24-Geometrical Optics
SECTION - I : STRAIGHT OBJECTIVE TYPE
[k.M-I: lh/ks oLrqfu"B izdkj
24.1 In the figure shown a person AB of height 170 cm is standing infront of a plane mirror. His eyes are at height 164 cm. At what distance from P should a hole be made in the mirror so that he cannot see the top of his head.
fp=k esa iznf'kZr ,d O;fDr AB ftldh Å¡pkbZ 170 cm gS] ,d lery niZ.k ds lkeus [kM+k gSA mldh vk¡[ks 164 cm Å¡pkbZ ij gSA niZ.k esa P ls fdruh nwjh ij Nsn cuk;k tk;s ftlls og viuk flj ugh ns[k ldrk A
(A*) 167 cm (B) 161 cm (C) 163 cm (D) noneof thesebuesalsdksbZugha
Sol.
6 cm A
3 cm
167 cm
B P
Figure in self explanatory. fp=k Lo;a gh le> esa vkuk gSA
24.2 is the image of a point object O formed by spherical mirror, then which of the following statement is incorrect
:
,dfcUnqor~oLrqOdkxksyh;niZ.k}kjkcuk;kx;kçfrfcEc gSrksfuEuesalsdkSulkdFkuxyrgS :
(A) If O and are on same side of the principal axis, then they have to be on opposite sides of the mirror.
;fnOoeq[;v{kds,dvksjgSarcosniZ.kdsfoijhrvksjgksrsgaSA
(B) If O and are on opposite side of the principal axis, then they have to be on same side of the mirror.
;fnOoeq[;v{kdsfoijhrvksjgSarcosniZ.kds,dvksjghgksrsgSA
(C*) If O and are on opposite side of the principal axis, then they can be on opposite side of the mirror as well.
;fnOoeq[;v{kdsfoijhrvksjgSarcosniZ.kdsfoijhrvksjgksldrsgSaA
(D) If O is on principal axis then has to lie on principal axis only.
;fn O eq[; v{k ij gS rc dsoy eq[; v{k ij gh gksrk gSA
I
Sol. O
V
= – u
I
If O and I are on same sides of PA . O
will be positive which implies v and u will be of opposite signs.
I
Similarly if O and I are on opposite sides, O
V
will be -ve which implies v and u will have same sign.
If O is on PA, I =
(O) I will also be on. P.A.
I
Sol. O
V
= – u
I
;fnOoPAds,dghrjQgSrks O /kukRedgksxkftldkrkRi;ZgSfdv oufoijhrfpUgdsgksaxsA
I
blhizdkj;fnOo,PAdsfoijhrrjQgSrks O _.kkRedgksxkftldkrkRi;ZgSfdvoudsfpUglekugkasxsA
V
;fnO,PA ij gS rks I= u (O) IHkh PAij gksxkA
24.3 An object and a plane mirror are shown in figure. Mirror is moved with velocity V as shown. The velocity of image is :
fp=k esa ,d lery niZ.k rFkk ,d fcEc iznf'kZr gSA niZ.k dks V osx ls fp=kkuqlkj pyk;k tkrk gS rks izfrfcEc dk osx gksxk &
(A*) 2 V sin (B) 2 V (C) 2V cos (D) noneof thesebuesalsdksbZugha
→ →
Sol
Vm = V0m (normal to plane mirror) ¼lery niZ.k ds vfHkyEcor~½
→ → → →
= – ( )
V Vm
→
V0 Vm
V V sin = – (0 – V sin)
V = 2V sin
24.4 Two plane mirrors are joined together as shown in the figure. Two point objects O1 and O2 are placed symmetrically such that AO1 = AO2. The image of the two objects is common if :
fp=kesafn[k ,vuqlkjnksleryniZ.k sadkstksM+kx;kgSAnksfcUnqfcEcO1vkSjO2leferh;blrjgj[kstkrs gSa fdAO1 =AO2AnksfcEcdkizfrfcEcmHk;fu"BgS;fn:
(A) = 60° (B*) = 90° (C) = 30° (D) = 45°
Sol. As AB is common and O1B = BI
O,BA and BAI are congruent
By symmetry AI is perpendicular to O1 to O2 and O1AB = BAI
BAI = 45°
and BAC = 90°
Sol. tSlk fd AB, mHk;fu"V gS vkSj O1B = BI
OBA rFkkBAIlokZxle gSA lefefr ls AI,O1O2ijvfHkyEc gSA
vkSj O1AB = BAI
BAI = 45°
vkSj BAC= 90°
24.5 The following figure represents a wave front AB which passes from air to another transparent medium and produces a new wave front CD after refraction. The refractive index of the medium is (PQ is the boundary between air and the medium). fn;kx;kfp=krjaxkxzABdksçnf'kZrdjrkgStksfdgokls,dvU;ikjn'khZek/;eesaços'kdjrkgSrFk viorZu ds i'pkr~u;k rjaxkxzCDmRiUu djrkgSA ek/;edk viorZukadgksxk %(PQ ek/;erFkk gokds e/;ijr gS)
(A)
cos1 cos4
(B)
cos4
cos1
(C*)
sin1 sin4
(D)
sin2 sin3
Sol. By snell's law: Lusy ds fu;e ls %
sini 2
sinr 1
for i = , r = and = 1.
i = 1 ds fy,, r = 4 vkSj1 = 1.
sin1
2 = sin4 .
