prism questions

8/12/2019 Prism questions

1/13

Page 1

CPP-8 Class - XI Batches - PHONON

PRISM

1. A ray of light is incident at angle ion a surface of a prism of small angleA& emerges normally from the oppositesurface. If the refractive index of the material of the prism is , the angle of incidence iis nearly equal to :(A)A/ (B)A/(2 ) (C*)A (D) A/2

Sol. i = i

iA

90

90

Ar=A

1 sin i = sin rsin i = sinA

For small angle i = A

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)

3

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

Ans. [(i) 1.5; (ii)38

]

Sol. i = 3 =3

180

=

60

390

i

ir30

= r iFor small angles = sin

r= sin ri = sin i

(i) Prism is in air

3

2sin i = 1 sin r

r= i3

2=

3

60 2

=

40

= r i=40 60

=

120

=

180

120=

3

2= 1.5

= 1.5(ii) Prism is in water

3

2sin i =

4

3sin r

sin r=3 3

2 4

sin i

r=9

8 60

=

3

160

= r i=3

160 60

=

480

=

180

480=

3

8

=3

8 Ans.


2/13

Page 2

3. A prism of refractive index 2 has refracting angle 60. Answer the following questions

(i) In order that a ray suffers minimum deviation it should be incident at an angle :(A*) 45 (B) 90 (C) 30 (D) None

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

(iii) Angle of maximum deviation is :(A) 45 (B) sin1( 2 sin 15) (C*) 30 + sin1( 2 sin 15) (D) None

Sol.r1 r2

i1 i2

60

r1+ r2= 60

(i) For minimum deviation

r1 r

2=

60

2

= 30

1 sin i1 = 2 sin 30 i1 = 45 Ans.(ii) at minimum deviation : r1= r2and 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= sin1

1

2

= 45

r1 = 60 45 = 15

1 sin i1 = 2 sin (15)

i1 = sin1( 2 sin 15)

and i2 = 90 max = i1+ i2A

= sin1( 2 sin 15) + 90 60

max = 30 + sin1( 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 face

AC.

A

B CAns. [n> 2 ]

Sol. 45 > cA

B C

45

sin 45 > sin c1

2>

1

n

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. [= 60]


3/13

Page 3

Sol. sin c =2

3< 0.86

60

60 6060

606030

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 30, has one of the refracting surfaces polished. A beam of

light incident on the other refracting surface will retrace its path if the angle of incidence is :(A) 0 (B) 30 (C*) 45 (D) 60

Sol. 1 sin = 2sin 3

30

30

sin =1

2

= 45

7. A prism (n= 2) of apex angle 90 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 60.Ans. [90]

Sol.

sin i= sin 30

sin i= 2 1

2

60 30

3060

i

i= i= 90

8. Light is incident normally on faceABof a prism as shown in figure.Aliquid of refractive index is placed on faceACof the prism. The prism ismade of glass of refractive index 3/2. The limits of for which total internal

reflection takes place on faceACis :

(A) >

2

3(B*) 3 (D) sin189

]

Sol. >C

> sin14 / 3

3/ 2

= sin18

a

3. For a prism of apex angle 45, it is found that the angle of emergence is 45 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= 45 45

r1=Cr2= 45 C

sin r2= sin (45 C)

sin r2= sin i2=1

2

sin r2=1

2 and sin C=1

1

2 = sin 45 cosC cos 45 sin C

1

= 2

11

1

2

= 211 24 = 1 21

= (5)1/2

PRISM



6/13

Page 6

4. The maximum refractive index of a material, of a prism of apex angle 90, for which light will be transmitted is :

(A) 3 (B) 1.5 (C*) 2 (D) None of these

Sol. r1= r2= Cr1+ r2=A= 90

909090

r1 r2So, r1= r2= 45 = C

sinC=1

n

=1

2n = 2

5. A prism of refractive index n1and another prism of refractive index n2are stuck together without a gap as shown in the figure. The angles ofthe prisms are as shown, n1and n2depend on , the wavelength of light

according to n1= 1.20 +4

2

10.8 10

and n2= 1.45 +4

2

1.80 10

where

is in nm. A B

D

C 70

60 40

20

n2

n1

(i) calculate the wavelength 0for which rays incident at any on the interfaceBCpass 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 atBCifn1 = n24

