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CPP-8 Class - XI Batches - PHONON

PRISM1. A ray of light is incident at angle i on a surface of a prism of small angle A & emerges normally from the opposite

surface. If the refractive index of the material of the prism is , the angle of incidence i is nearly equal to :(A) A/ (B) A/(2 ) (C*) A (D) A/2

Sol. i = i

i Aµ

90°90°

Ar = A1 × sin i = µ sin r

sin i = µ sin AFor small angle i = µA

2. Find the angle of deviation suffered by the light rays shown in figure for following twocondition the refractive index for the prism material is = 3/2.(i) When the prism is placed in air ( = 1)

3º

(ii) When the prism is placed in water ( = 4/3)

Ans. [(i) 1.5º; (ii) 3º8 ]

Sol. i = 3° = 3

180

= 60

3º90º

i

ir30°

= r – iFor small angles = sin

r = sin ri = sin i

(i) Prism is in air

32

sin i = 1 sin r

r = i · 32

= 3

60 2 = 40

= r – i = 40 60 = 120

=

180120 =

32

= 1.5°

= 1.5°(ii) Prism is in water

32

sin i = 43

sin r

sin r = 3 32 4

sin i

r = 98 60

=

3160

= r – i = 3

160 60 = 480

=

180480 =

38

= 38 ° Ans.

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3. A prism of refractive index 2 has refracting angle 60º. Answer the following questions(i) In order that a ray suffers minimum deviation it should be incident at an angle :(A*) 45º (B) 90º (C) 30º (D) None

(ii) Angle of minimum deviation is :(A) 45º (B) 90º (C*) 30º (D) None

(iii) Angle of maximum deviation is :(A) 45º (B) sin–1 ( 2 sin 15º) (C*) 30º + sin–1( 2 sin 15º) (D) None

Sol. r1 r2i1 i2

60°

r1 + r2 = 60°

(i) For minimum deviation

r1 – r2 = 602

= 30°

1 × sin i1 = 2 sin 30° i1 = 45° Ans.(ii) at minimum deviation : r1 = r2 and i1 = i2

min = (i1 + i2) – (r1 + r2)= (45 + 45) – 60= 90 – 60

min = 30° Ans.(iii) For maximum deviation : emergent ray should become parallel to emergent surface, for that :

r2 = c = sin–1 12

= 45°

r1 = 60° – 45° = 15°1 × sin i1 = 2 sin (15°)

i1 = sin–1 ( 2 sin 15°)and i2 = 90° max = i1 + i2 – A

= sin–1 ( 2 sin 15°) + 90° – 60°max = 30° + sin–1 ( 2 sin 15°) Ans.

4. At what values of the refractive index of a rectangular prism can a ray travel as shown in figure.The section of the prism is an isosceles triangle & the ray is normally incident onto the faceAC.

A

B CAns. [n > 2 ]Sol. 45° > c A

B C

45°sin 45° > sin c

12

> 1n

n > 2

5. The cross section of a glass prism has the form of an equilateral triangle. A ray is incident onto one of the facesperpendicular to it. Find the angle between the incident ray and the ray that leaves the prism. The refractive index ofglass is = 1.5.Ans. [ = 60º]

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Sol. sin c = 23 < 0.86

60°

60° 60°60°

60°60°30°

r (1)

(2)1.590°

> cAngle between ray (1) and (2) is 60° as shown in figure.