24.6 A man starting from point P crosses a 4 km wide lagoon and reaches point Q in the shortest possible time by the path shown in the figure. If the person swims at a speed of 3 km/hr and walks at a speed of 4 km/hr, then his time of journey is (µsalt water = 4/3) :
,d O;fDr P fcUnq ls pyuk çkjEHk djrk gS rFkk 4 fdeh- pkSM+s rkykc (Lagoon) dks ikj djds fcUnq Q ij fn[kk;s x;s iFkkuqlkj U;wure laHko le; esa igq¡prk gSA vxj O;fDr 3 fdeh-/?kaVk dh pky ls rSjrk gS rFkk 4
fdeh-/?kaVk, dh pky ls pyrk gS rks mldh ;k=kk dk le; gksxk (µsalt water = 4/ 3) :
(A*) 4hr, 10 min. (B) 4 hr and 30 min. (C) 3 hr and 50 min (D) 5 hr and 10 min.
(A*) 4?kaVs, 10feuV (B)4?kaVsrFk 30feuV (C) 3?kaVkrFkk50feuV (D)5?kaVsrFk 10feuV
Sol. As we know that light travels in a path such as to reach from one point to another in shortest possible time. tSlkfdgetkursgSafdizdk'kblizdkjpyrkgSfd;g,dfcUnqlsnwljsfcUnqrdigq¡pusesadelsdele; ysrkgSA
Therefore, the man must travel along that path on which light would have travelled in moving from P to
Q.
blfy, O;fDr dksml iFk ij pyuk pkfg,] ftl ij izdk'k fcUnq P ls fcUnq Q ij pyrk gSA By Snell's law ; Lusy ds fu;e ls %
sini 2
sinr 1
1
sin r = 2 . sin i
sin r =
3 4 4
5 . 3 = 5
r = 53º
AQ = 10 Km.
From P to A : P ls A rd
5
t1 = 3
From A to Q : A ls Q rd
t2 =
10 = 5
4 2
5
5 25
T = t1 + t2 = 3
+ 2 = 6
hr.
24 1
4hr 1 hr
=
6
hr =
= 4hr + 10 minutes Ans.
24.7 A mango tree is at the bank of a river and one of the branch of tree extends over the river. A tortoise lives in the river . A mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index of water is 4/3 and the tortoise is stationary)
,dvkedkisM+unhfdukjsfLFkrgSrFkkbldh,d'kk[kkunhdsÅijc<+hgqbZgSA,ddNqvkunhesajgrkgSA
,dvkedNq,sadsBhdÅijfxjrkgSAisM+lsfxjrsgq,vkedkRoj.kdNq,dksizrhrgksxk&¼ikuhdkviorZukad
4/3 gS rFkk dNqvk fLFkj gS½
(A) g (B) 3g
4
dksbZ ugh)
4g
(C*) 3
(D) None of these (mijksDr esa ls
Sol.
x xrel
x = x
1 rel
d2xrel d2 x
dt2 dt2
a = g
24.8 In the figure ABC is the cross section of a right angled prism and ACDE 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 AC is (given sin–1 (3/5) = 37º)
fp=k esa ABC ledks.k fçTe dk dkV {ks=kQy rFkk ACDE Xykl ifêdk dk dkV
{ks=kQy gSA dk eku ftlds fy, fdj.k AB lrg ij yEcor~ vkifrr gksrh gS ijUrq AC ls u xqtjs gS (fn;k gS sin–1 (3/5) = 37º)
A E
)q
B C D
(A) 37º (B*) < 37º (C) 53º (D) < 53º
Sol. A = 90° –
r2 = A = 90° – > C
6 / 5 4
cos > sinC =
3 / 2 = 5
(C is critical angle ØkfUrd dks.k gS )
4
< cos–1 5
= 37° .
B C
24.9 Refractive index of a prism is and the angle of prism is 60º. The minimum angle of incidence
of a ray that will be transmitted through the prism is :
fizTedkdks.k60ºgSvkSjviorZukad gSAfdj.kdkU;wurevkirudks.krkfd;gfizTelsgksdjxqtjxh:
(A*) 30º (B) 45º (C) 15º (D) 50º
Sol. r2 < C ; A – r1 < C
r1 > A – C
sinr1 > sin(A – C)
sini
> sin (A – C) A
sini > (sinA cosC – sinC cosA)
1 1 1
=
sini > 1 2
.
