NARAYANA EDUCATONAL INSTITUTIONS VIJ-I-ZONE. IIT - PHYSICS ASSIGNMENT (DT.08-11-11) TOPICS :
RAY OPTICS, PHYSICAL OPTICS, ROTATIONAL MOTION
========================================================================= = 01. A plan planee mirro mirrorr is place placed d with with its its plan planee at an angl anglee of 300 with the y-axis. Plane of the mirror is perpendicular to the xy plane and the length of the mirror is 3m. as insect moves along x-axis starting from a distant point with a speed 2cm/s. the duration of the time for which the insect can see its own image in the mirror is a 300s ! 200s c 1"0s d 100s #ol$ A #%&. 'n the figure figure shown shown the line line (%A) (%A) is normal normal to the the mirror mirror passing passing through through the end end point A. *y ray diagram it can !e shown that when the insect is to the left of (%) all its reflected rays will !e towards right of (%). #o it cannot see its image !ecause says are not reaching it. +hen the insect is the right of (%) its reflected rays will !e on !oth sides of the insect. ,hat means the insect is in the field of view of its image. #o it can see its image. image. #o it can see the image till it reaches the point (*) of the mirror from point (%). 3 ∴ 2t = ×100 0 ÷ cos-0 %r t 300 seconds.
02.
atch the following Column I
A *
Column II
iverging lens onverging lens oncave mirror onvex mirror
p
ocal length does not change on
4
dipping in water Always forms a virtual5 erect and
r
diminished image of a real o!6ect an form virtual5 erect and
s
magnified image of a real o!6ect an form real5 inverted and
t t
diminished image of a real o!6ect oca ocall leng length th cha chang nges es on on dip dippi ping ng in water.
#ol$
onceptual
03. 03.
A par parti ticl clee is dro dropp pped ed alon along g the the axi axiss from from a heig height ht
f
on a concave mirror of 2 focal (f) length f as shown in the figure. ,he acceleration due to gravity is g. ind the maximum speed of image. 3 1 3fg 2gf a gf ! c 2gf d 7 7
#ol$ * #%&. 'f the distance distance of the virtual virtual image image from from the the pole is is y 8 x is o!6ect o!6ect distance5 distance5 then then
1 y
1
1
x
f
− =− ⇒y=
fx f
−x 2
f dx 9ow5 = ÷ ÷ dt f − x dt 2 f f 2g − x ÷ ⇒ v' = ÷ f −x 2 dy
for v ' to !e maximum dv ' dx 07.
=0⇒x =
f 3
∴ ( v ' ) max =
3 7
3fg
A !eam of diameter d is incident on a glass hemisphere as shown in figure. 'f the radius of curvature of the hemisphere is very large in comparision to d5 then diameter of the !eam at the !ase of the hemisphere will !e
A 3 d / 7 #ol$
* d
#%&. :efraction at the curved surface
d/3
2d/3
µ 2 µ1 µ2 − µ1 − = v
u
:
1
PP is very small
AP1 1
=
*
P AP1 = d5 P1 = v
Paragraph for Question Nos.05 to 07 PASSAGE – 3
igure A shows two thin lenses & 1 and &2 placed in contact with a common optical axis. ,he lenses5 !ecause their nature is not specified are shown as dotted lines. ,hese lenses are made of the same material. An o!6ect of si;e 2 cm is
,he com!ination of lenses &1 and &2 !ehaves as A divergent lens of focal length 27 cm * convergent lens of focal length 7= cm divergent lens of focal length 7= cm convergent lens of focal length 27 cm 2
#ol$ * #%&. >iven magnification m =
3 2
=
v u
5 here u ?1 cm. ocal length of the com!ination is
istance !etween the o!6ect and the screen is A 10 cm * 120 cm 100 cm #ol$
1 f
1
1
v
u
= −
0.
#%&. >iven for second case
v u
=
=0 cm
3 5 also v − u = 20 . istance !etween the screen and lens D = v + u . 2
%n solving we get v 0 cm and u 70 cm. 0@.
