In second case

100 3 images if same size. Size of object A is 4 times that of B. If object A is placed at a distance of 50cm from the mirror, what should be the distance of B from the mirror?

Solution: For object A For object B should an object be placed to get in image on the wall?

Solution:

f = − 10 cm , V = − 35 cm

Distance of the object form wall

= 35 – 14 = 21 cm

Q: 6) An object is placed at a distance of 36cm form a convex mirror. A plane mirror is placed in between so that the two virtual images so formed coincide. If the plane mirror is at a distance if 24cm from the object, find the radius of curvature of the convex mirror.

Solution:

OP = u = − 36 cm

cm PI

V = = + 12

18 1 36

3 1 12

1 36

1 1 1

1 = + = + = − + =

∴ f u V

cm f = 18

∴

cm f

R = 2 = 2 × 18 = 36

∴

Q: 7) A convex mirror of focal length ‘f’ forms an image which is n

1 times the object. The distance of the object

which is n

1 times the object. The distance of the object from the mirror is

Solution:

u

− V

= +

= η

η 1

η V = − u

u V f

1 1 1 = +

∴

η

Q: 8) An object of size 7.5cm is placed in front of a convex mirror of radius of curvature 25cm at a distance of 40cm.

The size of the image should be Solution:

Q: 9) The image formed by a convex mirror of focal length 30cm is a quarter of the size of the object. The distance of the object from the mirror is

Solution:

Solution: On immersing a mirror in water, focal length of the mirror remains uncharged.

Q: 11) An object is 20cm away form a concave mirror with focal-length 15cm. If the object moves with a speed of 5m/s along the axis, then the speed of the image will be

Solution:

Q: 12) A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical when Sun-light falls normally in the mirror, it is focused at distance of 32cm form the mirror. If the tank filled with water

Solution: f 5cm 2

2 = =

=

For part PQ

0

1

L

u f L f

 

 

= −

( ²⁰ ) ³

5

5

₀

0

L = − L

 ×

 



−

−

−

= −

For part QR

0 2

2

L

u f L f  

 





= −

( ²⁰ ) ⁹

5

5

₀

0

2

L

L =

 ×

 



−

−

−

= −

1 3

2 1

=

∴ L L

•

CONCEPT OF NEWTON’S FORMULA (FOR A MIRROR)

In this formula, the object and image distance are expressed w.r.t. focus. Consider an object O kept beyond ‘C’

of a concave mirror, and whose image is formed at I with in C.

Let OF = x and IF = y From triangle OMC

( π α ) α

θ sin sin

sin

OM OM

OC =

= −

^{…………(1)}

And from triangle ICM

α θ sin sin

IM

IC = ………….(2)

Dividing equation (1) and (2) yields IP

OP IM OM IC

OC = = (since M is close to P)

(or)

y f

f x y f

f x

+

= +

−

−

y f f y x f x y f y x f f

x −

²

+ − = − +

²

−

(or)

x y = f

² y x f =

∴

1) As

x y = f

² (or)

x 1 y

α

(i.e.) The distance of object and image form the focus are inversely proportional to each other. In other words, the more the object distance (from the focus), the less will be the image distance (from the focus) and vice versa

2) If

x → 0 ; y → ∞

and if

x → ∞ ; y → 0

. If the object is at focus the image is a far off distance and vice-versa.

3) From

xy = f

²

; if x = f

, then

y = f

^.

Thus, if the object be at C, then image will also be at C (for a concave mirror) and if the object is at P, then the image will also be at P (for a convex mirror)

4) Since

f

² is necessarily +ve for both types of mirror, so x and y bear the same sign, which implies that both the object and the image always lie an the same side of focus.

A) GRAPH OF |x| Versus |y| :

Since

xy = f

² represents a rectangular hyperbola, existing in the first and third quadrant (

f

² being positive).

∴

The graph of |y| vs |x| will be a rectangular hyperbola existing only in the first quadrant.

