 öR

• R 9R      v

2     ç 2 ÷

2              è ø

The image will be formed at a distance do

3 R . This is equal to the focal length of the lens.

2

Tricky example: 4

A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other side of the lens. At what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order to have an upright image of the object coincident with it

[CBSE PMT 1998; JIPMER 2001, 2002]

(a) 12 cm                                 (b) 30 cm                                     (c) 50 cm                             (d) 60 cm

Solution : (c)      For lens u = 30 cm,

f = 20 cm

, hence by using 1 = 1 – 1 Þ     = 1 –     Þ v = 60 cm

f      v      u         + 20     v      – 30

The final image will coincide the object, if light ray falls normally on convex mirror as shown. From figure it is seen clear that reparation between lens and mirror is 60 – 10 = 50 cm.

Tricky example: 5

O

30 cm

60 cm

I

10 cm

A convex lens of local length 30 cm and a concave lens of 10 cm focal length are placed so as to have the same axis. If a parallel beam of light falling on convex lens leaves concave lens as a parallel beam, then the distance between two lenses will be

(a) 40 cm                                 (b) 30 cm                                     (c) 20 cm                             (d) 10 cm

d

Solution : (c)      According to figure the combination behaves as plane glass plate (i.e., F= ¥)

# Prism

#### Prism is a transparent medium bounded by refracting surfaces, such that the incident surface (on which light ray is incidenting) and emergent surface (from which light rays emerges) are plane and non parallel.

Commonly used prism :

• Refraction through a prism

A = r1 + r2 and i + e = A + d

sin i

i – Angle of incidence, e – Angle of emergence,

#### For surface

AC m =

;    A – Angle of prism or refracting angle of prism,

sin r1

r1 and r2 – Angle of refraction,

For surface AB m =

### (2)  Deviation through a prism

sin r2

#### sin e

d – Angle of deviation

#### For thin prism d

= (m – 1)A . Also deviation is different for different colour light e.g. m R < mV

so d R < d V .

mFlint > mCrown

so d F

• d C

 (ii) r = A and2 i = A + d m 2 (iii) m = sin isin A / 2 or m = sin A + d m2sin A / 2

Note : @              If d m = A

then m = 2 cos A / 2

### (3)  Normal incidence on a prism

#### If light ray incident normally on any surface of prism as shown

In any of the above case use m =

sin i

and d

= i A

### (4)  Grazing emergence and TIR through a prism

#### When a light ray falls on one surface of prism, it is not necessary that it will exit out from the prism. It may or may not be exit out as shown below

Note    :    @     For    the    condition   of    grazing   emergence.  Minimum    angle    of    incidence

 ë

imin = sin 1 éê

sinA cosAù .

 ú

û

### (5) Dispersion through a prism

#### The splitting of white light into it’s constituent colours is called dispersion of light.

• Angular dispersion (q ) : Angular separation between extreme colours e.

#### depends upon m and A.

= ωV ωR = (μV μR )A . It

#### (ii)   Dispersive power (w) :

w = q

m Vm R

where ìm

m V m R ü

d y           m y – 1

í y                     2     ý

 î
 þ

#### Þ It depends only upon the material of the prism i.e. m and it doesn’t depends upon angle of prism A

Note : @ Remember w Flint

• wCrown .

### (6)  Combination of prisms

Two prisms (made of crown and flint material) are combined to get either dispersion only or deviation only.

# Scattering of Light

#### Molecules of a medium after absorbing incoming light radiations, emits them in all direction. This phenomenon is called Scattering.

• According to scientist Rayleigh : Intensity of scattered light µ 1

l4

• Some phenomenon based on scattering : (i) Sky looks blue due to scattering.

#### (ii) At the time of sunrise or sunset it looks reddish. (iii) Danger signals are made from red.

• Elastic scattering : When the wavelength of radiation remains unchanged, the scattering is called
• Inelastic scattering (Raman’s effect) : Under specific condition, light can also suffer inelastic scattering from molecules in which it’s wavelength

# Rainbow

#### Rainbow is formed due to the dispersion of light suffering refraction and TIR in the droplets present in the atmosphere.

