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 cosC – 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 sinC = = 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
62 =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
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of same material as P are now added as shown in the figure. The ray will now suffer : (A) Greater deviation Q (B) No deviation P (C*) Same deviation as before R (D) Total internal reflection Two prism in pair will cancel effect of each other and third will give same deviation.
Page 8
CPP-10
Batches - PHONON
Class - XI
PRISM 1.
Sol.
A ray of light in air is incident on face AB of an irregular block made of material with refractive index 2 , as shown in figure. The face CD opposite to AB is a spherical surface of radius of curvature 0.4 m. From this face the refracted ray enters a medium of refractive index 1.514 and meets the axis PQ at point E. Determine the distance OE correct to two decimal places. Ans. [OE = 6.06 m] applying Snell's law at AB 1 × sin 45º =
A 45º P
O
E n=1.514
n=1 60º n= 2
B A
2 sin
= 30º BMN = 90 + = 120º DBM + BMN = 60 + 120 = 180º Þ therefore MN || BD Refraction curved surface CD.
C
45º n=1
C N O
E n=1.514
n= 2 60º B
D
M 90º
P
Q
Q
D
1.514 2 1.514 – 2 – = (OE ) (–) (40)
2.
Sol.
1.514 1.514 – 1.414 0.1 1 –O= = = OE 40 40 400
OE = 1.514 × 400 = 605.6 cm OE = 6.06 m
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 of white 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] (a) For red. A 60 = = 30º 2 2 sin i = µsin r = 1.52 × sin30º = 0.76 i = 49.7º Ans. (b) For red: 1 × sin(49.7º) = 1.52 sinr1 r1 = 30º r2 = 60º – r1 = 60 – 30 = 30º 1.52 × sin3º = 1 × sine e = 49.7º red = i + e – A = 49.7 + 49.7 – 60 = 39.4º For violet :
At min
r=
0.76 1 × sin 49.7º = 1.6 sin r1 r1 = sin–1 1.6
r1 = sin–1 (0.475) = 28.4º r2 = 60 – r1 = 60 – 28.4 = 31.6º Page 9
1 × sine = 1.6 × sin r2 = 1.6 sin 31.6 - 0.838 e = 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) 7.27 cm 180 = 12.68 cm 3.
In an experiment performed with a 60º prism where angle of minimum deviation for sodium light is 60º in air. The following experiment was done. When sodium light enters at one face at grazing incidence from a certain liquid, it emerges from the other face (in air) at 60º from the normal to edge of the prism. Are the observations correct ? Ans. [No] A sin sin 60º 2 µ= = = sin 30º sin 2
Sol.
3
For energing at 60º, sin60º =
3 sin i
60º 60º
4.
Sol.
1 sin i = Þ i = 30º 2 C = 30º µsin 90º
µl =
50º i 3
3 (not possible) 2
The following figure represents a wavefront AB which passes from air to another transparent medium and produces a new wavefront CD after refraction. The refractive index of the medium is (PQ is the boundary between air and the medium) : cos 1 cos 4 (A) (B) cos 4 cos 1 sin 2 sin 1 (C*) (D) sin 3 sin 4 µ=
AD sin 1 BD = AD sin AC 4
sin 1 = sin 4
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 the system ? Ans. [(a) i = 60º, (b) 60º]
A
D
60º
60º
i
B
60º
60º C
E
Page 10
Sol.
For minimum deviation i=e
60º
180 –120 2 r1 = 30º = r2 sin i = 4 sin r1
30º
r1 = r2 =
Thus
e
r1 r2
i
1 3× 2 i = 60º For minimum deviation but both lens together to radius the deviation to zero.
6.
Sol.
sin i =
60º
In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of so that light incident normally on the face AB does not cross the face BC is (given sin–1 (3/5) = 37º) : (A) 37º (B*) < 37º (C) 53º (D) < 53º Boundary conditions C = 90 – sin C = cos C 6/5 = cos 3/ 2 cos = 4/5 or sin = 3/5 = 37º For 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 transmitted through other face without total internal reflection is _________. 1 ] A = r1 + r2 For 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 energy at 90º. So for surface AB A 1 sin 90º = µ sin r1
Ans. [ 2 sin 1 Sol.
1 1 sin r1 = µ r1 = sin–1 µ
µ
Similarly for surface AC
r1 r2 i
1 1 sinr2 = µ r2 = sin–1 µ
A = r1 + r2
8.
B
C
1 A = sin–1 µ
Two mirrors, placed perpendicularly, form two sides of a vessel filled with water. A light ray is incident on the water surface at an angle and emerges 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 Page 11
Sol.
A C
D 45º
45º r
90– 90– B
From ABC 90 – + 45º + 90 – = + = From (1) and (2) we get = Hence =
9.
Sol.
45º...(1)
F
E
90–
From DEF 45º + + 90 + r = + r = 45º...(2)
r
O is a point object kept on the principal axis of a concave mirror M of radius of curvature 20 cm. P is a prism of angle = 1.8º. Light falling on the prism (at small angle of incidence) get refracted through the prism and then fall on the mirror. Refractive index of prism is 3/2. Find the distance between the images formed by the concave mirror due to this light : 2 (A) cm (B) cm 5 10 3 (C*) cm (D) cm 20 20
O1
O2
=1.8º
10 cm
20 cm
Due to prism two images of object O, O1 and O2 an formed Distance 6/n, O1, O2 is [(2)(µ–1)A]d =
cm 10
Here if we take v = – 30 cm f = – 10 cm then by mirror fomula 1 1 1 = f v v
v = – 15 cm m= –
v –15 0.1 = = u –30 2
Hence net length between I1 and I2 of object O1 and O2 respectively is Page 12
I1 I2 = m(O1 O2) = m × 10.
Sol.
1 = × = 10 2 10 20
Light travelling in air falls at an incidence angle of 2º on one refracting surface of a prism of refractive index 1.5 and angle of refraction 4º. The medium on the other side is water (n = 4/3). Find the deviation produced by the prism. Ans. [1º] 1 = i – r1 4º 2 = e – r2 = 1× 2 = i – r1 – r2 + e 2º e r2 = i + e – r1 – r2 r1 r2 = 2i – 4 1·sin2º =
2×
3 sin r1 2
2 =r 1 180 3
r1 = r2 = A – r1 =
4 3 180 4 4 8 – = 180º 3 180º 3 180º
4 3 sinr2 = sin e 3 2 9 8 3 5 4 =ee= So = – = 1º 8 3 180º 180º 180º 180º
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