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EXERCISE -I

1 .

90° – = 180 ° – 70° – 70° = 50°

2 .

3 . L1L2

I1 I2 I1'

Second Image I1' (image of I1) = 5 +1 = 6 m[from person]

4 .

vI1I2 = vI2

– vI1 = 0

5 .

Mˆˆ ˆv 4 i 5 j 8k ,

0ˆˆ ˆv 3 i 4 j 5k

Plane of the mirror is xy

I

ˆ ˆv 3i 4 j

and for IZkv

= 2 vMZ – vOZ = 2 × 8 – 5 =11

I

ˆˆ ˆv 3 i 4 j 11k

6 . Image formation by concave mirror when object isreal & placed at focus f, forms image at infinity butnot in other cases.

7 . For M1 : u = – 10 cmf = – 20 cm; h0 = – 0.1 cm

v = 200

10 20 =

200

10 v = 20 cm

m = – v

u

i

o

h

h =20

10

hi = 2 × – 0.1 = – 0.2 ; hi = –2mm

For m2 = 20

10

=2 hi = 0.2 cm = 2 mm

Distance between two images = 2 mm

8 . As convex mirror is placed at origin & reflectingsurface towards –ve x–axis.For u < 0 Real objects.For all cases of real objects, the image formationtowards right of the origin v is +ve and also betweenpole & focus.

9 .b

a

d

x-z planei

r

cFrom Snell's law : 1.5 sin i = 2 sin r

1.5 × 2 2

a

a b = 2 × 2 2

c

c d

a

c =

4

3

2 2 2 2a b c d 1 unit vector

1 0 .

1.014 1.02... 1.50

For composite transparent IIr slabes

=sin= constant sin = 1.6 sin xsin x = 5/8 sin.

1 1 . Initially observer is mirror for image formation ofobject O. As O is kept at the bottom, then it seessomewhat above the bottom.

Distance = 5 + 10

= 5 + 10 3

4

= 12.5 cm (in front)

Hence image from the mirror after reflection isalso at 12.5 cm.

UNIT # 11 (PART - I)

RAY OPTICS AND OPTICAL INSTRUMENTS

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1 2 . Image formation by mirror at a depth 'h' frommirror.The distance (apparent) between these two

will be = 2h

h

h

P

I

1 3 . Image position coincides with the object when anobject is placed on centre of curvature i.e.R = 0.5m when transparent liquid is filled thenphenomenon occurs when the pin is placed 0.4 mfrom mirror i.e. effective distance must be 0.5 again0.2 + 0.2 = 0.5 ; = 3/2

1 4 . From Snell's law 2 sin i = 1 sin 90°

i=c

30°

30°

30°

45°

30°

i' = 45°

Again 1. sin 45° = 2 sin r

r = 30° & 2 sin 30° = 1 sin r'

r' = 45°Hence = 180° – (30+45°) = 105°

1 5 . When a ray enters into different medium & againreenter in a same medium there is no deviationdue to refraction (overall) but bottom silvered glasssurface provides 180° deviation from the reflection.

1 6 . Maximum possible deviation = –C

c

c

Reflected ray)

Denser Rarer

1 7 . As given : r + r' = 90°

u2 sinr = u1sin r' 90°

r

r'rr

Denser(u ) 2

Rarer (u )1

u2 sin r = u1 sin (90–r)u2 sin C = u1 sin 90°

1

2

tan r

sin C =

1

2

(C: Critical angle) sin C = tan r; C = sin

–1 (tan r)

1 8 . An object is seen when reflected light from object

enters into eyes when refractive index exactly same

then no reflection from the object.

1 9 . As Speed of light in vacuum / air

= Speed of light in medium

= c

v velocity is different in different medium

From v = n

But frequency is the fundamental property. It never

change by changing the medium hence is also

changed as v is change.

2 0 .

From Snell's law

1. sin 45° = 2 sin r ; r = 30°

1 = 45° – 30° = 15°

= 180 – 2 × 30 = 120°

= 180 – 2 × 30 = 120°

& from Snell's law (refraction)

2 sin 30° = 1. sin r; r = 45°

4 = r – i = 45 – 30 = 15 so net = 1 + 2 + = 270°

2 1 . n1 > n2 But at here i > r not possible

2 2 . (i) For (i) 2

1

u

v – 1

=

2u 1

6

v1 = 3

2 × 6 × 2 = 18 cm

For plane surface : I1 object

dapperent= 2

1

n

n × dactual = 2

1

n

n × (18–R)

= 1

3 / 2 × 18 6 = 8 cm

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2 3 .3 / 2 4 / 3

v u =

3 / 2 4 / 3

10

3 1

2v 60

–

4

3u

3/24/3

So v = 2

ve3

So for any value of u, v is (–ve). So image is virtual.

2 4 . As given u2 > u1 > u3

u2 > u1 lens act as a concave lens for upperhalf portion of lens & u3 < u1

Lower half portion act as a convex lens.

2 5 . m = 2

1 = –2 =

v

u |v| = |2u|

(90–u) = 2 u u = 30cm

Hence lens location = –40 + 30 = – 10 cm

2 6 . u = – 30 cm; h0 = + 0.5 cm; f = 20 cm

0.5

v = uf

u f =

30 20

30 20

= 60 cm

m = v

u =

60

30

= –2

hi =mho = –2 × 0.5 = – 1cm from P axis1+ 0.5 = 1.5 cm below xy.

2 7 . From figure 1st behave as diverging and 2ndbehave as convergingIn 1st incident ray appear to pass at 5 cm distancepoint and after refraction it becomes parallel toprinciple axis i.e. the focal length f = – 5 cmand for second f = 5 cm

2 8 .

I1 I2

Final image (by lens) object I2

u = – 30, f = – 10

v =

uf

u f=

300

30 10 =

300

40 = – 7.5

2.5 cm in front of mirror.

2 9 .

(i) x

2I 0 xv m v (same direction)

(ii) m = –ve, m = v

u,

(v = –ve, u = +ve ) then Iyv ve

3 0 .1

v–

1

20 =

1

10 v = 20 cm

R for mirror = 2f = 2 × 60 = 120 cm x =20 + R = 20 + 120 = 140 cm

\\\\\\\\\\\\\\\\\ \\\ \\\\\ \\\\\\\\\\\\\\\\\\\\\

O

3 1 . Concave lens behaves as divergent lens anddiverges the beam of light as passes from lens.When the screen is brought close towards lens itreceive more number of rays hence light intensityincreases.

3 3 . Am = A1; 2

1A A

m

A2 = A1A2 A =

1 2A A

3 4 . X = u2 – u1 = 24

m = 1

1

v

u =3 v1 = 3u1 or u2 = 3u1

3u1 – u1 = 24 u1 = 12 cmDistance between object & screen isD = v1 + u1 = 3u1 + u1 = 4 × 12 = 48 cm

Hence f = 2 2D X

4D =

2 248 24

4 48

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=

48 24 48 24

4 48 = 2

72 24

4 48

= 9 cm

3 5 . red sin i = 1 sin r 45°

45°1.39 × 1

1.414 = sin r

sin r < 1 possible(refraction is possible)

green × sin i 1.44 × 1

1.41 >1 not possible

Same for blue 1.47 × 1

1.41 > 1

It also not possible.