+hich of the following is correct A &2 is convex lens of focal length 27 cm &2 is concave lens of focal length 7= cm #ol$ #%&. ocal length of the &1 is the lens &2 is
1 f
=
1 f1
+
1 f1
=
1 60
−
* &2 is concave lens of focal length 27 cm &2 is convex lens of focal length 7= cm
1 −40 . ocal length of the com!ination is 7= cm. ocal length of
1 f 2
0=.
&ight waves travel in vaccume along the y-axis. +hich of the following may represent the wavefront A x constant * y constant ; constant x + y + ; constant #ol$ * Belocity of light is always perpendicular to the wavefront. 0C.
Statement – 1: ,win
Co!mn – II
a
p
oncave lens in air
!
4
onvex lens in air
c
r
oncave mirror
d
s
onvex mirror
#ol$ 77. a r5 s
! p5 4
c p5 4
d r5 s
#%&. :eflected and refracted ray path depends that the incident light is !elow or a!ove the principal axis. 3
11.
A prism placed in air made up of flint glass is such that the any incident ray on one surface does not emerge from the second surface. ritical angle for flint glass is 3 0 in air. ,hen5 refracting angle A may !e a 3@0 ! "70 c @10 d @30
#ol$ #%&.or ray to not emerge from second surface5 r 2 E
⇒ ⇒
r2min
≥
A − r1max
≥ 5
*ut r1max = when i C00
⇒ ⇒
12.
A ≥ 2 A ≥ @20
STATE"ENT-1 : A ray is incident from outside on a glass sphere surrounded !y air as shown. ,his ray may suffer total internal reflection at second interface. S$&'# i#"$f%&$
Fi!" i#"$f%&$
*+%!! !,-$$ i#&i$#" %
STATE"ENT-2 internal reflection.
%i
: or a ray going from denser to rarer medium5 the ray may suffer total
#ol$ #%&. rom symmetry the ray shall not suffer ,': at second interface5 !ecause the angle of incidence at first interface e4uals to angle of emergence at second interface. Fence statement 1 is false. 13.
&ight from source falls on lens and screen is placed on the other side. ,he lens is formed !y cutting it long principal axis into two e4ual parts and are 6oined as indicated in column ''. Co!mn II Co!mn I
APlane of image move towards screen if G f G is increased
p #mall portion of each part near pole is removed. ,he remaining parts are 6oined
* 'mages formed will !e virtual
4 ,he two parts are separated slightly. ,he gap is filled !y opa4ue material
#eparation !etween image increases if G u G decreases
r ,he two parts are separated slightly. ,he gap is filled !y opa4ue material.
7
'nterference pattern can !e o!tained if screen is suita!ly positioned
s #mall portion of each part near pole is removed. ,he remaining parts are 6oined.
#ol$ A → p54H * → p545r5sH → r 5 sH → p
onceptual Pa##a$e : II ,he I#J apparatus is modified !y placing an isotropic transparent plate of high melting point 1." at room temperature and m in front of one of the slits. ,he refractive index of the plate is r =
t
=
2m m
its thic
l
0
-000 A . ,he separation !etween the slits is d 0.2 cm5 and separation !etween the slit and
=
the screen is 2 m. Assume that slits are of e4ual intensity. *ased on a!ove information5 answer the following 4uestions $ 17.
*efore insertion of plate5 mid-point of screen is location of central maxima5 then this point after the insertion of sla! is a a point of central maxima ! a point of maxima !ut not the central maxima c a point of minima d neither a point of maxima nor of minima
#ol$ #%&. +hen temperature of the plate is increased5 its refractive index increases and5 as a result5 the fringes cross at a particular point. Fere the num!er of fringes crossing through a point is due to change in refractive index of the plate. #hift in the location of a particular fringe due to change in refractive index is5
y=
( m2
-
m1 ) t
d
'0
'f intensity of uncovered slit is '0 and that of covered slit is 2 5 then intensity at mid-point of the
1".
screen5 is assume intensity remains same after light passing through plate é3 '0 ê ê ë2 a
1 ù ú 2ú û
3 '0 ´ 2 !
c 7'0
é ' 0 ê1ê ë d
1 2
#ol$ A D x = ( m 0 - 1) t = 10´10 @ m -
#%&. At %5 Df
=
2p l
´D x
=
2p ´10 ´10 -´10
-
@
-
@ =
10p 3
#o5 intensity at % is ' = '0 + =
1.
