Page 14

1. Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident onthe unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere willbe a :(A*) Parallel beam (B) Converging beam (C) Slightly divergent beam (D) Widely divergent beam

Sol. Refraction from surface 1 : 1

2 1v =

(2 1)( )R

u = 2

Surface 1 Surface 2

v1 = +2RReflection from silvered surface 2 : Since object is at pole so image also is at pole.

Refraction from surface 2 : 1

1 22v R

=

1 2( )R

v2 = Therefore the final emergent light will emergent as parallel beam.

2. A transparent sphere of radius R and refractive index µ is kept in air. At what distance from the surface of the sphereshould a point object be placed so as to form a real image at the same distance from the sphere ?

(A) R/µ (B) µR (C*) 1R (D) 1

R

Sol. This question can be solved by symmetry as is shown in figure. If d be the distance of object from first surface thenrefracted ray from first surface should become parallel.

1

( )d

=

( 1)( )R

d d 0 + 1d

= ( 1)

R

d = ( 1)

R

Ans.

3. A ray of monochromatic light is incident on the plane surface of separation between two media x & y with angle ofincidence 'i' in the medium x and angle of refraction 'r' in the medium y. The graph shows the relation between sin rand sin i :(A) The speed of light in the medium y is (3)1/2 times then in medium x.

30º

sin r

sin i

(B*) The speed of light in the medium y is (1/3)1/2 times then in medium x.(C) The total internal reflection can take place when the incidence is in x.

(D*) The total internal reflection can take place when the incidence is in y.Sol. From graph

sinsin

ri = tan 30° =

13

sin i = 3 sin rFrom snell's law :

µx sin i = µy = sin r

CPP-4 Class - XI Batches - PHONONREFLECTION

Page 15

= 3 sinx

c rv

= y

cv

sin r

3yv = vx

4. A point source is placed at a depth h below the surface of water (refractive index = ). The medium above the surfaceof water is air ( = 1). Find the area on the surface of water through which light comes in air from water.

Ans. [2

2 1h

]

Sol. u × sin = 1 × sin 90°

h

90°Rsin =

1

tan = 2

1

1

R = h tan = 2

1

1

Area = R2 = 2

2( 1)h

Area = 2

2( 1)h

Ans.

vy = 13 xv

5. Given that, v velocity of light in quartz = 1.5 × 108 m/s and velocity of light inglycerine = (9/4) × 108 m/s. Now a slab made of quartz is placed in glycerine asshown. The shift of the object produced by slab is :(A*) 6 cm(B) 3.55 cm(C) 9 cm(D) 2 cm

Sol. Normal shift = hact 1 r

d

here : hact = 18 cm

hr = hglycerine = 8

83 10

(9 / 4) 10

=

43

hd = hquartz = 8

83 10

1.5 10

= 2

Normal shift = 4 / 318 12

= 418 16

= 1183

= 6 cm

Normal shift = 6 cm Ans.

Page 16

6. Two refracting media are separated by a spherical interface as shown in the figure. P P' is he principal axis, µ1 and µ2are the refractive indices of medium of incidence and medium of refraction respectively. Then :(A*) If µ2 > µ1, then there cannot be a real image of real object

P P'µ1µ2(B) If µ2 > µ1, then there cannot be a real image of virtual object

(C*) If µ1 > µ2, then there cannot be a virtual image of virtual object(D) If µ1 > µ2, then there cannot be a real image of real object

Sol. Object is in medium µ1 at a distance x from pole

2 1( )v x

= 2 1

( )R

2v

= 1 2 1R x

v = 2

1 2 1( ) ( )Rxx R

= 2

21

11

RRx

Case-1 : Object is real x negative Let x = – d

So (1A) Image is real v positive 2

11 R

d

> 0

1 – 2

1

> Rd

d > 1

1 2( )R

(1B) Image is virtual v negative 2

11 R

d

< 0

1 – 2

1

< Rd

Case-2 : Object is virtual x negative Let x = –d

So (2A) Image is real v positive = 2

11 R

d

> 0

2

11

< Rd

d < 1

2 1( )R

(2B) Image is virtual v negative 2

11 R

d

< 0

2

11

> Rd

d > 1

2 1( )R

Looking at four casesIf : µ2 > µ1 Real object can not have real imageIf : µ2 < µ1 Virtual object can not have virtual image

7. A spherical surface of radius 30 cm separates two transparent media A and B with refractive Indices 4/3 and 3/2respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium Aso that the paraxial rays become parallel after refraction at the surface ?Ans. [240 cm away from the separating surface]

Page 17

Sol. Bµ

– Aµu =

–B Aµ µR

µ=4/3A

µ=3/2B

R = 30 cm

O – 4 / 3u =

3/ 2 – 4 / 3( 30)

µ = –240 cm.