3 6 . sin A = 1 × sin 90° = 1

sin A

& sin A

2 = 1 × sine =

1

sin A ×

A

2 = sine

90°

sine =

Asin

2A A

2 sin cos2 2

C = sin–1

1 Asec

2 2

3 7 . min= 2i – A; 60 = 2 × 60 – Å; A = 60°

=

min.Asin

2

Asin

2

=

60 60sin

2

60sin

2

= 3 / 2

1 / 2 = 3

3 8 . i = 50° , equilateral prism A = 60°; e = 40°Total angle of deviation = i + e – A = 50 + 40 – 60; = 30°Hence min < min < 30°

3 9 .fR

f =yR

f ; R g B (VIBGYOR)

y > z > x .

EXERCISE -II

1 .

OAB + 90 + = 180°; = 30°

2 .

(IR)45°

45°

OP

y

IR

(R.R)

y'(R.R)

Field of view of image is decided by the reflectedrays. As we can see that the reflected rays fromthe extreme end points of mirror at a height 'd'from P and below P all points on PY'. Hence (C).

3 . M2 M1

OI2 I1 I2

L/3

L

2L/3 2L/3

L+L/3

L/3

Distance between any to images must be 2nL/3

4 .

37°

37°

53°

v

x

10m

As we know the reflected ray = 2 mirror

R = 2× 2 × 9

= 18 × 2 = 36 vS cos 37° = v

vS = v sec 53° = RR sec 53°= 10 (sec 53°)2 × 18

= 10 ×

25

3

× 18× 2 = 1000 m/s

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In ACD : sin = f

302f

PB = R – R R

sec R2 3

PC = R CB = PC–PB = R–R + R

3

= R

3 =

2f

3

CB 2

f 3

1 1 .\\\\\ \\\\\\ \\ \\\\\\\\\\\\\\\\\

fA

u

u = – u, f = –f vA = uf

f u

Image of last end of rod vB = – f

Length = vB – vA = –f– uf

f u =

2f uf uf

f u

Length = 2f

u f. Hence [D]

1 2 . u = – u, f = –f vB = uf

f u

vC =

u b f

f u b

\\\\\\\\\\\ \\\\\ \\\\\\\\\\\\\\

CB

u

u b

b

vB – vC = uf

f u –

u b f

f u b

= ub uf

f u f u b

+

bf

f u b = b

2f

u f

1 3 . When observer in rarer

A

L

A

B

L

L

2L+ L

medium and object is in

denser medium then the

length of immersed

portion will be shorter

(apparent)

Hence length of image

of rod in water

= 2L + L –L = L + LHence length of image of rod seen by observer in

air is = L L

=

LL

. Hence [B]

1 4 . For parallel composite slab

2

1r=ai+bj

c

sin = constant

1

2 2

a

a b

=

2

2 2

c

c d

1 2ˆ ˆr r 1

5 . vI = 2vM – v0

= 2 × 0 – ˆ ˆ5 i 6 2t j

y

x

vI = ˆ ˆ5i 10 j ; vI0 = vI – v0

hence vIo = ˆ ˆ5i 10 j – ˆ ˆ5 i 10 j

[at t = 2 sec] = ˆ10i

6 .

vIG = 2vMG – vOG

vIG1 = 2v, vIG2= 4V .......... vIGn = 2nv

7 .

B

V

v sin

v cos

B

V

v sin

v cos

v cos

v sin

8 . Hence we can see from the images velocity inmirror 1 & II

nd vI2 = 2 vsin

9 . If the angle between mirrors is then the deviationfrom two plane mirrors will be = 2 –2 300 = 2 – 2 = 30°

Hence 360

30 = 12

Number of images = m–1 = 12 –1 = 11

1 0 .

A

B

f

C P

D

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2 2 2 2a b c d 1 1 2a c

1 5 . For parallel composite slab usin = constant

0 × sin i = (n) × sin 90°

0 sin 30° = 0 + 0

n 18

1 11

2 4n 18

1 1n 4

18 4n 2

upper layer of n=3

1 6 . From Snell's law 1.4 × sin x = 1 sin r

53°

liquid =1.4

sinx = 0.8x = 53°

1.4 × 0.8 = sin r > 1 (which is not possible)

Ray will not refrect but it has total internalreflection. Hence (C).

1 7 . Critical angle 3

2 sin c = 1; sin c =

2

3Geometrical path length travelled by the light inthe slab will be =2 +1 + 2 +1 =6m

1 8 . n1sinC = n2 sinC = 2

1

n

n

n1(denser)

n2(rarer)

C

TIR

1× sin = n1 sin (90–C) = n1cosC = n1

2221

n1

n

= sin–1

2 21 2n n ; sin < 2 2

1 2n n

1 9 . 1 =1; 2 = 3/2

2

v

–

1

= 2 1

R

B

dA

DC

E

x3

2v–

1

=

3/2 1

R

=

1

2R

v = 3R

ABE & CDE are similard

3R=

x

2R

x = 2

3d

2 0 . 1 2

1 2

1 1m m

u f u f

; 1

2

m1

m

u2 – f = –(u1– f) u1 + u2 = 2f

1 2u u

f2

2 1 . From Snell's law sin 2° = 2 sin r

= 2° (small angle)

3

22°

sin 2° 2° ; × 2° = 2 r°

r = 1° & 2 × 1° = 3 r'; r' =

02

3

2 2 . 2 = (2 + e) –

e = r1 + r2 =

sin 2 = sin r1 ,

sin ( – r1) = sin

m(sin .cos r1 – cos sin r) = sin

2

2

sin 2 cos sin 2sin 1 sin

2 2sin sin 2 cos sin 2 sin

2 2 2sin 2 2 cos 1

2 – sin

2 2 = 1 + 4 cos

4 + 4 cos

2

2 = 1 + 4cos

4 + 4 cos

2 + 4 sin

2 cos

2

2 = 1 + 4cos

2 (sin

2 + cos

2)

2 1cos

8

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2 3 .

3

90°

1

2 4

5

90°-r1

r2

1 sin 90° = 1 sin r1; 2 sin r2 = 1 cos r1

2cos r2 = 3 sin r3

4 sin r4 = 3 cos r3; 4 cos r4 = 5 sin r5

1. sin 90° = 5 cos r5

Square and add 1 + 2

2 + 4

2 = 1

2 + 3

2 + 5

2

2 4 . min = 38°; = 44°; i = 42° e = 62°

= i + e –A ; 44 = 42 + 62 –A A = 60°

min = 2i–A ; 38 = 2i– 60 i = 49°

2 5 . 2 < 1

1 = i + e–A

1

2

37° 53°50°

= 53° + 37°–A = 90–A

2 < 90–Å

50 + e –A < 90–A

e < 40°

2 6 . Required length of mirror = M1M2 So the mirror

should be the length of half of the size of man =

170

2 = 85 cm & it is placed such that its lower

edge at a half the height of eye level height.

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

M1

M2

E

80cm

160c

m170c

m

2 7 . If the angle between mirrors be – then the angle

of deviation after two reflections is = 2 – 2

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \

\\\\\\

\\\\\ \

\ \\\\\

\\ \\\\

\ \\\

=60°

240 = 2 × 180° – 2

2 = 360 – 240

2 = 120 ( = 60°) Hence [A].

m = 360

60

= 6 (even)

=60°

60°

60°

1

2

Number of images = n = m–1 = 6–1 =5

[Hence (B) & (C).

Reflected ray from 'M1' strikes r on mirror 'M2'.