3'0 2
+
æ10p ö cos ç ç ç è 3 ø 2 2 é3 1 2'0 ´( - 0.") = ' 0 ê ê 2 ë2
'0
+
2
'0 ´'0
'f the plate is heated so that it temperature rises !y 1005 then how many fringes will cross a particular point on the screen 9eglect the thermal expansion of plate 10000
a 10000
! 10 ?7
c "000
d
3
"
#ol$ * #%&. #o5 the num!er of crossed fringes5 n=
( m2
m1 ) t
-
l
1."´2´10 - ´10´2´10 -
=
=
1@.
-
-´10
-
-
10
-
@
7
A narrow monochromatic !eam of light of intensity ' is incident on a glass plate as shown in figure. Another identical glass plate is
a @ $ 1
! 7C $ 1
c 7 $ 1
d 1 $ C
#ol$ * #%&. &et ' 100 'ntensity of light after reflecting from 1st plate I 1 = 2" Amplitude A1
=
I1
= " units
After reflecting from 2nd surface intensity
=
@" × 2"
100 @"K of this light pass through 1 after reflection . @" @" 22"
∴ I 2 = A2
I min
×
22"
=
' m ax
1=.
7
1
100
=
=
1" 7
=
@" 7
1-
= 3.@" 2
" + 3.@" = ÷ = 7C$1 " − 3.@"
A particle is dropped along the axis from a height f/2 on a concave mirror of focal length f as shown in the figure. ,he acceleration due to gravity is g . +hat is the maximum speed of the image.
a
1 2
2 fg
!
3 =
2 fg
c
3 7
3 fg
d
fg
#ol$ #%&. 'f the distance of the vertical image from the pole is (y) and (x) is distance of the o!6ect from % then
1 y y
1
1
x
f
− =− =
fx f
−x 2
f dx = ÷ dx f − x dy 2 f f V1 = 2g − x ÷ ÷ 2 f − x dy
or B1 to !e maximum dV 1 =0 dT f x = 3 3 ( V 1 ) maximum = 3 fg 7
Pa%a$%a&' (o% )!e#t*on# No#. 1+ to 21
igure shows a convex lens of focal length 12cm lying in a uniform magnetic field of * of magnitude 1.2 , parallel to its principle axis. A particle having a charge 2 × 10 −3 C and mass 2 × 10−"
1C.
,he radius of the circular path of the particle is a 2 cm ! 7 × 10 −2 cm #ol$ mV 2 ×10 −" × 7.= = = 7 ×10−2 m #%&. r = −3 qB 2 × 10 ×1.2 20. #ol$ #%&.
c 7cm
,he distance of the image point on the axis of the particle from the lens. a 3cm ! 1=cm c @.2cm A 1 1 1 V 1 V
−
=
( −1= ) 1
−
=
1
12 1=
d =cm
d 17.7cm
12
⇒V =
1= ×12 2
= 3cm
21.
,he image of the particle is circle a!out the axis. ,he radius of the image is a 7cm ! =cm d 1cm d 9one of these
#ol$
* @
#%&.
m= rimage
V
3
=2 1= = m × r object
u
=
= 2 × 7 = =cm . 22.
A watch glass has uniform thic
#ol$ * #%&.
=R R2 = R + DR ∴ diverging 1 1 < 0 P = ( µ − 1) − + ÷ R R + DR R1
23.
'n an experiment to measure the focal length of a convex lens 5 the data for image distances v for different o!6ect distances u are plotted to o!tain the three graphs of 1v against u 5 2 1 v
against
1 u
5 and 3 v Lu against u . 't is possi!le to find the focal length directly5 without any
further calculations5 from which of these graphs. a 1 #ol$ 27.
! 2
c 3
d none
A5*5 Assuming the o!6ect for the optical entity given in column '' may !e either real or virtual match them will the type of image they can form given in column ' . Co!mn I A :eal 'mage
Co!mn II p onverging lens
*
4
iverging lens
r s t
oncave mirror onvex mirror Plane mirror
Birtual 'mage
agnified 'mage iminished 'mage #ol$ A-p545r5s5t *-p545r5s5t -p545r5s - p545r5s 2".