• Primary rainbow : (i) Two refraction and one (ii) Innermost arc is violet and outermost is red. (iii) Subtends an angle of 42o at the eye of the observer. (iv) More bright
• Secondary rainbow : (i) Two refraction and two

#### (ii)   Innermost arc is red and outermost is violet.

• It subtends an angle of Comparatively less

52.5 o at the eye. (iv)

# Colours

Colour is defined as the sensation received by the eye (rod cells of the eye) due to light coming from an object.

• Types of colours

Green (P)                                                                                                         Yellow (P)

Cyan (S)

white

Yellow (S)

Green (S)

Black

Orange (S)

Blue (P)                                               Red (P)

Magenta (S)

Blue (P)

Radish violet (S) (Mauve)

Red (P)

• Complementary colours :

Green and magenta Blue and yellow Red and cyan

• Combination :

Green + red + blue = White Blue + yellow = White

Red + cyan = White Green + magenta = White

• Complementary colours :

yellow and mauve Red and green Blue and orange

• Combination :

Yellow + red + blue = Black Blue + orange = Black

Red + green = Black Yellow + mauve = Black

#### (2)   Colours of object : The perception of a colour by eye depends on the nature of object and the light incident on it.

• Due to selective (i) Due to selective transmission.

• A rose appears red in white light because it reflects red colour and absorbs all remaining
• When yellow light falls on a bunch of flowers, then yellow and white flowers looks Other flowers
• A red glass appears red because it absorbs all colours, except red which it
• When we look on objects through a green glass or green filter then green and white objects will appear

Note : @A hot object will emit light of that colour only which it has observed when it was heated.

# Spectrum.

#### The ordered arrangements of radiations according to wavelengths or frequencies is called Spectrum. Spectrum can be divided in two parts (I) Emission spectrum and (II) Absorption spectrum.

• Emission spectrum : When light emitted by a self luminous object is dispersed by a prism to get the spectrum, the spectrum is called emission

#### (2)   Absorption spectrum : When white light passes through a semi-transparent solid, or liquid or gas, it’s spectrum contains certain dark lines or bands, such spectrum is called absorption spectrum (of the substance through which light is passed).

• Substances in atomic state produces line absorption Polyatomic substances such as

H 2 , CO2

and

#### (ii)   Absorption spectra of sodium vapour have two (yellow lines) wavelengths

D1 (5890 Å) and D2 (5896 Å)

#### Note : @ If a substance emits spectral lines at high temperature then it absorbs the same lines at low temperature. This is Kirchoff’s law.

• Fraunhoffer’s lines : The central part (photosphere) of the sun is very hot and emits all possible wavelengths of the visible light. However, the outer part (chromosphere) consists of vapours of different elements. When the light emitted from the photosphere passes through the chromosphere, certain wavelengths are absorbed. Hence, in the spectrum of sunlight a large number of dark lines are seen called Fraunhoffer

#### (i)   The prominent lines in the yellow part of the visible spectrum were labelled as D-lines, those in blue part as

F-lines and in red part as C-line.

#### (ii)    From the study of Fraunhoffer’s lines the presence of various elements in the sun’s atmosphere can be identified e.g. abundance of hydrogen and helium.

• Spectrometer : A spectrometer is used for obtaining pure spectrum of a source in laboratory and calculation of m of material of prism and m of a transparent

#### It consists of three parts : Collimator which provides a parallel beam of light; Prism Table for holding the prism and Telescope for observing the spectrum and making measurements on it.

The telescope is first set for parallel rays and then collimator is set for parallel rays. When prism is set in minimum deviation position, the spectrum seen is pure spectrum. Angle of prism (A) and angle of minimum deviation (d m) are measured and m of material of prism is calculated using prism formula. For m of a transparent liquid, we take a hollow prism with thin glass sides. Fill it with the liquid and measure (d m) and A of liquid prism. m of liquid is calculated using prism formula.