Hence retrace its path. Hence [D].

2 8 .

45°

45°

x45°

y

y = 2

sin x;

dy

dx =

2

cos x

dy1

dx

tan 45° = 1 = 2cosx

cos x = 1

2 x =

n

3

x =

1

3 and

2

3

y = 2

×

3

2

1 3sin

3 2

=

3

2 9 .

u1 = –60 & u2 =–90

1

1 1 1

60 v 60 v1 = 30

2

2

1 1 1v 36

90 v 60

v2 – v1 = 36 – 30 = 6cm

2 1

2 1

v v 6 1m

v v 30 5

for mirror vI = –m2v0

convex mirror makes virtual image.

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3 0 .

K1 K0 K

1|r i|

= 0 at K0 K0 = 1

2

=1 i

1

2

2 = ?; K = 1

2

; 1 :denser; K : increases

r=0, 3

-0, 2 = 3

1 = ? K = 1

2

2 :denser; 1 : rarer; K : decreases

2 sin i = 1sin 2

sin3

=

1

2

= 3

2= K1

At k1, r2

; so, 1 2 3 6

3 1 . 1 = 60°

5

3sin 30°=

4

3sin2

60°

60°

60°

30°

P

2

1

sin2 = 5

8

2 = sin–1

5

8

Also (C), TIR is at P ceases if

5

3 sin 60° = × sin 90°;

5 3

3 2

=

3 2 .

1

P C

2

O

x R

Real image can't be formed always virtual.

3 3 . v < 0

2 1 2 1

v u R

;

2 2 1 1 0v R u

(A) Virtual image is formed for any position of

0 if 2 < 1.

(B) Virtual image can be formed if x > R & 2 < 1.

3 4 . v > 0

2

v

= 2 1

R

+

1

u

> 0 and if 2 < 1

1 2

R

– 1

x

> 0; 1 2

R

> 1u

x

x > 1

1 2

R

(B) if 2 < 1,

Then virtual image is formed of x < 1

1 2

R

3 5 . 30cm

15cm 15cm

1 1 1

v 30 15 v = – 30 cm

1

v–

1

60 =

1

30 v = 60 cm

(B) Final image is at 60 cm from lens towards left

of it.

(C) final image is real

3 6 .

d = |f1| – 2|f2|

3 8 . m1m2 = 1 & d > 4f (uv method)

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4 0 . Height of object

= H of image 1 H of Image 2 = 1 2I I = 9 4

Height of object = 6 cm

m1 = 1I

O = 1v

u

9

6 =

1

1

v

u

u2 v1 3

2 v1 X =

3

2u1 – u1 = 1u

2 = X

D = 90 = u2 + u1 = 3

2 u1 + u1

u1 = 36 X = 36

2= 18

Distance between two position of lens = X = 18

Focal length f = 2 2D V

4D

=

2 290 18

4 90

= 21.6 cm

4 1 .

4 3 .1 2

1 2

0f f

(condition for achromatism)

1 : 2 = 2 : 1 (Given)

f1: f2 = –2 : 1 ......(i) (for fullfill condition)

1 2

1 1 1

f f 10

1

1 2

f1 11

f f 10

...(ii)

by (i) & (ii) f1 = 10cm & f2 = – 5cm

EXERCISE -II I

Fi l l in the blanks

2 . Rays starting from O will suffer single refractionfrom spherical surface APB. Therefore, applying

1

2

A+Ve

B

O

E

C

D

15cm

P

2

v

– 1

u

= 2 1

R

;

1.0

v –

2.0

15 =

1.0 2.0

10

1

v =

1

10 –

1

7.5 or v = – 30 cm

Therefore, image of O will be formed at 30 cm tothe right of P. Note that image will be vir tual.There will be no effect of CED.

3 . Rays falls normally on the face AB. Therefore itwill pass undeviated through AB.

r2 = 90 – 60 = 30°

2 = 2

2

sin i

sin r i2 = 45°

Deviation = i2 – r2 = 45° – 30° = 15°

(Deviation at face AC only)

4 . P = 2PL + PM

II case : 1

f =

2

f +

m

1

f = 2 1 2

R R

1

f=

2

R

F =

R

2 = 7 ...(i)

I case : P = 2PL + PM 1

F =

L

2

f + M

1

f

1

f =

2 1

R

+

1

F =

R

2 1 = 20...(ii)

By Dividing (i) and (ii)

2 1R 7

2 R 20

20 – 20 = 7

13 = 20 =20

13

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5 . (3h x)

(3h x)+2x

dx

dt=1cm/sec

=43

I3

I1

O

I2

Apparent distance of image by mirror from bird

= 3h x 2x

+ (3h–x)

= 2 3h x

+

2x

2 (3h–x) +

2x

Velocity of image

= 2 dx

0dt

+

2

dx

dt= 2× [0–(–1)] +

2

× (–1)

= 2 × 1 + 3

2 (–1) =

1

2 cm/sec.

Match the column1 . For (A) : All reflected rays pass through the two

mirror upto infinite reflections.For (B) : After IInd reflection ray will not passthrough any mirrorFor (C) : Only rays from object will be reflected bythe mirrors and no reflected rays will be furtherreflected.For (D) : Similarly as C

2 . m = f

f u

(A) Convex mirror u = –ve, f = +vem =

f 1

f f 2

(B) Concave m = f

f f

(C) Concave mirror u = – R (2f)

m = f f

1f 2f f

(D) Convex mirror u = – 2f

m = f

f 2f = f 1

3f 3

3 .

f

v (+ve)

u (+ve)

(A) Convex mirror Real object

virtual image between focus pole

(B) We can see from (v–u) graph, the image

distance v between pole & focus for

all position of real objects.

(C)x

2Ix ov m v ; vIx =

2f

uf x

vIx =

2R

uR 2x

; x object distance

(D) Mirror is convex

4 .

1

H HE at H H

2 2

2

H 3E at H H H

2 2

1

H 1F at H H 1

2 2

2

3H 1 3F at H H

2 2

5 . m = v f

f

=

v

f1 y = mx + c)

m = slope = 1

f (magnitude of slope of line)

Intercet on y axis c = –1 (unity)

Intercet on x-axis m = 0

0 = v

f–1 v = f (focal length)

7 . (A)

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Convex mirror(parallel to principle axisafter reflection passes from focus)

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(B) Diverging the ray

Concave lens

(C) Converge at focusConvex lens

(D) Reflects & intersectsactually

concave mirror

8 . (A) When object is at focus, image is also real

from equi biconvex lens.

1 2

1 1 11

f R R

For equi biconvex f= R

2 1 ...(i) As f

Hence final image position v decreases

hence m = v

u also decrease

(B) From (i) R f

Radius of curvature increases i.e. now the

object is not placed at focus and it is case

when object is between focus and optical

centre. Hence image formation is virtual.

(C) Glass slab provides shift of the object in the

incident ray direction. So for lens object

position becomes between focus and optical

centre.

(D) When medium is changed then relative .