20

9 10 = 0.25l0= 600 nm

n= 1.5

6. A parallel beam of light is incident on the upper part of a prism of angle 1.8 & 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. = ( 1)A= 0.9 = 0.9 180

rod

(10 )

y

cm = 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 6 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= 4, = 0.04]

Sol. For croun glass For flirit glass

Croun glass

flirit glass

A60blue = 1.51 blue= 1.77red = 1.49 red = 1.73


7/13

Page 7

y =1.51 1.49

2

= 1.5 y =

1.77 1.73

2

= 1.75

A

Al=

1

1

'y

y

= 1+

3

4

AAl

=

71

41.51

= 74

6

Al=

3/ 4

1/ 2=

3 2

4

=

3

2

A' =6 2

3

= 4

Net dispersion if system isv r = (v r)A ('v 'r)A'

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

= 0.04

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

Q

P

R

S(A) The ray PQmust be horizontal(B) The rayRSmust be horizontal(C*) The ray QRmust be horizontal(D) Any one of them can be horizontal

Sol. For minimumR

P

S

60

60 60

Q

= 30So, (1)1= (1)2Hence Ris || to box.

9. The faces of prismABCDmade of glass with a refractive Index nfrom dihedral anglesA= 90, B= 75, C= 135 & D= 60 (The Abbe's prism). A beam of light falls on faceAB& after total internal reflection from faceBCescapes through faceAD. Find the range ofnand angle of incidence of the beam onto faceAB, if a beam that has passed through theprism in this manner is perpendicular to the incident beam.

A

B

C

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

Sol. EBFH EHF= 180 = 75 = 105EHF r+ + 105 = 180 +r= 75

CDGF135 + 60 + 90 + r + 90 = 360 r= 15 = 45 r= 30Let C= Critical angle So > C

C < 45sinC < sin45

1

n 2

75

135

60

90

105

90

90+r

r

r

H

E

G

(90 )

A D

C

B

Snell's law

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

n

sin=2n

2sin = n> 2

nmax= 2sin= 2sin 90


8/13

Page 8

nmax = 2

sin >1

2 > 45

Range of 45 < < 90

Range of n 2 < n< 2

10. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prismsQandRof identical shapeand of same material as Pare 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.


9/13

Page 9

1. A ray of light in air is incident on faceABof an irregular block made ofmaterial with refractive index 2 , as shown in figure. The face CDopposite toABis 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 PQat pointE. Determine the distance OEcorrect to twodecimal places.

45

60B D

QPEO

CA

n=1 n=1.514

n= 2

Ans.[OE= 6.06 m]

Sol. applying Snell's law atAB1 sin 45 = 2 sin = 30BMN= 90 + = 120DBM + BMN= 60 + 120 = 180 thereforeMN||BD

45

60B D

QPEO

CA

n=1 n=1.514n= 2

M

90N

Refraction curved surface CD.

1.514

( )OE 2

( ) =1.514 2

( 40)

1.514

OE O=

1.514 1.414

40=

0.1

40=

1

400

OE= 1.514 400 = 605.6 cm OE= 6.06 m

2. A glass prism with a refracting angle of 60 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 (49.7) = 0.760; sin (31.6) = 0.520; sin (28.4) = 0.475; sin (56) = 0.832; = 22/7]Ans. [(a) 49.7; (b) 7.27; (c)f= 12.68 cm]

Sol. (a) For red.

At min r = 2A = 60

2= 30

sin i= sin r= 1.52 sin30 = 0.76i = 49.7 Ans.