8. A ray incident at a point at an angle of incidence of 60º enters a glass sphere of = 3 and it is reflected and refractedat the farther surface of the sphere. The angle between reflected an refracted rays at this surface is :(A) 50º (B*) 90º (C) 60º (D) 40º

Sol. 1 sin 60 = 3 sin r

90 =

60° 60°

r

sin r = 12

r = 30°= 180 – (60 + r) = 90°

9. A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence is small, thanthe displacement in the incident and emergent ray will be :

(A*) ( 1)t nn

(B) tn (C) 1

t nn

(D) None

Sol. sin = n sin rx = t sin r

x + y = t sin

y

x

n

rr

d

t

tsin

n

+ y = t sin

anddy = cos

sin cos

tn

+ d = t

or d = t 1–n

=( – 1)t n

n

10. A stationary swimmer S, inside a liquid of refractive index µ1, is at a distance d from a fixed point P inside the liquid. Arectangular block of width t and refractive index µ2 (µ2 < µ1) is now placed between S and P.S will observe P to be ata distance :

(A)

1

2

1td (B)

1

21td (C)

1

21td (D*)

1

2

1td

Sol. distance = (d – t) + t / 2

1

µµ

µ2

µ1

S

t

P

= d + t 1

2

–1µµ

Page 18

1. A ray of light falls on a transparent sphere with centre at C as shown in figure.The ray emerges from the sphere parallel to line AB. The refractive index of thesphere is :(A) 2 (B*) 3(C) 3/2 (D) 1/2

Sol. i2 = r1 (isoceles triangle CDE) r2 = 60º r1 + x2 = 60º

A D

E

B60º 60º

r

r1

C

r2

or r1 = 30º

µ =sin 60ºsin 30º =

3 / 21/ 2

= 3

2. A beam of diameter 'd' is incident on a glass hemisphere as shown. If the radius of curvature of thehemisphere is very large in comparison to d, then the diameter of the beam at the base of thehemisphere will be :

(A) 34 d (B) d

(C) 3d (D*) 2

3 d

Sol. For objects at infinity image will be formed at :

D E

R

2RC

A

3RB

r3/ 2V =

1/ 2R

or V = 3R.

from similer triangles ABC and ADE

'2DR

= 3DR

D' =23 D.

3. Rays incident on an interface would converge 10 cm below the interface if theycontinued to move in straight lines without bending. But due to refraction, therays will bend and meet some where else. Find the distance of meeting point ofrefracted rays below the interface, assuming the rays to be making small angleswith the normal to the interface.Ans. [25 cm]

CPP-5 Class - XI Batches - PHONONREFLECTION

Page 19

Sol. sin i = 2 210

xx

5/2

y x

x

10ir

i

sin r = 2 2

xy x

sinsin

ir =

52

= 2 2

2 210

x y

x

For small i and rx << 10 and x << y

52

= 10y

or y = 25 cm

4. Find the apparent depth of the object seen by observer A ?

Ans. [ 683

cm]

Sol. Apparent depth =1

1air

dµ +

2

1 2

dµ

=251.5 +

15(2.5) /(1.5)

=503 + 6 =

683

5. A point source of light at the surface of a sphere causes a parallel beam of light to emerge from the opposite surfaceof the sphere. The refractive index of the material of the sphere is :(A) 1.5 (B) 5/3 (C*) 2 (D) 2.5

Sol. µsin = 1 × sin 2µ sin = 2sin cos

C

2

2µ = 2 cos for is small

µ = 2

6. An object is placed 30 cm (from the reflecting surface) in front of a block of glass 10 cm thick having its farther sidesilvered. The final image is formed at 23.2 cm behind the silvered face. The refractive index of glass is :(A) 1.41 (B) 1.46 (C*) 100/ 132 (D) 1.61

Sol. BI1 = (10 + 20µ)

I1 I3 I2A BO 20cm

10cm

BI2 = (10 + 20µ)AI2 = (20 + 20µ)

AI3 = (23.2) + 10 = 33.2 = (20 × 20µ)1µ

µ =200132

7. The x-y plane is the boundary between two transparent media. Media-1 with z > 0 has refractive index 2 and medium– 2 with z < 0 has a refractive index 3 . A ray of light in medium-1 given by the vector A

= ˆˆ ˆ6 3 8 3 10i j k is

incident on the plane of separation. Find the unit vector in the direction of refracted ray in medium –2.