9 . r > y > g > b r < y < g < b

Cr > Cy > Cg > Cb (C : Critical angle)

1 0 . Critical angle

2 × sin C = 1. sin 90°

C = sin–1

1

2 =

6

A

r A-ri

1. sin 90° = 2 × sin r

sin r = 1

2 r =

6

0 < A – r < C for all rays refrection

(A) A = 15° A – R = ve hence (a)

(B) A = 45° 45– 30°= 15° < 30° for same rays

45° – 0° = 45° > 30° for same rays

(C) A = 70° : A–r 70° – 30° = 40° > C(30°)

70° - 0° = 70° > C All rays reflected back

(D) A = 50° : A–r = 50° – 30°

= 20° < C(30°) Refrection

50° - 0° = 50° > C (Reflection)

Comprehens ion # 2

1 . In this case x, H and remains same so required

length of mirror = H H x sin

H 2x

= 01 1 1 sin 45

1 2 1

=

2

3m

2 . In this case only x will change so equired length of

mirror = 01 1 2 sin 45

1 2 2

=

13

25

=

3

5 2m

3 . In this case x = 1–1 = 0 so length of mirror =

01 1 0 sin 45

1 2 0

=

1

2

Comprehension # 3

1 . On squeezing the lens, the focal length of lensedecreases.

2 . As v decreases, the magnification is also decreases.

3 . As turnip away from the lens, distance increasesi.e. u increases hence for same v, f has to decreasesso we should decrease the squeeze of lense.

Comprehens ion # 41 . u = –25, f = f, v = 2.5

2.5 = 25 f

25 f

f=

25

11

2 . Maximum focal length = distance of retina fromeye – lens = 2.5 cm

3 . Normal person power of eye–lens

P = 100

2.5 = 40 D; u = – 100 cm, v = 2.5

2.5 = 100 f

100 f

f =

100

41

Power =

100

100

41=41D

Required power of lens = 40 = 41 + P2

P2 = – 1D

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EXERCISE -IV (A)

1 . ABC & AB'C

x

0.2m

3m

A

B

B'C'

1m

Cx x 3

0.20 1

x = 0.2x + 0.6

0.8x = 0.6 3

x4

cm

x = 75 cm

2 .minH 40cm

80cm 20cm

Hmin

= 160 cm

& minH 80cm

80cm 20cm

Hmax

= 320 cm

3 .

f= –20 cm; 1 1 1

v 60 20

1 1 1 40v 30

v 60 20 60 20

v 30 1 1

mu 60 2

so image coordinates are (–30, 3/2)

At m2 :

1 1 1

v 70 20

1 1 1 50

v 70 20 70 20

v= 140

5

= – 28 cm

So magnification =

v 28 1 2

u 70 5

Object length = 3

cm2

image length = 3 2 3

cm2 5 5

4 . Minimum focal length of eye lense from Q.1= 25

11

Power = 100

25 / 11 =

100 11

25

= 44D

And u = – 100 cm, v = 2.5, f= 100

41,

Power = 41 DRequired power lens = 44D = 41 D + P2

P2 = 3D5 . For an average grown up person minimum distance

of object should be around 25 cm So (D).

Comprehens ion # 5

1 . The focal length varies with wave length.

2 . The combination will be free from chromaticobservation if dF = 0. Hence [A].

3 . Condition of achromatism

1

1f

+

2

2f

= 0

1

2

= 1

2

f

f

'

=

0

02

= 1

2

f

f

f2 = – 2f1. Hence [D]

4 . 1 = 0.02, 2 = 0.04 [Dispersive power]

1

2

= –1

2

f

f

0.02

0.04 =

1

2

f

f

f2 = –2f1

1 2

1 1 1

F f f

1 1 1 2 1 1

40 20 40 40 40

Hence f1 = 20, f2 = – 40. Hence [A]

5 . This aberration related with lenses not for mirrors.

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7 . Shift due to glass slab

\\\\

\\\\

\\\\

\\

6cm=

16 1

3 / 2

= 6 ×

1

3= 2cm

t = thickness of glass slab; = 3/2

Reflected final image concides with the object when

the effective distance of the object in equal to radius

of curvature of mirror u = 40 + 2 = 42 cm.

8 .

x = t1

1

=3cm

For mirror :

Object– I1 : u = –30, f=20

30 20v

30 20

= 12 cm = 33+9 = 42 cm

from object

9 .QC

Q

sinc

R

P

3 3 2 3

4 3 2

Q

C = 60°

3 3

4sin =

3

2sin 30°

11 1sin sin

3 3

1 0 . Let refrective index =

2

3.2

CC

From snell's law sin C = 1.sin90° = sin

C =

1

C 2

1 3.2tan

21

2–1 =

210

16

; 356

16 =1.17

2–1

1C

1 1 .R 1

sin cR d 1.5

R 1

R d 1.5

below axis of M2 : y =

3

5

x coordinate = (40–28) = 12

so coordinates = 3

12,5

4 . Speed of ballon = 20m/s

\\\\\\\\\\\\\\

15cm

20m/s

d = 20 × 4 –1

2× 10 × 16

d = 0

u =–15, R=–20, f=–10

v = ub 150

u f 5

=–30

v 30m

u 15

=–2, v

0=–20

vI = – m2v

0 = –4 × – 20 = 80

5 .4 2

x 45 40cm3 3

So total distance = 60 cm

f = – 20 cm

1 1 1

v u f

1 1 1

v 60 20

1 1 1 40

v 30v 60 20 60 20

So image after reflection from mirror

will be at d = 30 cm. So 10 cm above 'A'

After refraction at A : It will be at

10 90 453 / 2 cm

4 / 3 8 4

so distance from 'B' layer is

45 345 45

4 4

So final image after refraction at 'A' will be at :

3 2 4545

4 3 2 from 'A' surface

6 .1x

=6 x

1 = 6 ×

3

2x1

x2

and 2x

=4, x

2 = 4 ×

3

2

t = x1 + x

2 = 6 + 9 = 15cm

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1 1

1 d / R 1.5

1.5 > 1+

d

R

d 1

R 2

1 2 . vB = 16 cm/sec;

4

3

f

4 cm/sec

Bf B fv v v

–16 = Bv

– 4 ; Bv

= –12cm/sec

app act

obse object

v v

n n

act12 v

4 1 ; v

act = 9 cm/s

1 3 .

60°

rP

2r

r2r

C

Q

From snell's law1. sin 60° = sin r (at Pt 'P')again sin r = 1. sin 2 r (at Pt 'Q') sin 2r = sin 60°r = 30°

1. sin 60° = sin 30° 3

1 4 .n

ir r

R

From (1) surface : n 1 n 1

v R

;

nRv

n 1

1 n 1 n

R v 2R R

,

2 n n

R R v 2R

2 n(v 2R R)

R R(v 2R)

2v – 4R = nv – nR

On putting the value of v

nR n nR2 4R nR

n 1 (n 1)

2nR – 4nR + 4R = n2R – n2R + nR, 4

n3

1 5 . From lens maker's formula

1 2

1 1 1( 1)

f R R

we have 1 3 1 1

10.3 2 R R

(Here R1 = R and R

2 =–R) R = 0.3

Now applying 2 1 2 1

v u R

at air glass

surface, we get 1

3 / 2 1 3 / 2 1

v (0.9) 0.3

v1 = 2.7m

i.e., first image I1 will be formed at 2.7m from the

lens. This will act as the virtual object for glasswater surface.

Therefore, applying 2 1 2 1

v u R

at glass

water surface,

we have 2

4 / 3 3 / 2 4 / 3 3 / 2

v 2.7 0.3

v2 = 1.2 m

i.e. second image I2 is formed at 1.2m from the

lens or 0.4 m from the plane mirror. This will actas a virtual object for mirror. Therefore, third realimage I

3 will be formed at a distance of 0.4 m in

front of the mirror after reflection from it. Nowthis image will work as a real object for water-glassinterface.