B) GRAPH OF U Versus V :

Since

xy = f

²

( ^{u − f} )( ^{v − f} ) ^{= f}

²

∴

For a convex mirror, u is always negative and V is always positive. Further ‘f’ is also positive.

f f and y v x

u = = = −

∴ ,

We have, form

( ^{u − f} )( ^{v − f} ) ^{= f}

²

( ^{u + f} )( ^{y + f} ) ^{= f}

²

Or

[ ^{x −} ( ) ^{− f} ] [ ^{y −} ( ) ^{− f} ] ^{= f}

²

Evidently it is again an equation of a rectangular hyperbola with origin of coordinates shifted to the point (-f, -f) (see figure

↓

⁾

3) GRAPH OF v

1 VERSUS u 1

From mirror formula

f

u v

1 1 1 + =

Putting x

u1 =

and y

v =

1 , we have

y y

x 1

= +

It is the equation of a straight line having a slope +1 (or) -1 according as u and v bear the same

(or) opposite signs. The intercepts on x and y axis are each

( )

or f f

1

1 −

+

according as the

object and image are to the right (or) left of the mirror.

For a concave mirror:

u is always –ve

v can be positive (or) negative and f is –ve.

For a convex mirror u is always negative v is always positive and f is always positive.

•

CONCEPT OF CRITICLA ANGLE

When a ray of light is traveling form denser medium to rarer medium, it get refracted and the ray derivates away form the normal.

If we keep increasing angle of incidence then at an angle, the angle of refraction becomes 90⁰ . This is known as Critical –Angle (c).

When angle of incidence is increased, further the ray gets reflected back in the same medium. This phenomena is known as T.I.R.

According to Snell’s Law

C a

b

µ = sin i

 

 



 =

C a

b

i sin µ 1

⇒

C depends on colour and and temp

⇒

^{CRed > Cviolet}

∴

^{CRed < Cviolet}

⇒

^{If temp}

↑ C ↓

Q: 1) The sum (diameter d) subtends an angle

θ

radius at the pole of a concave mirror of focal length f. Find te diameter of the image of sun formed by mirror?

Solution:

f u v

1 1

1 + =

we get

f v = − 1

1

(

∴

u is very large so

1 0 u ≈

Or

v = − f

It means image is formed at focus Taking

' f '

as radius and using

f r and d

r when = =

= l l

θ

f d f or

d θ

θ = =

∴

•

REFRACTION AT SPHERICAL SURFACE:

From

∆

^le

s

OBC and IBC We have

i = α + β

And

β = ^r + ^r ( ) ^{or r} = β − ^r

From Snell’s law

For small angle of incidence I, we can write

sin i ≈ i and sin r ≈ r

After simplifying we get

This formula is derived for convex surface and for real Image

Analysis: Here two cases a rise. Refractive index is varying either as function of y (or) function of x.

Case-I:

µ = ^f ( ) ( ) ^{y i .e .}

Refractive index varies with y

At some height h angle of incidence is

θ

_y and refractive index is

^f ( ) ^y

from Snell’s Law

= θ

µ sin

constant

( ) ^{y y}

f θ

α sin

sin

1 × =

∴

^……..(1)

Slope of curve at A

( ^y )

dx

dy = tan 90 − θ dx y = dy

⇒ cot θ

From equation (1)

{ ( ) }

α α sin

sin

²

2

−

= f y dx

dy

Case-I:

µ = ^f ( ) ( ) ^{y i .e .}

Refractive index varies as function of x.

According to Snell’s Law

= θ

µ sin

constant

For initial refraction at the air medium interface

(

0

)

refracted ray.

Solution: We can use result derived above for which

( ) ^{y = 1 + y and α = 90}

⁰

Solution: For refraction at first face

R

For refraction at second face

1

•

PROBLEMS ON REFRACTION

1) A light ray is incident at an angle of incidence double that of refraction on one face of a parallel sided

As

r

the lateral displacement suffered by the light ray to the maximum value which it could have suffered?

Solution:

, 2

Q: 3) Light of a certain colour has 2000 waves in one millimeter of air. Find the number of waves of that light in one millimeter length of water and glass respectively?

Solution:

λ

_a

=

wavelength in air

m

=

λ

wavelength in medium

The number of waves of that light in a length of ‘d’ will be

m

t

_chloroform

m

/

beaker filled with wager to a height 6cm & placed over the block. Find the apparent shift of the position of the coin, when viewed from a point directly above it?