Page 20

Ans. [ r = 3 2 2 1 ˆˆ ˆ55 2 2

i j k (angle of incidence = 60º, r = 45º]

Sol. Unit vecto of in xy plane for incident ray and unit vector in xy plane for refracted will be same plane becuase they needto be in same plane.

Here component of incident ray in xy plane is ˆ6 3 5 3î j

Unit vector of incident ray in xy plane is 3 4 ˆ5 5

î j angle with x axis is 0 = 53º

Component of unit refractracted vection in xy plane is 12

Here refracted ray is 12

cosî + 12

sin j – 12 k

3 2 2 1 ˆ–55 2 2

î k

8. A ray of light passes through four transparent media with refractive indices µ1,µ2, µ3, and µ4 as shown in the figure. The surfaces of all media are parallel. If theemergent ray CD is parallel to the incident ray AB, we must have :(A) µ1 = µ2(B) µ2 = µ3

µ1 µ2 µ3 µ4

D

CB

A(C) µ3 = µ4(D*) µ4 = µ1

Sol. If CD is parallel to AB then µ1 = µy

9. An observer can see through a pin hole the top of a thin rod of height h, placed as shownin the figure. The beaker height is 3h and its radius is h. When the beaker is filled with aliquid upto a height 2h, he can see the lower end of the rod. Then the refractive index of theliquid is :

(A) 25 (B*) 2

5

2h

h

3h

(C) 23 (D) 2

3

Sol. tan =hh = 1 = 45º

hh

h

h

2h

E

Dh

hh

A

B

Ctan = 2

hh =

12

sin = 15

sin =12

Snell's lawµ × sin = 1 × sin

µ ×15

= 1 × 12

µ = 52

Ans.

10. A solid, transparent sphere has a small, opaque dot at its centre. When observed from outside, the apparent positionof the dot will be :(A) Closer to the eye than its actual position (B) Farther away from the eye than its actual position(C*) The same as its actual position (D*) Independent of the refractive index of the sphere

Sol. u = –R

O

µ1

v = ?

Page 21

1( )v R

= 1( )R

1v R

=

1R

v = –RHence dot will appear at its original position.

Page 22

1. Given that, velocity of light in quartz = 1.5 × 108 m/s and velocity of light in glycerine= (9/4) × 108 m/s. Now a slab made of quartz is placed in glycerine as shown. What isthe shift produced by slab ?(A*) 6 cm(B) 3.55 cm(C) 9 cm(D) 2 cm

Sol. Shift = t 1– rare

denser

µµ

µrare = µglycerine =8

8

3 109 / 4 10

= 43

µdense = µquarter =8

8

3 101.5 10

= 2

shift = 184 / 31–2

= 6 cm

2. A rectangular glass slab ABCD, of refractive index n1, is immersed in water of refractive index n2 (n1 > n2). A ray of lightin incident at the surface AB of the slab as shown. The maximum value of the angle of incidence max, such that the raycomes out only from the other surface CD is given by :

(A*)

1

21

2

11 sincossinnn

nn A

B

max

D

n2n1

C(B)

2

11

1 1sincossinn

n

(C)

2

11sinnn

(D)

1

21sinnn

Sol. Condition will be satisfied it on lateral surface T.I.R takes place. Forthat 9 – c 90 – c

sin sin(90 – c) [c = sin–1 2

1

nn

]

sin cot (c) [sin cos[sin–1 2

1

nn

]

Snell's law

CPP-6 Class - XI Batches - PHONONREFLECTION

Page 23

n2 sin = n1 sin 2

1

nn sin = sin cos [sin–1 2

1

nn

]

sin1

2

nn cos [sin–1 2

1

nn

]

max = sin–1 –11 2

2 1

cos sinn nn n

3. A light beam is traveling from Region I to Region IV (Referfigure). The refractive index in Regions I, II, III and IV are n0,