Hence applying 2 1 2 1

v u R

we get 4

3 / 2 4 / 3 3 / 2 4 / 3

v (0.8 0.4) 0.3

v4 = 0.54 m i.e. fourth image is formed to the

right of the lens at a distance of 0.54 m from it.Now finally applying the same formula for glass-air surface.

5

1 3 / 2 1 3 / 2

v 0.54 0.3

v5 = –0.9 m

i.e., position of final image is 0.9 m relative to thelens (rightwards) or the image is formed 0.1 mbehind the mirror.

1 6 .

\\\\

\\\\

\\\\

\\

1mm

(0,0)

15 45

f=10 f=10

Image formed by the lens 15 10

v15 10

= 30

(Right from lens)Reflection from mirror u =–15

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15 10v 30

15 10

;

( 30)m 2

15

Co–ordinate of Image = (30,0.3cm)Now again taking refraction from lens)

15 10v

15 10

= 30 (left from lens)

30m 2

15

Final co–ordinate of image = (–15cm, –6mm)

1 7 . For real image m = f

f 16

Virtual m = f

f 6

f f

f 16 f 6

f + 16 = –f–6

2f = – 22 f = 11cm

1 8 . A

O

BB'

C

A'1.5

15 20 10

x

v =uf 15 10

30u f 15 10

ABC & A'B'C are similar x 1.5 1

x10 30 2

Area = x2 = × 1

4=

4

cm2

1 9 . Given– Distance between two position of the lensfor image is formed

Candle

u v=(D-u)

Scre

en

D

x = u2 – u

1 = 0.8 m (Given)

D = u1 + v

1

m1 = 1

1 1

1

v3 v 3u

u

2 22

2

v 1 uv

u 3 3

x = v1–u

1 = 3u

1–u

1 = 0.8

u1 = 0.4

v1

= 3 × 0.4 = 1.2D = 1.2 + 0.4 – 1.6

2 2 2 2D x (1.6 ) (0.8)f

4D 4 1.6

= 0.3m = 30cm

2 0 .4mm

x

40cm

100cm

From similar triangles0.4 x

40 100 ; x = 1 cm

2 2 . A=60°i + e – 60 = 23° & e = i + 23°2i + 23°– 60° = 23°i = 30°

e = 53° 1

1sin r

2

1

4sin 60 r sin 53

5

1 1

3 1 4cos r sin r

2 2 5

2

1 1

3 1 41 sin r sin r

2 2 5

2

3 1 1 1 41

2 2 2 54

2 43 4 1 1 4

5

2

2 21 14 1

5 3

1 4411

4 25 3

129 43

75 5

2 3 . For critical angle sin (90–r) = 1. sin90

1cos r

90–r

60

rri

=30°90– (120–r) = (r –30)From snell's law sin (r–30°) = 1 sin30°

[sin r cos 30° – cos r sin30°] =1

2

2 1 3 1 1 1

2 2 2

2 3u 1 1

2

1 / 2

2 4 71

3 3

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EXERCISE -IV (B)

1 .AO OB

tan 30 , tan 301m 1m

AO + OB = AB = 2 tan30° = 2

3m

2 . From sine rule

O

120°2

1

20cm

I2

I1120°

30°

30°

20

sin120 sin 30

sin 30 120 20

cos 30 3

3 . (a)2

Im Jm2

vv v

u

vIm

= –1cm/sec Jm

11m / s v

100

vJm

= 100cm/sec = 1m/sec, vm = 20 m/sec

vJm

= vJ – v

m

vJ = v

Jm + v

m = 1 + 20 = 21m/sec

(b)v dm d f

mu dt dt u f

2

2

dm f du dm f 1 du

dt dt dt u f f dt(u f )

2

3dm 1 1( 1) 10

dt 10 10

4 .r

d–x

6'

i

r r

17'

11'

4"

x

10'8"=323

6'

2 2

2

4 x d x

3 36 (d x)32x

3

16x2[36 + (d–x)2] = 9(d–x)22 1024

x9

16 × 36x2 + 16x2(d–x)2 = 9(d–x)2x2 + 1024(d–x)2

16 × 362d

4+ 16

2 2 2 2 2d d d d d9 1024

4 4 4 4 4

4d 448

7 = 16

5 . As given in question

r 60°

60°

r

r r

||||||||

|||||

60°

(60–r)+(60–r)=1

3(60–r)+(180–2r)+(60–r)

120–2r=100–4r

3

2r

3=20 r=30°

1 sin 60 = × sin30° u 3

6 . Case- I

1 1 1 1 1

f ' u v 60 30

30 60

f '90

f' = –20cm

m

1 1 2

f ' f f

, 1 1 1

( 1)f 60

1 1 2( 1)

20 30 60

2( 1) 1 1

60 20 30

– 1 = 3

12

= 1.5

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Case- I I

When there is liquid in the mirror u = 2 f' &u = f'when there is no liquid in the mirror

1 1 1

u u 30 30

y

u(u 30)30

2u 30

u2 – 10x + (30)2 = 0

2 290 (10) 4(30)u

2

; u 45 15 5

But there u is +ve so that

u 45 15 5 & 45 15 5

u 2f f2 2

M

1 1 2

f f f

1

153 5

2

= 1

30 +

2 1

60

4 3 5

1 13 5 3 5

1 5 2 1 5 1

7 .

SinC = 1 1

2 / 3

1 3C Sin 60

2

Rcos60 x R, x 5cm

R x

8 . Image formed by partial reflection from watersurface at a distance 30cm below the air watersurface.

30cm

O

20cm

=43

f=40cm

water surface behaves l ike a plane mirrorsecondly image formed by refreaction from watersurface followed by reflection from mirror andagain refrection out of the water surface.

Distance of the object from mirror

= u = 20+4

3× 30 u =–60

v = 60 40

60 40

v = 24 cm

Now refrecation from water surface

1 4 / 3 1 4 / 3

v (20 24)

v =–33 cm

9 . 90– x – 360° –[135 + 90 + 90–r] x = 45 – r

r135°

90°

90°

x

For totally internal reflection

2 sin (45–r) = 1 sin 90°

sin (45–r) = sin30° r=15°

from snell'a law 1 sin = 2sin15°

sin v = 2 × 3 1

2 2

= sin–1

3 1

2

1 0 .1

Sinc

1 1C sin 45

2

Required area as shown in figure = 2

24 R2 R

2

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1 3 10

v cmv 10 3

from L3

Ray converges at d= 10

cm3

from L3

(b)

1 1 1 fxv

v x f x f

fx5

x f

1

1 1 1

v v f

1

1 1 1

fxv f5

x f

1

1 1 1

fxv f5

x f

1 1

fu5 v u 5 v

u f

5u 5f fu

u f

1

1 u f 1 f u f

fxv 5u 5f fu f 5u 5f fx5

x f

1 x f f u

f 5x 5f fx 5x 5f fu

Putting the value after solving 50

x cm3

1 4 . Initially the system will act as simple concave mirrorgiven

v = –60, m=–2 u = –30

1 1 1

v u f

1 1 1

60 30 f

f = – 20 cm R = 40 cmIn second case : lens + mirror systemThen P

meq = P

L + P

L + P

M

Pmeq

= equivalent power of the mirrorP

L = power of the lens

PM = power of the mirror

Pmeq

=L L M

1 1 1

f f f

L

1 1 2

f 3 R

L

1 1 2 2 2 2 3

f 3 R 3R R 3

Pmeq

= meq

m

10 1 3Rf

3R f 10

feqm

= – 12 cm

1 1 . 1

1

n 1 n 1 2nRv

v 2R R 2n 1

Now u2 = R-V

1=

2nR 4n 1

R R2n 1 2n 1

2

1 h 1 n

4n 1v 2RR

2n 1

2

4n 1v 2R

3n 1

Total shift = 3R – |v2|

= 3R 4n 1

2R3n 1

=

n 1R

3n 1

1 2 .I v f

m0 u f u

; rel = 0

2

2dI f du2

dt dt(f u)

140

2 4 210 10

8

5

cm/sec

1 3 . (a)

When a parallel beam comes from '' from left,after refraction from L

1 it forms image at 10 cm

from L1.