6. A prism having refractive index 2 and refracting angle 30º, has one of the refracting surfaces polished. A beam oflight incident on the other refracting surface will retrace its path if the angle of incidence is :(A) 0º (B) 30º (C*) 45º (D) 60º

Sol. 1 sin = 2 sin 3

30°30°

sin = 12

= 45°

7. A prism (n = 2) of apex angle 90º is placed in air (n = 1). What should be the angle of incidence so that light ray strikesthe second surface at an angle of incidence 60º.Ans. [90º]

Sol. × sin i = µ sin 30º

sin i = 2 × 12

60º 30º30º 60º

i

i = i = 90º

8. Light is incident normally on face AB of a prism as shown in figure. Aliquid of refractive index µ is placed on face AC of the prism. The prism ismade of glass of refractive index 3/2. The limits of µ for which total internalreflection takes place on face AC is :

(A) µ > 23 (B*) µ < 4

33

(C) µ > 3 (D) µ < 23

Sol.3/ 2µ

=1

sin C

µ = 3 /2 × 32

30º60º

3/2

µ

=3 3

4 µC

3 34

9. The wavelength of light in vacuum is 6000 Å and in a medium it is 4000 Å. The refractive index of the medium is :(A) 2.4 (B*) 1.5 (C) 1.2 (D) 0.67

Sol. µ =60004000

= 1.5

10. Ref. index of a prism (A = 60º) placed in air (n = 1) is n = 1·5.(i) If light ray is incident on this prism at an angle of 60º. Find the angle of deviation. State whether this is a minimumdeviation.(ii) A light ray emerges from the prism at the same angle as it is incident on it. Determine the angle by which the raysis deflected from its initial direction as a result of its passage through the prism :

Given : sin–1 13

= 30º, sin–1 0.4 = 25º, sin–1 0.6 = 37º.

Ans. [(i) 37º, This deviation is not minimum; (ii) 38º = m = 2 sin–1(3/4) – 60º]

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Sol.1

sin 60ºsin r = 1.5

sin r1=3

2

=13

r1 = sin–1 13

60º

60º

r1r2

120º

e

i2 = (60 – r1)sin e = 1.5 + sin (60 – r1)

= 1.5 (sin 60º cosr1 – cos60º sin r1)

sine = 2 6 – 3

4 3e = sin–1 (0.4) = 25

= i + e – A= 60 +25 – 60

i e, hence this is not minimum deviation.

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1. A beam of white light is incident on hollow prism of glass. Then :(A*) The light emerging from prism gives no dispersion(B) The light emerging from prism gives spectrum but the bending of all colours isaway from base(C) The light emerging from prism gives spectrum, all the colours bend towards base,the violet the most and red the least(D) The light emerging from prism gives spectrum, all the colours bend towards base,the violet the least and red the most.

Sol. Surface and hollow prism behaves like slabs thus there is no disperism.

2. A triangular glass wedge is lowered into water ( = 4/3). The refractiveindex of glass is g = 1.5. At what angle will the beam of light normallyincident on AB reach AC entirely ?

Ans. [ > sin–1 89 ]

Sol. > C

> sin–1 4 / 33/ 2

= sin–1 8a

3. For a prism of apex angle 45º, it is found that the angle of emergence is 45º for grazing incidence. Calculate therefractive index of the prism :(A) (2)1/2 (B) (3)1/2 (C) 2 (D*) (5)1/2

Sol. r1 + r2 = A = 45º 45º

µ

r1 = Cr2 = 45 – C

sin r2 = sin (45 – C)

µ sin r2 = sin i2 = 12

sin r2 =12µ and sin C =

1µ

12µ = sin 45º cosC – cos 45º sin C

1µ = 2

11–µ

– 1µ

2µ = 2

11–µ 2

4µ = 1 – 2

1µ

µ = (5)1/2

PRISMCPP-9 Class - XI Batches - PHONON

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4. The maximum refractive index of a material, of a prism of apex angle 90º, for which light will be transmitted is :(A) 3 (B) 1.5 (C*) 2 (D) None of these

Sol. r1 = r2 = Cr1 + r2 = A = 90º

90º90º90º

r1 r2So, r1 = r2 = 45º = C

sinC =1n =

12

n = 2

5. A prism of refractive index n1 and another prism of refractive index n2are stuck together without a gap as shown in the figure. The angles ofthe prisms are as shown, n1 and n2 depend on , the wavelength of light

according to n1 = 1.20 + 4

210.8 10

and n2 = 1.45 +

4

21.80 10

where

is in nm. A B

D

C 70º

60º 40º

20º

n2

n1

(i) calculate the wavelength 0 for which rays incident at any on the interface BC pass through without bending atthe interference.