Solution:

Q: 6) The time taken by light to cover a distance of 9mm in water is____

Solution:

C

_w

10 m / s

this surface is_____

Solution:

•

REFRACTION AT SPERICAL SURFACES

Q: 1) Sunshine recorder globe of 30cm diameter is made of glass of refractive index

µ = 1 . 5

. A ray enters the globe parallel to the axis. Find the position from the centre of the sphere where the ray crosses the principal axis?

Solution:

For first refraction (Rares to deuces)

5

For second refraction (douses to rarer)

( ) ^cm

Q: 2) A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find the amplitude of oscillation of the image of the particle?

Solution: When the particle is

At R

When the particle is at left extreme position

cm

∴

Amplitude of oscillation of image 2

Q: 3) A point object is moving with velocity 0.01m/s on principal axis towards a convex lens of local-length 0.3m when object is a distance of 0.4m form the lens, find

a) Rate of charge of position of the image and b) Rate of charge of lateral magnification of image Solution: Differentiating the

a) Equation

u

cm

Q: 4) Find the position of the image formed by the lens combination given in figure?

Solution: Image formed by first lens

( ) 10

The image formed by the first lens server as the object for the second.

This is at a distance of (15 – 5)cm i.e. 10cm to the right of the second lens. It is a virtual object

Now

10

10 0

The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.

30

The final image is formed 30cm to the fight of the third lens.

Q: 5) Two Plano-concave lenses of glass of refractive index 1.5 have radii of curvature 20cm and 30cm respectively.

They are placed in contact with the curved surfaces towards each other and the space between them is filled with a liquid of refractive index 5/2. Find the focal length of the

combination.

Solution: For first plano concave lens

R cm

For second plano concave lens

R cm

The focal length of the liquid lens is given by

(

⁻

)

_^{ +} _^

Q: 6) Given the object image and principal axis find the positions and nature of the lens Solution: First join the object and image

If object and image points are

Above the principal axis and image point is higher, then the lens is convex and is present between the image and object points.

Other wise the lens is concave.

Q: 7) For the given positions of the objects and the image in figure determine the location and the nature of the lens used?

Solution:

Q: 8) A ray of light passes through a medium whose refractive index varies with distance as





 

  +

= a

1 x µ

0

µ

^{. If the}

ray enters the medium parallel to the x-axis, what will be the trajectory of the ray and what will be the time taken for the ray to travel a distance a?

Solution: The ray enters normally and proceeds along a straight line. At a distance ‘x’ in the medium consider a slab of thickness “dx”. Velocity of the light ray at this point is

Time taken to cross the distance ‘dx’ is

 

 

  +

×

=

 

 

  +

=

= a

x C

dx a x C dx V

dt dx 1

1

µ

0

µ

∴ Total time of travel is

∫

^_^{ + }_^

= ^a

a x C

t dx

0

0 1 µ C

0a 2 3 µ

=

Q: 9) A fish is rising up vertically inside a pond with velocity 4cm/s and notices a bird which is diving vertically downward and in velocity appears to be 16cm/s (to the fish). What is the velocity of the diving bird, if R. I of water is 4/3?

Solution: Vbf =Vb −

( )

−Vf f

b V

V +

=

16

V V_b

=

∴ µ

4

16 = V

_b

+

V 12 3 4 =

= 12

V

b V 12 9 cm/s

4 3× =

=

Q: 10) Solar rays are incident at 45⁰ on the surface of water

( µ = ^{4 / 3} )

. What is the length of the shadow of a pole of length 1.2m erected at the bottom of the pond, if the pole is vertical assuming that 0.2m of the pole is above the water surface?

Solution: Applying Snell’s law at point c

θ 3sin 45 4 sin 1× ⁰ =

2 4 sin θ = 3

Here AE =CD=0.2m BE BE CE

BC = =

= 1

tanθ

( )

²

1 sin

BE BE

= + θ

( )

²

2 1 4

3

BE BE

= +

m BE = 0 . 625

∴ The length of shadow = AB = AE + EB = 0.2+0.625 = 0.825

Q: 11) In a lake, a fish rising vertically to the surface of water uniformly at the rate of 3m/s, observes a bird diving vertically towards the water at a rate of 9m/s vertically above it. The actual velocity of the dive of the bird is_______

( µ = ^{4 / 3} )

Solution:

D R

D A

.