This image will behave as a object for 'L2'

So object distance form 'L2' = + 5 cm

So image formed at v :1 1 1

v 5 10

1 1 1 1

v 10cmv 5 10 10

Therefore L2 will form a image at +10 cm from L

2

on the right side. This image will act as object for L3

Object distance for L3 = + 5 cm

final image at v = ?

1 1 1

v u f

1 1 1

v 5 10

1 1 1

v 5 10

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Now if object is placed at distance = x then

eq

1 1 1

v u f , ( v = – 60 )

1 1 1 1 1 1 4

60 u 12 u 60 12 50

u=–15 cmSo object displaced = 15 cm

1 5 .

\\\\\\\\\\\\\

f =20cm1

d=10cm

f =20cm2

=3/2

0,I

1 2 1 2

1 1 1 d

f f f f f

L m

2 1

f f 2 2

20 120

2

120 60f

14 7

1 1 1 1 1 1 120u

v u f 120 u 20 7

120 50 7 1 110

7 7 50 u 20

u =

100

19cm

1 6 . (i) Given that r2=C

1r 45 C & sin e = n sin

(45°–C)sin e = n (sin 45 ° cos C - cos 45° sin C)

1 2 2

1 1

1e sin n n n

2

(ii) to pass refracted ray undeviatedr

2 = 0 r

1 = 45° and sin e = n sin 45°

1 11.352e sin sin 0.956

2

1 7 .

Now d

2 f d2 f

EXERCISE -V-A

1 . The formula for the number of images formed bytwo inclined mirrors; inclined at angle is

360N 1

360If even

whereas

360N

360If odd

For = 60°, N = 360

60=6

Hence, N = 6 –1=5

3 . The working of optical fibres is based on thephenomenon of total internal reflection.

4 . The resolving power of an astronomical telescope

is =

D

1.22 where D is the diameter of the objective

lens. In order to improve the resolution of anastronomical telescope; one has to increase thediameter, i.e., linear aperture of the astronomicaltelescope.

5 . The image formed by the objective of a compoundmicroscope is real and enlarged as object liesbetween focus and centre of curvature.

6 . The number of images formed when the anglebetween the mirrors is 90°

N = 360

60–1=4–1=3

7 . As ray AB is incident at an angle of 45° and for thisray to suffer total internal reflection.

45°

B

45°

C

A

Here 45° > C sin 45° > sinC

1 1

n 2n2

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8 . A silvered lens behaves as a mirror. For the image

to be of same size as of object must the object be

held at the centre of curvature of mirror when

silvered lens 1 m 1 1 m

1 1 1 1 2 1

f f f f f f

1

1 1

f R m

1 2 Rf

f R 2

Therefore

2 11 2 2

f R R R

The radius of curvature of this mirror

= 2f =

R=30

1.5= 20 cm

9 . Radius of circle

2

t 12 36t tan c cm

16 711

19

tC

1 0 . Let dots can be resolved by eye at maximum distance

R then 1.22 d

D R

–3 –3

–9

3 10 1 10DdR

1.22 1.22 500 10

= 4.92 m ~ 5m

1 1 . Power of lens =

2 1

1 2

1 1

R R

In air power is – 5D

–5 = (1.5 – 1)

1 2

1 1

R R ...(i)

In other medium power is P

1 2

1 1P 1.5 1.6

R R ...(ii)

Dividing (ii) by (i)

P 0.1 1P 1D

5 1 0.5

1 2 . The relation between angle of minimum deviation

and refractive index is D=(–1)A

1 3 . P1=

1

115D

f , P2 = 2

1

f =5D

When thin lenses are kept in contact with each other,the effective focal length of the combination will be

eff 1 2

1 1 115 5 10

f f f

eff

1f 0.1m 10cm

10

1 5 . By using Snell's law we get 1 sin

221 sin

3 sin2 =

1

3

sin = 1

3 = sin–1

1

3

1 6 . Incident angle from given plane

1 = cos–1

A ·k

A

= cos–1

–10

20 = cos–1

–1

2 =

3

Now by snells law n1 sin

1 = n

2 sin

2

sin2 =

1

2

n

n sin1 =

2 3 1

23 2

2 = 45°

1 7 . Velocity of image = –m2V0 = –

2ƒ

ƒ uV0

= –

220

20 ( 280) (15) = –

1

15 m/s

1 8 . Focal length of a lens is given by

1 2

1 1 1( 1)

f R R

1 1

f f

R B R Bf f

Focal length will increase

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1 9 . Apparent shift =

1h 1

2

Kerosene

1

Water

h2

h1

Total apparent shift =

1 2

1 2

1 1h 1 h 1

2 0 .

1 1 1 1 1 240f

f v u 12 ( 240) 21

Shift =

1 1 1h 1 1 1

µ 1.5 3 cm

To get image at fi lm, lens should form image at

distance = 12 – 1 35

3 3;1 1

v u

21 3 1

240 35 u u = –560 cm

2 2 . From trignometry

3min

R-3mm

R

3cm

R2 = (3)2 + (R–0.3)2 R 15 cm

8

lens 8

3 101.5

2 10

from Lens maker formula,

L

s 1 2

1 1 1 1 11 0.5

f R R 15 30

f = 30cm

EXERCISE -V-B

1 . The ray diagram is shown in figure. Therefore, theimage will be real and between C and O.

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

C

Normal at P willpass through C

image

C

2 . Let us say PO = OQ = X

1=1.01=1.5

QP

O

+ve

Applying 2 1 2 1

v u R

Substituting the values with sign

1.5 1.0 1.5 1.0

X X R

(Distances are measured from O and are taken aspositive in the direction of ray of light)

2.5 0.5

X R X = 5R

3 . R1 = – R, R

2 = + R,

g = 1.5 and

m = 1.75

g

m 1 2

1 1 11

f R R

R1 R2

Substituting the values,

we have 1 1.5 1 1 1

1f 1.75 R R 3.5R

f = +3.5 RTherefore, in the medium it will behave like aconvergent lens of focal length 3.5R. It can beunderstood as,

m >

g, the lens will change its

behaviour

4 . The lens Maker's formula is :

L

m 1 2

1 n 1 11

f n R R

where nL = Refractive index of lens and

nm = Refractive index of medium.

In case of double concave lens, R1 is negative and

R2 is positive.

Therefore, 1 2

1 1

R R

will be –ve.

