Prism questions

CPP-8

Batches - PHONON

Class - XI

PRISM 1.

Sol.

2.

Sol.

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 i=i r =A A 1 × sin i = µ sin r 90° sin i = µ sin A i A 90° For small angle µ  i = µA

Find the angle of deviation suffered by the light rays shown in figure for following two condition the refractive index for the prism material is  = 3/2. (i) When the prism is placed in air ( = 1) (ii) When the prism is placed in water ( = 4/3) 3º Ans. [(i) 1.5º; (ii) ] 8 i = 3° =

3  = 180 60

=r–i For small angles = sin    r = sin r i = sin i (i) Prism is in air 3  sin i = 1 sin r 2  3  3  = r=i· = 40 2 60 2    180 3  =r–i= = = = = 1.5° 40 60 120 120 2   = 1.5° (ii) Prism is in water 4 3  sin i = sin r 3 2 3 3 sin r = sin i 2 4 9  3 r=  = 8 60 160 3   180 3  =r–i= = = = 160 60 480 480 8 3   = ° Ans. 8

3º 90º 30° i

3º

r i

Page 1

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

(iii) Angle of maximum deviation is : (A) 45º (B) sin–1 ( 2 sin 15º)

(C*) 30º + sin –1( 2 sin 15º) (D) None

(D) None

60°

Sol.

i1

r1 r2

i2

r1 + r2 = 60°

(i) For minimum deviation 

r1 – r2 =

60 = 30° 2

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 :  1  r2 = c = sin–1   = 45°  2



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.

Sol.

5.

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. Ans. [n > 2 ] 45° > c A sin 45° > sin c 45° 1 1 > n 2 B C n> 2

A

B

C

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

Page 2

60° 2 sin c = < 0.86 3

Sol.

90°

1.5 (2) 60° 60° 60° 30° 60° 60° r A prism having refractive index 2 and refracting angle 30º, has one of the refracting surfaces polished. A beam of (1) light incident on the other refracting surface will retrace its path if the angle of incidence is : (A) 0º (B) 30º (C*) 45º (D) 60º  > c Angle between ray (1) and (2) is 60° as shown in figure.

6.

Sol.

1 sin  = 2 sin 3

30°

1

sin  =

30°

2

 = 45° 7.

Sol.

8.

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 strikes the second surface at an angle of incidence 60º. Ans. [90º]  × sin i = µ sin 30º 

sin i = 2 ×

 

i= i = 90º

1 2

i

60º 30º

30º 60º

Light is incident normally on face AB of a prism as shown in figure. A liquid of refractive index µ is placed on face AC of the prism. The prism is made of glass of refractive index 3/2. The limits of µ for which total internal reflection takes place on face AC is : (A) µ >

3 2

3 3 4 3 (D) µ < 2

(B*) µ <

(C) µ > 3  3/ 2  1   = µ   sin C

Sol.

µ 60º



µ=

=

9.

Sol.

3 3 /2 × 2

30º 3/2

3 3 3 3  µC 4 4

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 µ=

6000 4000

= 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 minimum deviation. (ii) A light ray emerges from the prism at the same angle as it is incident on it. Determine the angle by which the rays is deflected from its initial direction as a result of its passage through the prism : 1 Given : sin–1 = 30º, sin–1 0.4 = 25º, sin–1 0.6 = 37º. 3 Ans. [(i) 37º, This deviation is not minimum; (ii) 38º = m = 2 sin–1(3/4) – 60º] Page 3

sin 60º = 1.5 sin r1

Sol.



sin r1=

=

3 2 1 3

 r1 = sin–1

60º

 1     3

e 60º

i2 = (60 – r1) sin e = 1.5 + sin (60 – r1) = 1.5 (sin 60º cosr1 – cos60º sin r1)  sine =

2 6 –3 4 3

r1

r2 120º

e = sin–1 (0.4) = 25 = i + e – A = 60 +25 – 60 i  e, hence this is not minimum deviation.

Page 4

CPP-9

Class - XI

Batches - PHONON

PRISM 1.

Sol.

2.

Sol.