(b) For red:1 sin(49.7) = 1.52 sinr1r1= 30

r2 = 60 r1= 60 30 = 301.52 sin3 = 1 sinee= 49.7

red = i+ eA= 49.7 + 49.7 60 = 39.4For violet :

1 sin 49.7 = 1.6 sin r1r1= sin1

0.76

1.6

r1 = sin

1(0.475) = 28.4 r2 = 60 r1= 60 28.4 = 31.6

PRISM



10/13

Page 10

1 sine= 1.6 sin r2= 1.6 sin 31.6 - 0.838e= sin1(0.838) = 56.97

violet = i+ eA= 49.7 + 56.97 60 = 46.67 Angular width of spectrum

=violet red= 46.67 39.4

= 7.27

(c) length =f

= (100) 7.27180

cm

= 12.68 cm

3. In an experiment performed with a 60 prism where angle of minimum deviation for sodium light is 60 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 60 from the normal to edge of the prism. Are the observations correct ?Ans. [No]

Sol. =sin 2

sin2

A

=

sin60sin30

= 3

For energing at 60,

60

60

50 i

3

sin60 = 3 sin i

sin i=1

2 i= 30

C = 30 sin 90

l =

3

2 (not possible)

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

(A) 14

coscos

(B) 41

coscos

(C*) 14

sinsin

(D) 23

sinsin

Sol. =BD

AC =1

4

sin

sin

AD

AD

=1

4

sin

sin

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

prismABC?(b) Find the angle through which the prismDCEshould be rotated about the edge

passing through point Cso that there should be minimum deviation from the

system ?

i

A

BC

E

D

6060

6060

Ans.[(a) i= 60, (b) 60]


11/13

Page 11

Sol. For minimum deviation

i

60

r1 r2

30

ei = e

r1 = r2=180120

2

r1 = 30 =r2Thus sin i = 4 sin r1

sin i= 3 12

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

6. In the figure ABCis the cross section of a right angled prism and BCDEis the crosssection of a glass slab. The value of so that light incident normally on the faceABdoesnot cross the faceBCis (given sin1(3/5) = 37) :(A) 37 (B*) < 37(C) 53 (D) < 53

Sol. Boundary conditions

C = 90

C

sin C = cos

6 / 5

3/ 2= cos

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

7. The refractive index of a prism is . the maximum angle of the prism for which a ray incident on it will be transmitted

through other face without total internal reflection is _________.

Ans. [ 1 12 sin

]

Sol. A = r1+ r2ForAto be max r1and r2both should be more r1is max if light is incident onABat angle and r2is max of light energyat 90.So for surfaceAB

1 sin 90 = sin r1

sin r1 =1

r1= sin1

1

i

A

r2r1

B C

Similarly for surfaceAC

sinr2 =1

r2= sin1

1

A = r1 + r2

A =sin11

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


12/13

Page 12

Sol.

45 45

90

90

90

A

B

CD

E

F

r

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

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

= r

Hence =

9. Ois a point object kept on the principal axis of a concave mirrorMof radius of curvature 20 cm. Pis a prism of angle = 1.8. 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) 320

cm

Sol.

=1.8O1

O210 cm 20 cm

Due to prism two images of object O, O1and O2an formedDistance 6/n, O1, O2is [(2)(1)A]d

=10

cm

Here if we take v = 30 cmf= 10 cm

then by mirror fomula

1 1

v v =

1

f

v = 15 cm

m= v

u=

15

30=

0.1

2

Hence net length betweenI1andI2of object O1and O2respectively is


13/13

Page 13

I1I2= m(O1O2) = m 10

=

1

2

10

=

20

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

Sol.

1 = i r1 4

2r1 r2

r2e

2 = e r2=1 2= i r1 r2+e

= i+ e r1 r2= 2i 4

1sin2 =3

2sin r1

22

180 3

= r1

r1 =

4

3 180

r2 =A r1 =4

180

4

3 180

=

8

3 180

3

2sinr2 =

4

3sin e

9 8

8 3 180

= e e=

3

180

So=

5

180

4

180

= 1

prism questions

Documents