Solutions of Optics & Modern Physics. Lesson 26th to 30th

99 Pages·1.94 MB·English

by  DC Pandey

Solutions
of
Optics &
Modern Physics
Lesson 26 to 30
th th
By DC Pandey

26. Reflection of Light
Introductory Exercise 26.1
1. Since c= 1 where c is the speed of light w=1.5¥1011 rad/s
m e
0 0 fi 2pn=1.5¥1011
1
in vacuum hence unit of is m/s. 1.5
m e fi n= ¥1011 Hz
0 0 2p
2. Hence w 1.5¥1011
Speed of the wave v= =
B =2¥10-7 Tsin[500x+1.5¥1011t] k 500
y
Comparing this equation with the standard
=3¥108 m/s
wave eqution B =B sin[kx+ wt]
y 0
Let E be the amplitude of electric field.
2p 0
k=500m-1 fi k=
l Then E =cB =3¥108 ¥2¥10-7
0 0
fi l=2p m =60 V/m
k
Since wave is propagating along x-axis and
2p p
fi l= m = metre B along y-axis, hence E must be along z-axis
500 250
fi E=60 V/m sin[500x+1.5¥1011t]
Introductory Exercise 26.2
1. Total deviation produced From figure
1 q=90∞-i
fi d=360∞-2[i+90-i]
N i
1 i =180∞
2i 90° 2 Hence rays 1 and 2 are parallel (anti-
–
0° parallel).
8
1 q q
90° 90° 2. v =2 m/s for plane mirror v =2 m/s.
0 i
180°–2q
Velocity of approach =v +v =4 m/s.
0 i
N
2
3. In figure, AB is mirror, G is ground, CD is
d=180∞-2i+180∞-2q pole and M is the man. The minimum height
to see the image of top of pole is =EN
d=360∞-2(i+ q)
2
BC¢ 2
L' tanf= = =1
m L q A CC¢ 2
8 q f=45∞
N
2 m
So, NK =4¥ tan45∞=4 m
K N1 ff B Hence in minimum height
=6m + 4m =10m
f
M C C' 6 m In DAC¢C
pole 4
tanq= =2
= 4 m
2
E 2 m C
2 m D
In DL¢LA we get,
=EK + KN =6+ KN LL¢
=tanq
LA
Now in DNKB,
LL¢
NK fi =2
=tanf fi NK = KBtanf
4
KB
fi LL¢=8 m
=4tanf
Maximum height =CA + LL¢=8+8=16 m
In DBC¢C we get,
Introductory Exercise 26.3
1 -1+2
1. Here f =-10 cm (concave mirror) fi =
v 10
(a) u=-25 cm
fi v=10 cm
Using mirror formula,
Hence, image is virtual, erect and two time
1 1 1
+ = of the object.
v u f
1
2. Here u=-3 m, f =- m,
1 1 1 1 1
fi = - =- + 2
v f u 10 25
we have,
1 -5+2
fi = 1 1 1
v 50 (a) = -
v f u
50
fi v=- =-16.7 cm 1 1
3 fi =-2+
v 3
Hence image is real, inverted and less
fi v=-0.6 m
height of the object.
As ball moves towards focus the image
(b) Since u=-10 cm,
moves towards -• and image is real as the
Hence object is situated on focus of the distance decreases by focal length image
image formed at •. become virtual which moves from + • to
(c) u-5, f =-10 zero.
1= 1 -1 =- 1 + 1 (b) The image of the ball coincide with ball,
v f u 10 5 when u=-R=-1 m
3
1
Using h=ut+ gt2 4. Since the incident rays and reflected rays
2
are parallel to each other therefore mirror is
2h 2¥2 plane mirror.
fi t= =
g 9.8
5. Let us solve the first case :
=0.639 s
Similarly again images match at t=0.78 s.
q M
3. Since image is magnified, hence the mirror q
B
is concave.
q
-v -v a a 2q q
Here, m= fi =5 a
u u A A' F C
fi v=-5u …(i)
Let distance between mirror and object is x. 40 cm
20 cm
Since image is formed at a distance 5 m
20 cm
from mirror
By applying the geometry we can prove that,
v=-(5+ x) …(ii)
40
PA¢=v= cm
From Eqs.(i) and (ii), we get
3
-(5+ x)=-5x
Further, in triangles ABP and PA¢B¢ we
fi 4x=5 have,
AB A¢B¢
fi x=1.25 =
40 (40/3)
Hence mirror is placed at 1.25 m on right
side of the object by mirror formula AB 2
\ A¢B¢= = cm
1 1 1 3 3
+ = ,
v u f Similary, we can solve other parts also.
we have 6. Simply apply :
1 1 1 1 1 1
=- - = =
f 6.25 1.25 v u f
-6.25 I -v
fi f = , and m= = for lateral magnification. If
6 o u
6.25 magnitfication is positive, image will be
Hence R=2f fi R=- =-2.08 m
3 virtual. If magnification is negative, image
will be real.