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For the lens to be diverging in nature, focal length

f should be negative or L

m

n1

n

should be positive

or nL > n

m but since n

2 > n

1 (given), therefore the

lens should be filled with L2 and immersed in L

1.

5 . O

i iA D

r r

r

EB

C Fi i

Divergence angle will remain unchanged becausein case of a glass slab every emergent ray is parallelto the incident ray. However, the rays are displacedslightly towards outer side.(In the figure OA|| BC and OD|| EF)

6 . The ray diagram will be as follows :

S

D

E

G

H

J

I

HI = AB = d; DS = CD = d

2

Since, AH = 2AD GH = 2CD = 2 d

2= d

Similarly, IJ = d GJ = GH + HI+ IJ = d + d + d = 3d

7 . Rays come out only from CD, means rays afterrefraction from AB get total internally reflected atAD.From the figure :

max

r1

r2

A D

CB

n1

n2

r1 + r

2 = 90° r

1 = 90° – r

2

(r1)max

= 90° – (r2)min

and (r2)min

= C

[for total internal reflection at AD]

where 2

C

1

nsin

n or

21C

1

nsin

n

(r1)max

= 90° – C

Now applying Snell's law at face AB :

1 max max max

2 1 max C C

n sin sin sin

n sin(r ) sin(90 ) cos

or sin max

= 1

2

n

ncos

C

max

= 11

C

2

nsin cos

n

1 21 1

2 1

n nsin cos sin

n n

8 . Applying Snell's law at B and C,

iBiC

sin i = constant or 1 sin i

B =

4 sin i

C

But AB||CD iB = i

C or

1 =

4

9 . Figure (a) is part of an equilateral prism of figure(b) as shown in figure which is a magnified imageof figure (c).

Therefore, the ray will suffer the same deviation infigure (a) and figure (c).

1 0 .

R

KM

T

S

P

riQ

N

PQ = QR = 2h i =45

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ST = RT = h = KM = MN

So, KS = 2 2h (2h) = h 5

sin r = h 1

h 5 5

sin i sin 45 5

sin r 21 / 5

1 1 .1 2

1 1 1( 1)

f R R

For no dispersion

1d

f

=0 1 2

1 1d 0

R R

R1 = R

2

1 2 . d=0.2 tan 30° 0.2

3

2 330

d 0.2 / 3

0.2m30°

=2 3

30°

Therefore, maximum number of reflection are 30.

1 3 . Image formed by convex lens at I1 will act as a

virtual object for concave lens. For concave lens

I2

26 cm 4cm

I1

1 1 1

v u f

1 1 1

v 4 20

v = 5 cm

Magnification for concave lens v 5

m 1.25u 4

As size of the image at I1 is 2cm. Therefore, size of

image at I2 will be 2 × 1.25 = 2.5 cm.

1 4 .

Glass

Water

Air

r

r

i1

90°2

Applying Snell's law ( sin i = constant) at 1 and 2,we have

1 sin i

1 =

2 sin i

2

Here, 1 =

glass, i

1 = i;

2 =

air = 1 and i

2 = 90°

g sin i = (1) (sin 90°) or g

1

sin i

1 5 . Critical angle

1C

1sin

Wavelength increases in the sequence ofVIBGYOR.According to Cauchy's formula refractive index ()decreases as the wavelength increases. Hence therefractive index will increase in the sequence ofROYGBIV. The critical angle

C will thus increase

in the same order VIBGYOR. For green light theincidence angle is just equal to the critical angle.For yellow, orange and red the critical angle willbe greater than the incidence angle. So,thesecolours will emerge from the glass air interface.

1 6 . During minimum deviation the ray inside the prismis parallel to the base of the prism in case of anequilateral prism.

1 7 . When the object is placed at the centre of theglass sphere, the rays from the object fall normallyon the surface of the sphere and emergeundeviated.

18. Distance of object from mirror 33.25

15 40cm1.33

Distance of image from mirror 25

15 33.8cm1.33

For the mirror, 1 1 1

v u f

1 1 1

33.8 40 f

f = –18.3 cm

Most suitable answer is (c)

1 9 . Let focal length of convex lens is +f, then length of

concave lens would be –3

2f. From the given

condition. 1 1 2 1

30 f 3f 3f f = 10 cm

Therefore, focal length of convex lens = +10cmand that of concave lens =–15cm

2 0 . Refraction from lens : 1

1 1 1

v 20 15

v = 60cm + ve directioni.e. first image is formed at 60 cm to the right oflens system.Reflection from mirror : After reflection from

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the mirror, the second image will be formed at adistance of 60 cm to the left of lens system.

Refract ion f rom lens : 3

1 1 1

v 60 15 +ve

direction or v3 = 12cm

Therefore, the final image is formed at 12 cm tothe left of the lens system.

2 1 .

f

r

r = f tan r f r2 f2

2 2 . Since <c, both reflection and refraction will take

place. From the figure we can see that anglebetween reflected and refracted rays is less than180° –2.

AirWater

2 3 . Since object and image move in opposite direction,the positioning should be as shown in the figure.Object lies between focus and centre of curvaturef < x < 2f

\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\rightleft

Image

Object

2 4 . In the position of minimum deviation angle of

refraction r = A

2

2 5 . n0sin = 0n

8

sin90° sin =

1

8 = sin–1

1

8

2 6 . v2 – u2 = 2as; v2 = 2(g)(7.2); v = 12 m/s

12.8m

7.2m

Xb/f

= µx

(B / f )dX

dt = µ

dx

dt

4

3 (12m/s) 16 m/s

2 7 .1 1 1 1 1 1 1 1

and v 30 15 v 30 v 10 15

1 1 1

v 15 10 v=6

2 8 . Here normal is along j

Angle between incident ray and normal

1 ˆ ˆ ˆi 3 j j 32cos 301 1 2

2 9 . Real image v = +8 m

m = v 1

u 3 u = –24 m

f = 24 8uv

6 mu v 24 8

for plano convex lens

1 1 1 1 3 1

u 1 1f R 6 2 R

R = 3 m

30.(P) at prism surface ray moving towards normal so(µ

2 > µ

1)at block surface ray moving away from

normal so (µ3 < µ

2)

(Q) No deflection of ray on both surface; so µ1 = µ

2

= µ3

(R) At prism surface ray moving away from normal soµ

2 < µ

1. At block surface ray movign away from

normal so µ3 < µ

2 but since on total internal reflection

not takes place on prism surface

µ1 sin 45° < µ

2 sin 90° µ

1 < 2µ

2

(S) Total internal reflection takes place

so µ1sin45° > µ

2 sin 90° µ

1 >2µ

2

M C Q1 . For total internal reflection to take place :

Angle of incidence, i > critical angle, c

sin i > sin c sin 45 ° >

1

n

1 1

n2

n 2 n > 1.414

Therefore, possible values of n can be 1.5 or 1.6in the given options.

2 . At point O (1) sin60° = 3 sin r r=30°

At point L : TIR occurs : because 3 sin 45° > 1

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At point M : 3 sin (30°) = 1 sin 60

Match the column2 . (A)–pr, (B) –qst, (C) –pr t, (D) –qs

For (p) : 2 1 2 3 3 1

and

For (q) : 1 2 3

For (r) : 3 2 1

For (s) : 3 2 1

For (t) : 2 3 1

SUBJECTIVE

1 .

4

1 2

10.8 10n 1.20 and

n2 = 1.45 +

4

2

1.80 10Here, is in nm.