= . µ

y

= y ' µ

y y = µ

∴ ' ' y x h = +

∴

y x h = + µ

Differentiating

dt dy dt

dx dt

dh= +µ

dt µdy +

=3 9

( ) ^{m s}

t d

y

d 4 . 5 /

3 / 4

6 =

=

Q: 12) A convex lens of focal length 0.2m is cut into two halves each of which is displaced by 0.0005m and a point object is placed at a distance of 0.3m form the lens, as shown in figure. The position of the image is ________

Solution:

u v f

1 1 1 = −

u f v

1 1 1 = +

m

u=−0.3

f = 0 . 2

3

. 0

1 2 . 0

1

1 = −

v

m v=0.6

Q: 13) A pole 5m high is situated on a horizontal surface. Sun rays are incident at an angle 30⁰ with the vertical. The size of shadow in horizontal surface is______

Solution:

30 5

tan ⁰ = BC

m

BC 3

30 5 tan

5

⁰

=

=

Q: 14) The Sun subtends an angle

α = 0 . 5

⁰ at the pole of a concave mirror. The radius is curvature of concave mirror is R = 1.5m. The size of image formed by the concave mirror is_____

Solution: As Sun is at infinity image is formed at the focus of mirror

1) 0.5 108

2

1× 0×

=

180 105 5 . 2 0

1× × ×

= π

= 0.654 cm u POQ= = D^S

∠ α Or

=

S i

D

D

magnification

u f u v =

=

u

R D

D

S i

= 2

S

i

D

u D = R ×

2

•

MAGNIFICAIOTN IN CASE OF CURVED SURFACE:

Consider an extended object “QO” placed

⊥

^1r to the principal axis at the point O. The image of the point ‘O’

is formed at I. Image of extended object is ‘IQ’

∴ From

∆

^les

POQ and PIQ '

we can say that PO

= OQ

tanθ1 and

I P

Q I ' tan θ

₂

=

But since

θ

₁

and θ

₂ are very small, we can approximate

Also

µ

₁

sin θ

₁

= µ

₂

sin θ

₂ Sign conversion

2 2 1

1

θ µ θ

µ = h

₀

= + OQ

PI IQ PO

OQ '

.

. ₂

1 µ

µ =

∴

h

_i

= − IQ '

( ) ( )

( ) ^{v h i}

u u

h = −

∴ µ

₁

−

⁰ ₂

.

u=−PO

u V h

i h

2 1

0

µ

= µ

^V =+^PI

•

LATERAL MAGNIFICATION

The lateral magnification is dx

m_L = dv

Which can be obtained by differentiating equation

( )

R u

v

1 2 1

2 µ µ µ

µ − = −

Thus − ₂² . + ₂¹ =0 u du dv v

µ µ

•

REFRACTION AT PRISM

A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle.

a) Calculation of angle of Derivation:

In passing through the prism, ray KL suffers two refractions and has tweed through an

LQPN = δ

. (with is angle of direction)

In

∆ PLM

^,

In Quadrilateral ALOM,

As

∠ L + ∠ M = 180

⁰ (

∴

Sum of 4 angles of a Quad = 360) If

' µ '

is the refractive index of the material of the prism, then according to Snell’s Law

1 Putting the above two in equation (4) we get

( ^{u r} + ^{u r} ) − ^A

=

1 2

δ = µ ( ^r

1

+ ^r

2

) − ^A δ = ( − ¹ ) ^A

∴ δ µ

(when refracting angle are small)

This is the angle through which a ray derivate on passing through a thin prism of small refracting angle A.

Prism formula:

In minimum derivation position i1 = i2 = i and r1 = r2 = r

And from equation (4)

( ) ^{i i A}

m

= + −

δ

A

•

MAXIMUM DEVIATION:

For

i

₁

= 90

⁰

For surface CD

2

CONCEPT: Sometimes a part of a prism is given and the keep on thinking whether how should we proceed?

To solve such problems first complete the prism then solve as the problems of prism are solved

Defects in Image formed by leuses:

The defect in the lens on account of which it does not form a white point image of white point object is defined as Aberration.