Ans. [0 = 600 nm, n = 1.5]

Sol. n1 = 1.2 + 4

2

10.8 10

and n2 = 1.45 + 4

2

1.8 10

the incident ray will not deviate at BC ifn1 = n2

4

20

9 10 = 0.25 l0 = 600 nm

n = 1.5

6. A parallel beam of light is incident on the upper part of a prism of angle 1.8º & R.I.3/2. The light coming out of the prism falls on a concave mirror of radius ofcurvature 20 cm. The distance of the point (where the rays are focused afterreflection from the mirror) from the principal axis is :(A) 9 cm(B*) 1.5 7 mm(C) 3.14 mm(D) None of these

Sol. = (µ – 1) A = 0.9º = 0.9 – 180

rod

(10 )ycm = 0.9

180

y= 0.9 × 180

× 10 cm

y = 1.57 mm

7. The refractive indices of the crown glass for blue and red lights are 1.51 & 1.49 respectively and those of the flint glassare 1.77 & 1.73 respectively. An isosceles prism of angle 6º is made of crown glass. A beam of white light is incident ata small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that thereis no deviation of the incident light. Determine the angle of the flint glass prism. Calculate the net dispersion of thecombined system. Ans. [A = 4º, = 0.04]

Sol. For croun glass For flirit glass

Croun glass

flirit glass

A

60ºµblue = 1.51 µblue= 1.77µred = 1.49 µred = 1.73

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µy =1.51 1.49

2

= 1.5 µy =1.77 1.73

2

= 1.75

AAl = –1

–1

'yy

µµ

= 1+34

AAl =

7 –14

1.5 –1=

74

6Al =

3/ 41/ 2 =

3 24

= 32

A' =6 2

3

= 4

Net dispersion if system isv – r = (µv – µr)A – (µ'v – µ'r)A'

= (0.02)6º – (0.04)4º= 0.12º – 0.16= –0.04º

8. An equilaterial prism is kept on a horizontal surface. A typical ray of light PQRS is shownin the figure. For minimum deviation :

Q

P

R

S(A) The ray PQ must be horizontal(B) The ray RS must be horizontal(C*) The ray QR must be horizontal(D) Any one of them can be horizontal

Sol. For minimumR

P

S

60º

60º 60º

Q = 30º

So, (1)1 = (1)2Hence R is || to box.

9. The faces of prism ABCD made of glass with a refractive Index n from dihedral anglesA = 90º, B = 75º, C = 135º & D = 60º (The Abbe's prism). A beam of light falls on faceAB & after total internal reflection from face BC escapes through face AD. Find the range ofn and angle of incidence of the beam onto face AB, if a beam that has passed through theprism in this manner is perpendicular to the incident beam.

A

BC

DAns. [r + = 75º; = 45º (geometry), c < 45º 2 > n > 2 , 45º < < 90º (snell's law)]Sol. EBFH EHF = 180º = 75º = 105º

EHF r + + 105º = 180º +r = 75º CDGF 135º + 60º + 90º + r + 90º – = 360º – r = 15º

= 45º r = 30ºLet C = Critical angle So > C

C < 45ºsinC < sin45º

1n <

12

n > 2

75º

135º

60º

90º

105º

90º 90+r

r

rH

E

G

(90 – )

A D

C

B

Snell's law

1 × sin = n × sin r = n + sin 30º = 2n

sin =2n

2sin = n > 2

nmax = 2sin = 2sin 90º

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nmax = 2

sin >12

> 45º

Range of 45º < < 90º

Range of n 2 < n < 2

10. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shapeand of same material as P are now added as shown in the figure. The ray will now suffer :(A) Greater deviation

P

Q

R

(B) No deviation(C*) Same deviation as before(D) Total internal reflection

Sol. Two prism in pair will cancel effect of each other and third will give same deviation.