(a) The incident ray will not deviate at BC if n1 = n

2

4 4

02 20 0

10.8 10 1.80 101.20 1.45 ( )

4

20

9 100.25

2

0

3 10

0.5or

0 = 600 nm

(b) The given system is a part of an equilateralprism of prism angle 60° as shown in figure.At minimum deviation

1 2

60r r 30 r(say )

260°

70°CD

60° 40°

20°

A B

i 1

sin in

sin r

sin i = n1 sin 30°

4

2

10.8 10 1 1.5 3sin i 1.20

2 2 4(600)

( = 0 = 600 nm)

i = sin–1 (3/4)

2 . Incident ray

ˆˆ ˆA 6 3i 8 3 j 10k ˆˆ ˆ(6 3i 8 3 j) ( 10)k

z yQ'

xO

–10k

PA

63 i +

83 j

Q

zy

x

QO PQ (As shown in figure)

Note that

QO is lying on x-y plane.

Note, QQ' and Z-axis are mutually perpendicular.Hence, we can show them in two-dimensional figureas below.

z

PA

Q Q'O

1= 2

r=12= 3

R

k

q

Vector A makes an angle i with z-axis, given by

1

2 2 2

10i cos

(10) (6 3 ) (8 3 )= cos–1

1

2

i = 60°Unit vector in the direction of QOQ' will be

2 2

ˆ ˆ6 3i 8 3 j 1 ˆ ˆq (3 i 4 j)5

6 3 8 3

Snell's law gives

3 sin i sin 60

sin r sin r2

3 / 2 1

sin r3 / 2 2

r = 45°

Now, we have to find a unit vector in refrated ray's

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direction OR. Say it is r whose magnitude is 1.

Thus, ˆˆ ˆr (1 sin r )q (1 cos r )k 1 ˆˆ[q k]2

1 1 ˆˆ ˆ(3 i 4 j) k52

1 ˆˆ ˆr (3 i 4 j 5k )

5 2

3 . Applying

2 1 2 1

v u RFirst on plane surface

1

1.5 1 1.5 10

AI mR

AI1 = –(1.5 mR)

Then on curved surface

1 1.5 1 1.5

(1.5mR R) R

[v = because final image is at infinity]

1.5 0.5

(1.5m 1)R R

3 = 1.5m + 1 3

2m=2

4m

34 . (a) Rays coming from object AB first refract from

the lens and then reflect from the mirror.Refraction from the lens :u =–20 cm, f = +15cm

Using lens formula 1 1 1

v u f;

1 1 1

v 20 15 v = +60cm and linear magnification,

m1 =

v 603

u 20 i.e., first image formed by

the lens will be 60cm from it (or 30cm from mirror)towards left and 3 times magnified but inverted.Length of first image A

1, B

1 would be 1.2 × 3 =

3.6 cm (inverted).

A

B

Optic axis of lens

Optic axis of mirror

B1

A1

Reflection from mirror: Image formed by lens(A

1,B

1) will behave like a virtual object for mirror

at a distance of 30cm from it as shown. Thereforeu=+30cm, f=–30cmUsing mirror formula,

1 1 1

v u f or

1 1 1

v 30 30 v =–15cm

& linear magnification, 2

v 15 1m

u 30 2

i.e. final image A'B' will be located at a distance of

15cm from the mirror (towards r ight) and sincemagnification is +½ , length of final image would be

1

3.6 1.8cm2

A'B' = 1.8 cmPoint B

1 is 0.6 cm above the optic axis of mirror,

therefore, its image B' would be (0.6)

1

2= 0.3cm

above optic axis. Similarly, point A1 is 3 cm below

the optic axis, therefore, its image A' will be

1

32

=1.5 cm below the optic axis as shown below:

0.3cm

1.5cm

B'

A'A'B'=1.8cm

15cm

Total magnification of the image.

m = m1 × m

2 = (–3)

1 3

2 2

A'B' = (m)(AB)

3

2(1.2) =–1.8 cm

Note that, there is no need of drawing the raydiagram if not asked in the questions.

5 . (i ) When angle of prism is small and angle ofincidence is also small, the deviation is given by=(–1)A

Flint

Crown

A1

A2

Net deviation by the two prism is zero.So,

1 +

2 = 0 or (

1–1) A

1 + (

2–1)A

2 = 0 ...(i)

Here, 1 and

2 are the refractive indices for crown

and flint glasses respectively.

Hence,

1

1.51 1.49

2 = 1.5

and 2 =

1.77 1.731.75

2A

1= Angle of prism for crown glass = 6°

Substituting the values in Eq. (i) we get(1.5 – 1)(6°) + (1.75 – 1)A

2 = 0

This gives A2 = –4°. Hence, angle of flint glass

prism is 4° (Negatice sign shows that flint glass prismis inverted with respect to the crown glass prism.)(ii) Net dispersion due to the two prisms is

1 1 2 2b r 1 b r 2( )A ( )A

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= (1.51 – 1.49) (6°) + (1.77 – 1.73) (–4°) = –0.04°

6 . For refraction at first surface

2 1 2 1

1v R ...(i)

For refraction at second surface,

3 2 3 2

2 1v v R ...(ii)

Adding equation (i) and (ii), we get

3 3 1

2v R or

32

3 1

Rv

1

3

Therefore, focal length of the given lens system is

3

3 1

R

7 . Applying Snell's law on face AB,

(1) sin 45° ( 2 ) sinr 1

sin r2

or r = 30°

i.e., ray becomes parallel to AD inside the block.

Now applying,

2 1 2 1

v u Ron face CD,

1.514 2 1.514 2

OE 0.4Solving this equation, we get OE = 6.06 m

8 . Differentiating the lens formula 1 1 1

v u fwith respect to time,

we get 2 2

1 dv 1 du. . 0dt dtv u

(as f = constant)

2

2

dv v du

dt dtu ...(i)

Further, substituting proper values in lens formula,we have

1 1 1

v 0.4 0.3(u=–0.4m, f=0.3m) or v = 1.2 m

Putting the values in equation (i)Magnitude of rate of change of position of image

= 0.09 m/sLateral magnification

m =v

u

2

dv duu. v.

dm dt dtdt u

2

( 0.4)(0.09) (1.2)(0.01)

(0.4)= – 0.3/s

Magnitude of rate of change of lateral magnification = 0.3 /s.

9 . Critical anglest at 1 and 2

1 1= 2

3= 3 2

i

2=2i

1

1 11C

2

1sin sin 45

2

2

1 13C

2

3sin sin 60

2

Therefore, minimum angle of incidence for totalinternal reflection to take place on both slabs shouldbe 60°.

imin

= 60°

1 0 . (a) At minimum deviation, r1 = r

2 = 30°

From Snell's law 1

1

sin i

sin r or

1sin i3

sin 30

1

3sin i

2or i

1 = 60°

(b) In the position shown net deviation suffered bythe ray of light should be minimum.Therefore, the IInd prism should be rotated by 60°(anticlockwise)

B,D E

A C

1 1 .f

mf u

25

20m 4

20 25

and

50

20 2m

20 50 3

25

50

m 4 36

m 2

1 2 .1 1 1 R

f 10mv u f 2

1st condition 1

1

3 1 1u 50m

25 u 10

2nd condition 2

2

7 1 1u 25m

50 u 10

Velocity = u 25 18

3km / ht 30 5