#### (5)   Direct visionspectroscope: It is an instrument used to observe pure spectrum. It produces dispersion without deviation with the help of n crown prisms and (n – 1) flint prisms alternately arranged in a tabular structure.

For no deviation n(m – 1)A = (n – 1) (m ‘-1)A‘ .

Example: 36       When light rays are incident on a prism at an angle of 45o, the minimum deviation is obtained. If refractive

index of the material of prism is           , then the angle of prism will be                                             [MP PMT 1986]

(a) 30o                             (b) 40o                         (c) 50o                             (d) 60o

Solution: (d)

m = sin i Þ

= sin 45 Þ sin A =

= 1 Þ

A = 30 o Þ A = 60 o

sin A

2

sin A                2               2       2

2

Example: 37       Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of

prism is (cos 41o = 0.75)

(a) 62o                             (b) 41o                         (c) 82o                            (d) 31o

[MP PET/PMT 1988]

sin A + d m                  sin A + A

 2
 î
 þ

Solution: (c)         Given d

= A , then by using m =           2     Þ m =           2    =

sin A = 2 cos A

ìsin A = 2 sin A cos A ü

m                                                                                            A

sin 2

sin A

2

sin A                 2 í                         2          ý

2

Þ 1.5 = 2 cos A Þ 0.75 = cos A Þ 41o = A Þ A = 82o .

2                      2                2

Example: 38       Angle of glass prism is 60o and refractive index of the material of the prism is 1.414,then what will be the angle of incidence, so that ray should pass symmetrically through prism

(a) 38o 61′                          (b) 35o 35′                      (c) 45o                            (d) 53o 8′

Solution: (c)         incident ray and emergent ray are symmetrical in the cure, when prism is in minimum deviation position.

Hence in this condition m =

sin i

A

Þ sin i = m sin æ A ö Þ sin i = 1.414 ´ sin 30 o =

 ç     ÷
 è     ø

2

1 Þ i = 45 o

sin 2

Example: 39       A prism (m = 1.5) has the refracting angle of 30o. The deviation of a monochromatic ray incident normally on

its one surface will be (sin 48 o 36′ = 0.75)

(a) 18o 36′                          (b) 20o 30′                      (c) 18o                            (d) 22o 1′

[MP PMT/PET 1988]

Solution: (a)        By using

m = sin i Þ 1.5 = sin i  Þ sin i = 0.75 Þ i = 48 o 36′

sin A                   sin 30

Also from d = i A Þ d = 48 o 36′-30 o = 18 o 36′

Example: 40       Angle of a prism is 30o and its refractive index is

and one of the surface is silvered. At what angle of

incidence, a ray should be incident on one surface so that after reflection from the silvered surface, it retraces its path                                                                                                                    [MP PMT 1991; UPSEAT 2001]

(a) 30o                             (b) 60o                         (c) 45o                            (d)

Solution: (c)         This is the case when light ray is falling normally an second surface.

sin-1

Hence by using m =

sin i  Þ

sin A

=     sin i

sin 30 o

Þ sin i =

´ 1 Þ i = 45 o

2

Example: 41       The refracting angle of prism is A and refractive index of material of prism is

cot A . The angle of minimum

2

deviation is                                                                                                                           [CPMT 1992]

(a)

180 o – 3 A

(b)

180o + 2 A

(c)

90 oA

(d)

180o – 2 A

Solution: (d)        By using

sin A + d m

m =           2

sin A

2

Þ cot A =

2

sin A d m

2

sin A

2

cos A

Þ          2

sin A

2

sin A + d m

=              2

sin A

2

Þ sinæ 90 – A ö = sinæ A + d m ö Þ 90 – A =

A + d m Þ d

= 180 – 2A

 2

ç               ÷           ç               ÷

è               ø           è       2     ø

• 2 m

Example: 42 A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of the angle of the prism. The angle of deviation is                                                                                [MNR 1988; MP PMT 1999; Roorkee 2000; UPSEAT 2000]