A ray of light is incident normally on the diagonal face of a right? " angled prism µ = ÷ as shown. ,he deviation suffered !y the ray 3
is sin3@M = A
C0 M
3
÷
" 12@M * C0M
3@M
101M @7M. =
#ol$ 2.
?>,-" #%&-9'& #ome optical components are given in olumn '. ,he values of lateral magnification are listed in olumn ''. atch all the possi!le magnification values from olumn '' with the appropriate component in olumn '. 'ndicate your answer NNN. . Co!mn A Co!mn , A oncave mirror p 2 + 3 *
#ol$ 2@.
onvex mirror onvex lens oncave lens
4
+
r
−
s
?>,-" #%&-9'& A convex lens of focal length 1" cm is split into two halves and the two halves are placed at a separation of 120 cm. *etween the two halves of convex lens a plane mirror is placed hori;ontally and at a distance of 7 mm !elow the principal axis of the lens halves. An o!6ect A* of length 2 mm is placed at a distance of 20 cm from one half lens as shown in figure.
−
3 2 2 3 3 2
f. 1 " c m
f. 1 " c m
* 2 m m A
2 0 c m
7 m m
120 cm
,he final image of the point A is formed at a distance of
n 3
mm from the principle axis.
#ol$
etermine the value of n. ?>,-" #%&-9'&
2=.
#,A,JJ9, ? ' $- ,he mirage is not only formed in hot desert !ut it is also formed in a very cold regioncalled looming. #,A,JJ9, ? '' $ - ,he total internal reflection ta
#ol$
*
Passage – I
'n a modified I#J the region !etween the screen and slits is immersed in a li4uid whose
" − T ÷ until it reaches 2 7
refractive index varies with time as µ l =
a steady state value "/7. A
glass plate of thic
C
2C.
,he position of central maxima at , 0 from the point % is a 1.= mm
! 3.mm
#ol$
30.
,he time when central maxima reaches % is a 1#ec
! 2#ec
#ol$
31.
O#peed of the central maxima when it is at % is a 2 × 10 −3 m / s
#ol$
! 3 ×10 −3 m / s
c @.2 mm
d 17.7 mm
c 3#ec
d 7#ec
c 7 × 10 −3 m / s
d " ×10 −3 m / s
*
2+ / 1 #%& . µl 1P air
= ( 1P − t ) air µl + µ st
( s1 p − s2 p ) air =
yd D
1 − ÷ d µ l tD T − 7 y = ÷ d 10 − T 2 Dt y T = 0 = − ÷ " d
y =
tD µ q
dy Belocity v
dt
=
Dt
d ( 10 − T )
2
I0 at , 7sec 32.
A dou!le convex lens forms a real image of an o!6ect on a screen which is fixed. 9ow the lens is given a constant velocity 1 m/s along its axis and away from the screen. or the purpose of forming a sharp image always on the screen5 the o!6ect is also re4uired to !e given an appropriate velocity. ,he velocity of the o!6ect at the instant the si;e of the image is half the si;e of the o!6ect. a 1 m/s
! 2 m/s
c 3 m/s
d 7 m/s
#ol$ #%&.
vi / l
= m2 vo / l
uuv uv
vo
= vl 1 −
1 m2
10
33.
,wo spherical mirrors one convex and other concave are each of same radius r. ,hey are co-axially arranged at a distance 2 r from each other. A small portion in the form of circle of radius (a) is cut out from convex mirror near the pole. % A radius of 1st image of hole is 3 % * radius of second image of hole is 11 6 image distances for the second image is 11 .% difference in radii !etween 1st and 2nd image is 33
#ol$ A5*5 #%&. :eflection from m2
1 v1
+
1 2
2
= ⇒ v1 =
2 3
O1
1 3 :adius of 1st image 1 = % / 3 :eflection at convex
agnification m1 =
v1 u
=
f u−f
=
−3 11
2
=
3 % 11 3
=
M2
M1
istance from convex mirror 2 −
m=
O2
2
2 3
=
4 3
% 11
37. STATE"ENT–1 A dou!le convex lens made of material of :'
µ1 is placed inside two µ3 as shown in fig. µ2 > µ1 > µ3 . A wide µ 2
li4uids of :' µ 2 and parallel !eam of light is incident on lens from left. ,he lower half will give rise to a convergent !eam and upper half of lens will give rise to a divergent !eam. 3
µ1
STATE"ENT–2 'f a lens is immersed in li4uid. ,he focal length would change.