Thus mirror is concave mirror of radius of
curvature 2.08 m.
4
AIEEE Corner
Subjective Questions (Level 1)
¢
1. Here v=39.2 cm, hence v=-39.2 cm Hence the images distance are 2nb, where
n=1,2,K . Ans.
and magnification m=1
5. Suppose mirror is rotated at angle q about
fi h =h =4.85
i o its axis perpendicular to both the incident
Hence image is formed at 39.2 cm behind ray and normal as shown in figure
the mirror and height of image is = 4.85 cm.
y
2. From figure, angle of incident =15∞ I R
N
N Incident ray
90°
Reflacted ray Mirror i i
1
15q0°50° x
Horizontal (a)
15°
15° y
IV
R'
Let reflected ray makes an angle q with the i–2q
I
horizontal, then
i–q q i–q
q+15∞+15∞=90∞ fi q=60∞
x
3. A C q
(b)
O''' O'' O' O'''
m
30 cm c m m In figure (b) I remain unchanged N and R
0 c c
3 o o shift to N¢ and R¢.
50 cm 1 150 cm 70 cm
From figure (a) angle of rotation =i,
B 40 cm D
From figure (b) it is i-2q
Since mirror are parallel to each other •
Thus, reflected ray has been rotated by
image are formed the distance of five closet
angle 2q.
to object are 20 cm, 60 cm, 80 cm, 100 cm and
6. I is incident ray –i=30∞=–r
140 cm.
1.6m
4. The distance of the object from images are A B
2b,4b,6b..... etc.
A C
0°3
2 b 3 0
° 20 cm
R
O''' O'' b b O' O''' I
b A' x P B'
1 4b
4b
From D PA¢A, we get
B D
5
x 1 1 1
=tan30∞ fi x=20tan30∞ Using mirror formula, + =
20 v u f
AB 160cm 1 1 1
No. of reflection = = fi = -
x 20cm¥ tan30∞ v f u
=8 3 ª14 fi 1=- 1 + 1 = -165+11
v 11 16.5 16.5¥11
Hence the reflected ray reach other end
after 14 reflections. 16.5¥11
fi v=- =-33 cm
5.5
7. The deviation produced by mirror M is
1
Hence the image is formed at 33 cm from the
=180∞-2a
pole (vertex) of mirror on the object side the
M
1 image is real, inverted and magnified. The
Z' absolute magnification
I
A a 1 ΩvΩ 33
|m|= = =2
180°–2a90°–aa R2 ΩuΩ 16.5
q 90°–Rf1 ff180°–2q Hence size of image is hi =2¥ h0
=2¥6=12 mm.
C R
9. Here u=-12 cm, f =+ =+10 cm
2
and the deviation produced by mirror M is
2
=180-2 Using mirror formula
1 1 1
Hence total deviation + =
v u f
=180-2a +180-2f
we get
=360-2(a + f)
1 1 1 1 1
= - = +
In D ABC we get, v f u 10 12
90-a + q+90-f=180 6+5
=
fi a + f=q 60
60
Hence deviation produces =180-2q. fi v= cm =5.46 cm
11
R 22
8. Here f =- =- =-11 cm
2 2 The image is formed on right side of the
60
vertex at a distance cm. the image is
Object height h =6 mm
0 11
u=-16.5 cm virtual and erect the absolute magnification
ΩvΩ
(a) The ray diagram is shown in figure is given by |m|=
ΩuΩ
B Ω 60 Ω 5
fi |m|=Ω Ω=
Ω11¥(-12)Ω 11
A'
A f
Q m<1
Hence image is de-magnified.
B' u = 16.5 cm
Height of image h =|m|¥ h
i 0
6
5 45
fi h = ¥9= =4.09 mm fi v=3¥u
i 11 11
fi v=3u and v is +ve
The ray diagram is shown in figure
By mirror formula,
1 1 1 1 1 1
B + = fi - -
B' v u f 3u u 12
1-3 1
A 12 cm A' F fi =- fi u=8 cm
5/11cm 3u 12
(O)