(a) 45o                             (b) 39o                         (c) 20o                            (d) 30o

Solution: (d)        Given that

A = 60 o

and i = e = 3 A = 3 ´ 60 = 45 o

4        4

By using i + e = A + d Þ 45 + 45 = 60 + d Þ d = 30 o

Example: 43 PQR is a right angled prism with other angles as 60o and 30o. Refractive index of prism is 1.5. PQ has a thin layer of liquid. Light falls normally on the face PR. For total internal reflection, maximum refractive index of liquid is

(a) 1.4

(b) 1.3

(c) 1.2

(d) 1.6

Solution: (c)       For TIR at PQ q < C

From geometry of figure q = 60 i.e. 60 > C Þ sin 60 > sin C

Þ  3

2

• mLiquid Þ m

m Pr ism

Liquid              2

Pr ism

Þ m Liquid

<  3 ´ 1.5 Þ m

2

Liquid

< 1.3 .

Example: 44 Two identical prisms 1 and 2, each will angles of 30o, 60o and 90o are placed in contact as shown in figure. A ray of light passed through the combination in the position of minimum deviation and suffers a deviation of 30o. If the prism 2 is removed, then the angle of deviation of the same ray is                                              [PMT (Andhra) 1995]

• Equal to 15o
• Smaller than 30o
• More than 15o
• Equal to 30o

Solution: (a)

d = (m – 1)A

as A is halved, so d is also halves

Example: 45       A prism having an apex angle 4o and refraction index 1.5 is located in front of a vertical plane mirror as shown in figure. Through what total angle is the ray deviated after reflection from the mirror

Solution: (c)

• 176o
• 4o
• 178o
• 2o

d Pr ism = (m – 1)A = (1.5 – 1)4 o = 2o

d Total

= d Pr ism + d Mirror = (m – 1)A + (180 – 2i) = 2o + (180 – 2 ´ 2) = 178 o

Example: 46       A ray of light is incident to the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If m is the refractive index of the material of the prism, the maximum value of the base angle for which

light is totally reflected from the hypotenuse is                                                                            [EAMCET    2003]

-1 æ 1 ö

-1 æ 1 ö

1 æ m – 1 ö

-1 æ 1 ö

(a)

sin

ç m ÷

(b)

tan

ç m ÷

(c)

sin ç   m   ÷

(d)

cos

ç m ÷

è     ø                                             è     ø                                       è           ø                                      è     ø

Solution: (d)        If a = maximum value of vase angle for which light is totally reflected from hypotenuse.

(90 – a) = C = minimum value of angle of incidence an hypotenuse for TIR

sin(90 – a) = sin C = 1 Þ a = cos -1 æ 1 ö

m                    ç m ÷

è     ø

Example: 47 If the refractive indices of crown glass for red, yellow and violet colours are 1.5140, 1.5170 and 1.5318 respectively and for flint glass these are 1.6434, 1.6499 and 1.6852 respectively, then the dispersive powers for crown and flint glass are respectively                                                                                  [MP PET/PMT 1988]

(a) 0.034 and 0.064        (b) 0.064 and 0.034     (c) 1.00 and 0.064               (d) 0.034 and 1.0

mv mr

1.5318 – 1.5140

m m

1.6852 – 1.6434

Solution: (a)

w Crown =                   =

= 0.034 and

w Flint =  v                r  =

= 0.064

m y – 1

(1.5170 – 1)

– 1

1.6499 – 1

 m
 y

Example: 48 Flint glass prism is joined by a crown glass prism to produce dispersion without deviation. The refractive indices of these for mean rays are 1.602 and 1.500 respectively. Angle of prism of flint prism is 10o, then the angle of prism for crown prism will be                                                                                                      [DPMT 2001]

(a)

12o 2.4′

(b)

12o 4′

(c)

1.24 o

(d) 12o

Solution: (a)        For dispersion without deviation

AC = (m F – 1) Þ A(1.602 1) Þ

A = 12.04 o = 12o 2.4′

AF            (m C – 1)       10     (1.500 – 1)