02n

, 06n

and 08n

, respectively. The angle of incidence for

which the beam just misses entering Region IV is Figure :

Region I Region II Region III Region IV

n0—2n0—6

n0—8

0 0.2 m 0.6 m

n0

(A) sin–1 34

(B*) sin–1 18

(C) sin–1 14

(D) sin–1 13

Sol. n0sin = 0

2n

sin1 = 0

6n

sin2 sin = 16 sin2

2 (c)3-4 sin2 0

0

/ 8/ 6

nn sin2

34

sin1 36 4 sin

18

min = sin–1 18

4. A hemispherical portion of the surface of a solid glass sphere (µ = 1.5) of radius r is silvered to make the inner sidereflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The lightfrom the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at theunsilvered part. Locate the final image formed.Ans. [At the pole of reflecting surface of the sphere]

Sol. Refraction at sngle surface no. (1)

O3r

r r

µ=3/2µ=1

Surface(1)

Surface(2)

1

32v

– 1

(–2 )r =

3 –12

r

v1 Reflection from silvered concave surface (2)

2

1v +

1

= –2r v2 = –

2r

Refraction from surface (1) emerging out:

3

1v –

3 / 2–(3 / 2)r =

1 – 3/ 2(– )r v3 = – 2r

This means taot final image is formed at pole of refleting surface (2)

Page 24

5. In ray of light (GH) is incident on the glass-water interface DC at an angle i.It emerges in air along the water-air interface EF (see figure). If the refractiveindex of water µw is 4/3, the refractive index of glass µg is :

(A) 34sin i (B*) 1

sin iGlassµ = ?0

airµ = 1a

Waterµ = 4/3w

i

H

G BA

CD

EF

(C) 4sin3

i (D) 43sin i

Sol. Refraction at CD:µg × sini = µw × sin r

Refraction at F2Fµw × sin r = 1× sin90º

µg sin i = 1 × sin 90º

µg =1

sin i

6. A parallel beam of light is incident normally on the flat surface of a hemisphereof radius 6 cm and refractive index 1.5, placed in air as shown in figure-(i).Assume paraxial ray approximation :(A*) The rays are focussed at 12 cm from the point P to the right, in thesituation as shown in figure-(i)(B) The rays are focussed at 16 cm from the point P to the right, in the situation as shown in figure-(i)(C) If the rays are incident at the curved surface (figure-(ii)) then these

are focused at distance 18 cm from point P to the right.(D*) If the rays are incident at the curved surface (figure-(ii)) then these

are focused at distance 14 cm from point P to the right.Sol. Figure (i) Figure (ii)

(1) (2)

P

µ=3/2

R=6

P

µ=3/2

R=6

Surface (1) :1

3/ 2 1–v =

3/ 2 –1

Surface (1):1

3/ 2 1–v =

3/ 2 –1( 6)

v1 . v1 = + 18 cm

Surface (2):2

1v –

3/ 2

=1– 3/ 2

(–6) r2 = + 12 Surface (2):2

1v –

3/ 2( 12) =

1– 3/ 2

option (A) and option (B) are correct. v2 = +8 cm

7. A point source of light is placed at a distance h below the surface of a large deep lake.(a) Show that the fraction f of the light energy that escapes directly from the water surface is independent of h and is

given by 21 1 12 2f nn where n is the index of refraction of water..

(Note : Absorption within the water and reflection at the surface; except where it is total, have been neglected)

(b) Evaluate this ratio for n = 4/3. Ans. [(b) (4 – 7 ) / 8]Sol. n + sin = 1 × sin 90º 90º

R

y

hr

sin =1n

Page 25

cos = 2

11–n

Note:Fore sphere

dA = (2R cos) (Rd)dA = 2R2cosd

For area of out side sphere, varies from (90 – ) to 90º

Aarea of side water =90

2

90–2 cosR d

= 2R2 90

90–cos d

= 2R2 [sin 90 – sin (90 – )]= 2R2 [1 – cos]

Area of side water = 2R2 [1– 2

11–n ]

Praction:

F =Area of sphere outside water

Total area of sphere

F =2

2

24

RR

2

11– 1–n

F =12

2

11– 1–n

=

12

– 12n

2( –1)n

(b) n = 4/3 F = 4 – 7

8

8. A ray of light is incident from air to a glass rod at point A. The angle of incidence being 45º.The minimum value of refractive index of the material of the rod so that T.I.R. takes place atB is_____.