Axial Chromatic Aberration: The variation of the image distance form the lens with the colour measures axial chromatic Aberration.

Lateral Chromatic Aberration: The variation in the size of the image with colour measures the lateral transverse chromatic aberration.

Concept of Dispersion of Light:

Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent colours on passing through a prism. (accuse due to wavelength).

The band of seven colours so obtained is called the visible spectrum.

The order of colours from the lower end of spectrum is “VIBGYOR”.

Violet colour deviates through maxi. Value and red colour deviates through the minimum angle.

Causes of Dispersion: Each colour has in own wavelength according to Cauchy’s formula R.I. of a material depends on wavelength

( ) λ

...

4

2 + +

+

= λ λ

µ B C

A

For a prism of small refracting angle of deviation is

Angular dispersion: It is the angle in which all colous of white light are contained dv = deviation of violet colour

dr = deviation of red colour

( ) θ ∴

Angular dispersion = dv - dr

As

^d

V

= µ (

V

− 1 ) ^A

Dispersive Power (w): Ratio of angular deviation to the mean position produced by the prism.

( )

Note: Single prism produces both deviation and dispersion simultaneously. It cannot give deviation without dispersion (or) dispersion without deviation. However a suitable combination of two prisms can do so.

Dispersion of light occurs because velocity of light in a material depends upon its colour There is no dispersion of light refracted through a rectangular glass slab.

Combinations of prim:

I) Deviation without dispersion:

Net dispersion = 0, Net deviation

≠

⁰ Necessary condition

(

δV ⁻δR

)

⁺

(

δV^{1 −}δR¹

)

⁼⁰

(

µV ⁻µR

)

^A+

(

µV^{' −}µ ^'R

)

^A'=0

In this Situation

Net deviation = 1⁺ '⁼

(

⁻1

) (

⁺ '⁻1

)

'⁼

(

⁻1

)

⁺

(

^{+ −}1

)

_^− ' ₋⁻ ' _^

The prism which produces deviation without dispersion is called achromatic prism.

II) Dispersion without deviation:

Net deviation = 0, Net dispersion

≠

⁰

In this situation

If ω >' ω, the resultant dispersion is negative. (i.e.) opposite to that produced by the first prism.

This prism which produces dispersion without derivation is called direct vision prism.

Q: 1) A prism having an apex angle 4⁰ and refractive index 1.5 is located in front of a vertical plane mirror. Through what total angle is the ray deviated after reflected form the mirror?

Solution: δprism ⁼

(

µ⁻¹

) (

^{A= 1.5−1}

)

^{40 =20}

Q: 2) A container contains water upto a height of 20cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius ‘r’ floats centrally on the water. The ceiling of the from is 2.0m above the wager surface.

(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15cm.

(b) Find the maximum value of ‘r’ for which the shadow of the ring is formed on the ceiling

( µ

w

^{= 4 / 3} )

? Solution: a) Using Snell’s Law

r i sin sin = µ

2 2

2

2

200

1 20 15

15 3

4

× + + =

×

x x

cm x 3

=800

∴

∴ Radius of shadow =

3 15+800 cm

3

=845

= 2 . 81 m

b) For shadow to be formed angle of incidence should be less than critical angle using Snell’s Law

0 2

2 max

max

1 sin 90

20 3

4 = ×

+ r V

r

2 2

max 2

max 9 9 20

16r = r + ×

2 2

max 9 20

7r = ×

m cm

r 20 0 . 2268

7 9

max

= × =

Q: 3) Monochromatic light falls on a right angled prism at an angle of incidence 45⁰. The emergent light is found to slide along the face AC. Find the refractive index of material of prism?

Solution:

r

₂

= θ

C ………..(1)

0 2

1

+ r = A = 90

r

cos

1

1 = r

µ

^{and 1 2}

2 1

1 1 cos

1

sinr = − r = − µ Snell’s law AR

1

0

sin

45 sin

1 − µ r

2

0 1

1 45

sin =µ −µ

1 2

1

₂

−

= µ

2 1

1 = µ2 − 5 . 2 1

2 = 3 = µ

5 .

= 1 µ

In document Optics (Page 27-58)