#ol$
*
3".
,he refractive index of the medium within a certain region5 > 0 > 0, changes with y. A thin light ray travelling in the xdirection stri
#ol$
11
Y
#%&. *y #nell)s law # !i# α = &'#! "%# " 0 At ) = 0, α = 0 ,# = #0
# ( ) ) !i# α = #0 !i#0
#0
0
α
= # ( ) ) ( !i# α ) = # ( ) )
#( ))
(R5)
P
α 4
( R − ) ) R
> #0
,he material with the greatest
= 1 "' #m% = 2 1 1 !i# α = 00 − α ) = &'! −1 ( ÷ 2 2 −1 1 aximum angular si;e of arc is &'! ÷ 2 refractive index changes from #0
3.
STATE"ENT – 1
A '++i#* &)+i#($ -%vi#* "-$ !%m$ i#$"i& $#$*) %! % !+i(i#* &)+i#($ '# % *iv$# !uf%&$ -%! % +$!!$ !$$( -'7$v$ STATE"ENT – 2
8' i! ('#$ 7-$# "-$ &)+i#($ i! $f'mi#* u$ '++i#* S'+9
D Pa%a$%a&' (o% )!e#t*on# No#. 0 to +
A cylindrical glass rod of radius 0.1 m and :' 3 lies on a hori;ontal plane mirror. A hori;ontal ray of light going perpendicular to the axis of rod is incident on it
O2
O1
M 3@.
At what height from plane mirror should the ray !e incident so that it emerges from the rod at a height 0.1 m a!ove the mirror. 2 1 A 1. &m * 1 &m 3 3 2 1 6 &m &m 3 3 #ol$ A #%&. rom fig i = 2 i !i#i = 3 !i# ⇒ i = 600 2 B
i
A
H R
M
δC
D
i
i P
:
M
12
-−R 2 ⇒ - = 10 + 3 ≡ 1. &m R 3 3=. At what distance a second identical rod !e placed on the mirror such that emergent ray from the second rod is in line with incident ray on 1 st rod. A 21." cm * 31." cm 11." cm 71." cm #ol$ * #%&. or the ray to retrace path in second glass rod with centre O2 !i#i =
O1O2
= 2( R + )
"%#i = O1O2
R
⇒=
10 3
10 = 2 ÷+ 20 3
3C.
eviation suffered !y ray in second rod is A 300 * 40 ;0 600 #ol$ #%&. rom fig δ = 2 ( i − ) since the emergent ray from the second rod is in line with the incident ray on 1 st rod.
δ = 2 ( 60 − 30 ) = 600 70.
,he image of a small real o!6ect placed perpendicular to the optical axis is erect and smaller than o!6ect in si;e. 1 concave lens 2 convex lens 3 concave mirror 7 convex mirror #ol$ 1 #%&. *asic concept. 71. ,he radius of curvature of each surface of an e4ui convex lens is R = 72cm . :efractive index of the glass = 1.2" . 'f the final image forms after four internal reflections inside the lens for paraxial incident !eam calculate the distance of the image from the lens. So: 7 #%&. #olving for each reflection we get distance from the pole of the lens. 72 R
=
2 ( µ n + µ − 1)
=
2 ( 7 ×1.2" + 1.2" −1 )
= 7cm.
2. A !lac< spot B is mar
2 2 ?1 A 2 sin ÷ 3÷ 3
A sin ?1 #ol$
A
B
C
: )$
2 ÷ 3
A 2 sin ?1
13
SOL. ,o satisfy the conditions r C and r ′ C A r L r ′ 2C 2 sin ?12/3 ∴ ∴ d
QQQ
17