45ºAir

A

BAns. [ 3/ 2 ]

Sol. r = 90 – = 90 – r qc

45ºAir

Ar

1×sin45º= ×sin

sin =

n r

r1

2n

sin (90 – r) sin c

cos r 1n

1 – sin2r 1n 1 – 2

12n ³ 2

1n

n (3/ 2) , nmin = (3/ 2) Ans.

9. A man observes a coin placed at the bottom of a beaker which contains twoimmiscible liquids of refractive indices 1.2 and 1.4 as shown in the figure. Aplane mirror is also placed on the surface of liquid. The distance of image(from mirror) of coin in mirror as seen from medium A of refractive index 1.2 byan observer just above the boundary of the two media is :(A) 18 cm (B*) 12 cm(C) 9 cm (D) None of these

Page 26

Sol.

A

B

1.2

1.4

coin

7cm

3cm

Image of coin due to AB interface is at depth 4 1.21.4

from interface AB.

For light moving from nod A to air image is at 7 11.2 3

1.4 1.2

from mirror and nod A interface.

Now final image will from observer will be at

7 11.2 31.4 1.2

× 1.2 + 3 = [6 + 3 ] + 3 = 12 cm

10. A container contains water upto a height of 20 cm and there is a point source at the centre of the bottom of thecontainer. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the watersurface. (a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm. (b) Find the maximum value ofr for which he shadow of the ring is formed the ceiling. Refractive index of water = 4/3.

Ans. [(a) 16960 m = 2.8 m; (b) 3

5 7 m = 22.6 cm]

r'2m

20cm

Sol. µsin q = 1·sinr'q = 37º

4 33 5 = sinr'

sinr' = 45 r' = 53º

tan 53º =2x

43 =

2x x =

83 m

Net r adius will be

15 8100 3

= 2.8 m

Shadow will form till critical incidence

C = sin–1 34

i.e., tan = 20r

sin =34

tan =37

r =37

× 20 cm or 3

5 7 m

Page 27

1. A small rod ABC is put in water making an angle 6º with vertical. If it is viewed paraxiallyfrom above, it will look like bent shaped ABC'. The angle of bending CBC') will be in

degree … 43wn

:

(A*) 2º

A

6º

C'

C

B

(B) 3º(C) 4º(D) 4.5º

Sol. =v

OC , = 'v

OCO

C

C'6º

6ºv

xx/µ

a =

'OCOC =

1µ

= µ = 43

(6º) = 8º

– = 2º

2. If the observer sees the bottom of vessel at 8 cm, find the refractive indexof the medium in which observer is present.

Ans. [ 1615

]

Sol. µsel.=108

4 / 3µ =

108 µ =

4 83 10 =

1615

3. A man starting from point P crosses a 4 km wide lagoon and reachespoint Q in the shortest possible time by the path shown. If theperson swims at a speed of 3 km/hr and walks at a speed of 4 km/hr,then his time of journey (in minutes) is ?

3km

Q

6km

4km

P

LAGOON(Salt water lake)

Ans. [250]

Sol. sin 1 =35

2

2

sinv

=1

1

sinv

4

3

6

5

2sin4

=315

3

CPP-7 Class - XI Batches - PHONONREFLECTION

Page 28

sin 2 =45

distance travelled in water = 5 km

time taken in water =53 ×60 hr

= 100 min

distance travelled in ground =2

6 =

63/ 5

time =10 60

= 150 min.

150 + 100 = 250 min.

4. The figure shows a parallel slab of refractive index n2 which is surrounded by media of refractiveindices n1 and n3. Light is incident on the slab at angle of incidence (0). The time taken by the rayto cross the slab is 't1' if incidence is from 'n1' and it is 't2' if the incidence is from 'n3'. then assumingthat n2 > n1, n2 > n3 and n3 > n1. then value of t1/t2 :(A) = 1 (B) > 1(C*) < 1 (D) Cannot be decided