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1. A ray of light in air is incident on face AB of an irregular block made ofmaterial with refractive index 2 , as shown in figure. The face CDopposite to AB is a spherical surface of radius of curvature 0.4 m. Fromthis face the refracted ray enters a medium of refractive index 1.514 andmeets the axis PQ at point E. Determine the distance OE correct to twodecimal places.

45º

60ºB D

QPEO

CA

n=1 n=1.514n= 2

Ans. [OE = 6.06 m]

Sol. applying Snell's law at AB 1 × sin 45º = 2 sin = 30ºBMN = 90 + = 120ºDBM + BMN = 60 + 120 = 180º Þ therefore MN || BD

45º

60ºB D

QPEO

CA

n=1 n=1.514n= 2

M

90ºN

Refraction curved surface CD.

1.514( )OE –

2(– ) =

1.514 – 2( 40)

1.514OE – O =

1.514 –1.41440 =

0.140 =

1400

OE = 1.514 × 400 = 605.6 cm OE = 6.06 m

2. A glass prism with a refracting angle of 60º has a refractive index 1.52 for red and 1.6 for violet light. A parallel beam ofwhite light is incident on one face at an angle of incidence, which gives minimum deviation for red light. Find.(a) The angle of incidence(b) Angular width of the spectrum(c) The length of the spectrum if it is focussed on a screen by lens of focal length 100 cm.[Use : sin (49.7º) = 0.760; sin (31.6º) = 0.520; sin (28.4º) = 0.475; sin (56º) = 0.832; = 22/7]Ans. [(a) 49.7º; (b) 7.27º; (c) f = 12.68 cm]

Sol. (a) For red.

At min r =2A

= 602

= 30º

sin i = µsin r = 1.52 × sin30º = 0.76i = 49.7º Ans.

(b) For red:1 × sin(49.7º) = 1.52 sinr1 r1 = 30º

r2 = 60º – r1 = 60 – 30 = 30º1.52 × sin3º = 1 × sine e = 49.7º

red = i + e – A = 49.7 + 49.7 – 60 = 39.4ºFor violet :

1 × sin 49.7º = 1.6 sin r1 r1 = sin–1 0.761.6

r1 = sin–1 (0.475) = 28.4º r2 = 60 – r1 = 60 – 28.4 = 31.6º

PRISMCPP-10 Class - XI Batches - PHONON

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1 × sine = 1.6 × sin r2 = 1.6 sin 31.6 - 0.838e = sin–1 (0.838) = 56.97º

violet = i + e – A = 49.7 + 56.97 – 60 = 46.67º Angular width of spectrum

= violet – red= 46.67 – 39.4

= 7.27º(c) length = f

= (100) 7.27180

cm

= 12.68 cm

3. In an experiment performed with a 60º prism where angle of minimum deviation for sodium light is 60º in air. Thefollowing experiment was done. When sodium light enters at one face at grazing incidence from a certain liquid, itemerges from the other face (in air) at 60º from the normal to edge of the prism. Are the observations correct ?Ans. [No]

Sol. µ =sin

2

sin2

A

= sin 60ºsin 30º = 3

For energing at 60º,

60º

60º

50º i

3

sin60º = 3 sin i

sin i =12

Þ i = 30º

C = 30º µsin 90º

µl =3

2(not possible)

4. The following figure represents a wavefront AB which passesfrom air to another transparent medium and produces a newwavefront CD after refraction. The refractive index of themedium is (PQ is the boundary between air and the medium) :

(A) 1

4

coscos

(B) 4

1

coscos

(C*) 1

4

sinsin

(D) 2

3

sinsin

Sol. µ =BDAC =

1

4

sinsin

ADAD

=1

4

sinsin

5. In the figure two triangular prisms are shown each of refractive index 3 .(a) Find the angle of incidence on the face AB for minimum deviation from the

prism ABC ?(b) Find the angle through which the prism DCE should be rotated about the edge

passing through point C so that there should be minimum deviation from thesystem ?