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

A triangular glass wedge is lowered into water ( = 4/3). The refractive index of glass is g = 1.5. At what angle  will the beam of light normally incident on AB reach AC entirely ? 8 Ans. [ > sin–1 ] 9  > C

 4/3 8 > sin–1   = sin–1    3/ 2  a 3.

Sol.

For a prism of apex angle 45º, it is found that the angle of emergence is 45º for grazing incidence. Calculate the refractive index of the prism : (A) (2)1/2 (B) (3)1/2 (C) 2 (D*) (5)1/2 r1 + r2 = A = 45º 45º r1 = C r2 = 45 – C sin r2 = sin (45 – C) µ µ sin r2 = sin i2 =

1 2

1

sin r2 =

2µ

and sin C =

1 µ

1 2µ

= sin 45º cosC – cos 45º sin C

1 = µ

1  1  1 – 2  –  µ  µ

2 4 1 1 1– 2  2 = 1 – 2 = µ µ µ µ

µ = (5)1/2 Page 5

4. Sol.

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 r1 = r2 = C r1 + r2 = A = 90º 90º So, r1 = r2 = 45º = C 90º 90º r r 1

2

1 1 sinC = = n 2 n= 5.

D

2

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

C

70º

n2

n

20º

1 10.8 104 1.80 104 according to n1 = 1.20 + and n = 1.45 + where  2 60º 40º 2 2 A B is in nm. (i) calculate the wavelength 0 for which rays incident at any on the interface BC pass through without bending at the interference. Ans. [0 = 600 nm, n = 1.5]

Sol.

n1 = 1.2 +

10.8 104 1.8  104 and n = 1.45 + 2 2 2

the incident ray will not deviate at BC if n1 = n 2 9  10 4 = 0.25  l0 = 600 nm  20

n = 1.5 6.

Sol.

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 of curvature 20 cm. The distance of the point (where the rays are focused after reflection from the mirror) from the principal axis is : (A) 9 cm (B*) 1.5 7 mm (C) 3.14 mm (D) None of these  = (µ – 1) A = 0.9º = 0.9 –

 rod 180

y     = 0.9  × 10 cm   y= 0.9 × (10cm)  180  180

y = 1.57 mm 7.

Sol.

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

µblue= 1.77 µred = 1.73

60º A Croun glass

Page 6

µy =

1.51  1.49 = 1.5 2

µy =

µ ' y –1 A = Al µy –1

1.77  1.73 = 1.75 2

= 1+

7 –1 A 4 = Al 1.5 – 1

=

3 4

7 4

6 3/ 4 3 2 3 = = = Al 1/ 2 4 2

62 =4 3 Net dispersion if system is v – r = (µv – µr)A – (µ'v – µ'r)A' = (0.02)6º – (0.04)4º = 0.12º – 0.16 = –0.04º

A' =

8.

Sol.

9.

Sol.

An equilaterial prism is kept on a horizontal surface. A typical ray of light PQRS is shown in the figure. For minimum deviation : (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 60º For minimum R  = 30º S Q So, (1)1 = (1)2 Hence R is || to box. P 60º 60º

R Q

S

P

B The faces of prism ABCD made of glass with a refractive Index n from dihedral angles A = 90º,  B = 75º,  C = 135º &  D = 60º (The Abbe's prism). A beam of light falls on face AB & after total internal reflection from face BC escapes through face AD. Find the range of n and angle of incidence  of the beam onto face AB, if a beam that has passed through the prism in this manner is perpendicular to the incident beam. A Ans. [r +  = 75º;  = 45º (geometry), c < 45º 2 > n > 2 , 45º <  < 90º (snell's law)] EBFH EHF = 180º = 75º = 105º EHF r + + 105º = 180º +r = 75º CDGF 135º + 60º + 90º + r + 90º – = 360º – r = 15º B  = 45º r = 30º 75º C Let C = Critical angle So > C (90 – ) 135º C < 45º sinC < sin45º r 105º

C

D

E

1 1 < n > n 2

H

2 90º

Snell's law 

1 × sin  = n × sin r = n + sin 30º =

n  sin =  2sin  = n > 2 nmax = 2sin = 2sin 90º

n 2

r 90+r

90º

2

A

G

60º D

Page 7

nmax = 2 

sin  >

1 2

 > 45º

Range of   45º <  < 90º Range of n 

10.

Sol.

2