Sol. n1 sin = n2 sin 1n1 n2 n3

sin 1 =1

2

nn sin

n3 sin = n2 sin 2

n1 n2 n3

sin 2= 3

2

nn sin

But n3 > n1 sin 2 > sin 1thus t2 > t1

and1

2

tt < 1

5. A pole of length 2.00 m stands half dipped in a swimming pool with water level 1 m higher than the bed (bottom). Therefractive index of water is 4/3 and sunlight is coming at an angle of 37º with the vertical. Find the length of the shadowof the pole on the bed : Use sin–1 (0.45) = 26.8º, tan(26.8º) = 0.5.Ans. [1.25 m]

Sol. sin37º =43 sin r 37º

r4_3 =µ

35 =

43 sin r

sin r =920

r = 26.8

Length of shadow=34

+ 1 × tan26.8

= 0.75 + 0.5= 1.25 m

Page 29

6. A ray of light is incident from air to a glass rod at point A. The angle of incidence being 45º. Theminimum value of refractive index of the material of the rod so that T.I.R. takes place at B is_____ :

(A) 1.2 (B*) 1.5

(C) 1.7 (D) 2.3Sol. For surface B

A

B

45º

90–

µ sin (90 – ) = 1 sin 90ºµ cos = 1

cos =1µ ...(1)

For surface A1 ·12

= µ sin

sin =12µ ...(2)

From (1) and (2)

tan =

121µ

µ

= 12

13

2tan =12

µ =1

cos = 32

= 32

= 1.5 Ans.Ans.

7. In the figure shown a slab of refractive index 32 is moved toward a stationary observer. AA

point 'O' is observed by the observer with the help of paraxial rays through the slab. Both'O' and observer lie in air. The velocity with which the image will move is :

(A) 2 m/s towards left (B) 43 m/s towards left

(C) 3 m/s towards left (D*) Zero

Sol.µ = 3/2

Oobs x

Shift in the position of image is independent of the position of slab hence due to movement of slab object do not gainany velocity.

8. A beam of light is incident on a spherical drop of water at an angle i. Find the angle between the incident ray & theemergent ray after one reflection from internal surface. Is this possible by total internal reflection ?

Ans. [ = 2i – 4 sin–1 3 sin4

i

+ , No]

Sol. For first deviation S1 = i – ri

i

For second deviation S2 = – 2rFor third deviation S3 = i – rTotal deviation will be

Page 30

= S1 + S2 + S3= 2i + – 4r ...(1)

and from Snell's law1.sin i = µ sin r

sin i =43 sin r

= sin–1 3 sin4

i

...(2)

Hence from (1) and (2)

= 2i + – 4 sin–1 3 sin4

i

9. A glass hemisphere of refractive index 4/3 and of radius 4 cm is placed on aplane mirror. A point object is placed on axis of this sphere at a distance 'd'from as shown. If the final image is formed at infinity, then find the value of'd' in cm.Ans. [3 cm]

Sol. Image of object O due to refraction at AB is at du from surface ABNow due to ACB surface.

1v – 3

(4 )du

=

1 – 4 / 3–4 A B

I'

dO d×µ

4cmr

Cv =– 9

(12)(3 )d

d

Now suppose d > 9 then image is formed below the mirror at distance v' = – 9

(12)(3 )d

d from mirror

Now this image of mirror will act as object for refraction again at surface ACB (light goes from air to water)So,

43v –

( – 9)(12)(3 )

dd =

4 / 3 –14

= 1

12For final image to be at infinity v =

So,–( – 9)

12(3 )d

d =1

12 – d + 9 = 3+d

= 2d – C d = 3 cm

10. An observer observers a fish moving upwards in a cylindrical container of cross section area 1 m2 filled with water upto a height of 5 m. A hole is present at the bottom of the container having cross section area 1/1000 m2. Find out thespeed of the fish observed by observer when the bottom hole is just opened. (Given : The fish is moving with thespeed of 6 m/s towards the observer, µ of water = 4/3) Ans. [4.4975 m/s]

Sol. Velocity of liquid from the hole is 10 m/sec. So using equation of continuityA1V2 = A2V2

600cm__sec

v2

x

v = gh2

5m

Page 31

U2 =1

10 m/sec = 1 cm/sec

y =xµ

dydt =

1µ

dxdt

Here x decreases so dxdt is – 601 cm/sec

34

× (–601) = – 450.75 m/sec

This dydt is velocity of image from surface so velocity of image will be 450.75 – 1

= 449.75 cm/sec = 4.4975 m/sec