i

A

B C E

D

60º60º

60º60º

Ans. [(a) i = 60º, (b) 60º]

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Sol. For minimum deviation

i

60º

r1 r230º

ei = e

r1 = r2 = 180 –120

2 r1 = 30º = r2Thus sin i = 4 sin r1

sin i = 3 × 12

i = 60ºFor minimum deviation but both lens together to radius the deviation to zero. 60º

6. In the figure ABC is the cross section of a right angled prism and BCDE is the crosssection of a glass slab. The value of so that light incident normally on the face AB doesnot cross the face BC is (given sin–1 (3/5) = 37º) :(A) 37º (B*) < 37º(C) 53º (D) < 53º

Sol. Boundary conditionsC = 90 –

C sin C = cos

6 / 53/ 2 = cos

cos = 4/5or sin = 3/5 = 37ºFor T.I.R., < 37º

7. The refractive index of a prism is µ. the maximum angle of the prism for which a ray incident on it will be transmittedthrough other face without total internal reflection is _________.

Ans. [ 1 12 sin

]

Sol. A = r1 + r2For A to be max r1 and r2 both should be more r1 is max if light is incident on AB at angle and r2 is max of light energyat 90º.So for surface AB

1 sin 90º = µ sin r1

sin r1 =1µ r1 = sin–1

1µ

i

µ

A

r2r1

B C

Similarly for surface AC

sinr2 =1µ r2 = sin–1

1µ

A = r1 + r2

A = sin–1 1µ

8. Two mirrors, placed perpendicularly, form two sides of a vessel filledwith water. A light ray is incident on the water surface at an angle andemerges at an angle after getting reflected from both the mirrors inside.The relation between and is expressed as :(A*) = (B) > (C) < (D) All are possible, depending upon

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Sol.

45º 45º

90–

90–

90–

A

B

CD

E

Fr

From ABC From DEF90 – + 45º + 90 – = 45º + + 90 + r =

+ = 45º...(1) + r = 45º...(2)From (1) and (2) we get

= rHence

=

9. O is a point object kept on the principal axis of a concave mirror Mof radius of curvature 20 cm. P is a prism of angle = 1.8º. Lightfalling on the prism (at small angle of incidence) get refractedthrough the prism and then fall on the mirror. Refractive index ofprism is 3/2. Find the distance between the images formed by theconcave mirror due to this light :

(A) 25 cm (B) 10

cm

(C*) 20 cm (D) 3

20 cm

Sol.=1.8º

O1

O210 cm 20 cm

Due to prism two images of object O, O1 and O2 an formedDistance 6/n, O1, O2 is [(2)(µ–1)A]d

= 10

cm

Here if we take v = – 30 cmf = – 10 cm

then by mirror fomula

1 1v v =

1f

v = – 15 cm

m = – vu =

–15–30 =

0.12

Hence net length between I1 and I2 of object O1 and O2 respectively is

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I1 I2 = m(O1 O2) = m × 10

= 12

× 10

= 20

10. Light travelling in air falls at an incidence angle of 2º on one refracting surface of a prism of refractive index 1.5 andangle of refraction 4º. The medium on the other side is water (n = 4/3). Find the deviation produced by the prism.Ans. [1º]

Sol. 1 = i – r1 4º

2ºr1 r2

r2e

2 = e – r2 = 1× 2 = i – r1 – r2 + e

= i + e – r1 – r2 = 2i – 4

1·sin2º =32

sin r1

2×2

180 3

= r1

r1 =4

3 180

r2 = A – r1 =4

180º

– 43 180º

=

83 180º

32

sinr2 =43 sin e

9 88 3 180º

= e e =

3180º

So = 5

180º

– 4

180º

= 1º