(b) Since image is real
v
10. Here f =-18 cm fi m=- =3 fi v=-3u
u
1 1 1
Let distance of object from vertex of concave By using + = , we get
mirror is u. Since image is real hence image v u f
and object lie left side of the vertex. 1 1 1 -4 1
- - =- fi =-
v 1 e u 12 3u 12
Magnification m=- =
u 9 fi u=16 cm
fi v=-u -v 1 u
(c) Here m= = fi v=-
9
m 3 3
1 1 1 1 1 1
By mirror formula, + = , we have fi - - =-
v u f u/3 u 12
1 1 1 10 1 4 1
- - =- fi - =- fi - =- fi u=48 cm
u/9 u 18 u 18 u 12
1 1 1
fi u=180 cm (left side of the vertex). 13. We have + =
v u f
11. Here u=-30 cm, since image is inverted.
uf
fi v= at u= f, v=•
Hence the mirror is concave. u- f
1 -v u
m= = fi v=- The variation is shown in figure
2 u 2
1 1 1 v(m)
Using mirror formula, + = , we get
v u f
0.5
2 1 1 -3 1
- - = fi = 0.25
u u f u f u(m)
u 30 0.25 0.5
fi f =- =- =-10 cm
3 3
Hence mirror is concave of focal length
10 cm. Hence focal length if assymtote of the curve.
24 When u< f, Image is virtual. It means v is
12. Here f =- cm =-12 cm
2 negative.
(a) Since image is virtual When u=2f
v
fi m= fi v=mu v=2f
u
uÆ0, vÆ0
7
14. Here f =21 cm fi R=2f =42 cm Let v be the distance of the image from pole
(vertex) of convex mirror.
Since the object is placed on C. Hence its
1 1 1
image by concave mirror is formed on C. This Using + = , we get
v u f
image acts as a virtual objet for plane mirror
1 1 2 xR
the distance between plane mirror and - = fi v=
virtual object =21 cm. v x 12 2x+ R
Hence plane mirror forms its real image in For concave mirror
front of plane mirror at 12 cm. È xR ˘ È2R2 +5xR˘
u¢=-Í2R+ ˙ =-Í ˙
15. Let u is the object distance from vertex, v is 2x+ R 2x+ R
Î ˚ Î ˚
the image distance for vertex and f is the
R
focal length then distance between object v¢=-(2R-x) and f¢=-
2
and focus is u- f and distance between
1 1 1
image and focus is v- f ie, Using + = , we get
v¢ u¢ f¢
(u- f)(v- f)=uv-(u+v)f + f2 …(i)
1 (2x+ R) 2
1 1 1 - - -
Using + + , we get (2R-x) (2R2 +5xR) R
v u f
fi 4R3 -2x2R+8xR2
uv=(u+v)f …(ii)
=8R3 +16xR2 -10x2R
Putting the value of uv in RHS of Eq. (i), we
get fi 4R3 +8xR2 -8x2R=0
(u- f)(u- f)=(v+u)f -(v+u)f + f2 fi 4R[R2 +2xR-2x2]=0
(u- f)(v- f)= f2 fi 2x2 -2xR-R2 =0
Hence proved. Q Rπ0
2R±2 3R [1± 3]
16. Let object is placed at a distance x from the
fi x= = R
convex mirror then for convex mirror 4 2
R Ê1+ 3ˆ
u-x and f =+ fi x=Á ˜R
2 ËÁ 2 ¯˜
Objective Questions (Level 1)
¢
1. When convergent beam incident on a plane 2. When an object lies at the focus of a concave
mirror, then mirror forms real image mirror u=- f focal length of a concave
mirror is negative.
Plane mirror
Using mirror formula
1 1 1
+ =
I v u f
O
we get,
Virtual object 1 1 1
- =- fi v=•
v f f
8
v
also magnification m=- =•. 6. From figure
u
1
Hence, correct option is (c) •,•. 2
3. Total deviation, d=d + d
1 2 q
q
9
0
°–
qq N1 70 q° 70°
180°
2q
20° + q
a a
180°–2a 20+ q=70∞
q=70∞-20∞
N
2 q=50∞
=180-2q+180-2a Here (1) and (2) are paralledl 11 to each
other.
but a =90-q
Hence the correct option is (a) =50∞.
fi d=180-2q+180-2(90-q)
7. The radius of curvature of convex mirror
fi d=180∞
R=+60 cm.
Hence, option (a) is correct.
R
4. A concave mirror cannot from a virtual Its focal length f = =+30 cm
2
image of a virtual object.
v 1
Magnification m= =
Hence option (a) is correct.
u 2
5. For a concave mirror for normal sign u
fi v=
convention if u=- f fi v=• 2
1 1 1
and at u=-•, v=- f Using mirror formula, + = ,
v u f
graph between u and v is
1 1 1
v we get, - - =
u/2 u 30
-3 1
fi =
u 30
fi u=-90 cm
u
u
v= =-45 cm
2
Hence distance between A and B is
=90-45
The dotted lines are the asymptotes
=45 cm
(tangent at •) of the curve.
Hence the correct option is (c).
Hence correct option is (b).