= 1
B C
or;k i > 300 .
24.10 For a prism kept in air it is found that for an angle of incidence 60°, the angle of refraction 'A', angle of deviation ' ' and angle of emergence 'e' become equal. Then the refractive index of the prism is gokesaj[ksfizTedsfy,tcvkirudks.k60°gksrkgS]rcfizTedkviorZudks.k'A',fopyudks.k''rFk fuxZr dks.k e cjkcj ik;s tkrs gSA rc fizTe dk viorZu xq.kkad gS &
(A*) 1.73 (B) 1.15 (C) 1.5 (D) 1.33
Sol. Given i = 60° fn;k gS
= i + e – A A = = e
= i
( e = A) and rFkk = i = e
sin A m
2
= sin A 2
Here angle of deviation is minimum ( i = e)
sin 60 60
;gka fopyu dks.k U;wure gS ( i = e)
2
sin60 / 2 =
Ans.
24.11 As shown in the figure The observer 'O' sees the distance AB as infinitely large. If refractive index of
1
liquid is 1 and that of glass is 2, then 2 is :
çs{kd'O',nwjhABdksvuUr:ilsyEckns[krkgSAtSlkfdfp=kesaiznf'kZrgSA;fnnzodkviorZukad1rFk dkapdkviorZukad
2, gks rc
1
2 gS:
(A*) 2 (B) 1/2 (C) 4 (D) Noneof thesebuesalsdksbZugha
Sol. Using formula of spherical surface taking 'B' as object
xksyh;lrgdklw=kmi;ksxesaykusijtcfdBdks,dfcEcdhHkkafrfy;kx;kgSA
2 1 = 2 1
(R being the radius of the curved surface)
(2R) R
2 1 = 2 1
(R oØh;lrg dhf=kT;k gS)
(2R) R
1
2 = 2
24.12 A transparent cylinder has its right half polished so as to act as a mirror. A paraxial light ray is incident from left, that is parallel to principal axis, exits parallel to the incident ray as shown in the figure. The refractive index n of the material of the cylinder is :
,dikjxE;csyudsnk;svk/ksHk xdksikWfy'kfd;kx;kgS]ftlls;gniZ.kdhHkkafrO;ogkjdjrkgSAv{kds utnhd]eq[;v{kdslekUrjfdj.ksablijck;havksjlsvkifr gksrhgSrFk vkifr fdj.k sadslekUrjghfp=k uqlkj ckgj fudyrh gSA csyu ds inkFkZ dk viorZukad n gSa &
(A) 1.2 (B) 1.5 (C) 1.8 (D*) 2.0
Sol. For spherical surface xksyh; lrg ds fy,
using
n2 –
v
n1 =
u
n2 n1
R
mi;ksx esa ysus ij
n –
2R
1 n 1
= R
n = 2n – 2
n = 2.
24.13 In the figure shown, the image of a real object O is formed at point . AB is the principal axis of the mirror. The mirror must be: fp=kesan'k Z;svuqlkjokLrfodfcEcOdkizfrfcEcfcUnqijcurkgSAABniZ.kdheq[;v{kgSAniZ.kgksukpkfg;s &
(A) concave & placed towards right of (B) concave & placed towards left of
(C) convex & placed towards right of (D*) convex & placed towards left of .
(A) vory o ds nk¡;h rjQ j[kk gS (B) vory o ds ck¡;h rjQ j[kk gS
(C) mÙkyodsnk¡;hj[kkgS (D*) mÙky o ds ck¡;h rjQ j[kk gS
Sol. The image is erect hence,mirror must be between object & image. Virtual image of real object is diminished, hence mirror is convex.
fcEc mYVk gS] blfy, niZ.k fcEc vkSj izfrfcEc ds chp gksuk pkfg,A okLrfod fcEc dk vkHkklh izfrfcEc NksVk gksrk gS
blfy,niZ.kmÙkygksxkA
24.14 Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces w.r.t. each other is:
nks CykWd ftudk izR;sd dk nzO;eku m gS] ?k"kZ.kjfgr Vscy ij j[ks gSA m u d k s n k s n w l j s n z O ; e k u k s a l s f p = k k u q l k j ck¡/k x;kgSAf?kjfu;k¡ojLlh;k¡gYdhgSA;fnoLrqOVscyijfLFkjvoLFk esaj[khgSAnksuksaCykWddhlrgAB oCDdks ijkofrZrlrgcuk;kx;kgSAbulrgksijcu sokysnksuksizfrfcEcdk,dnwljsdslkis{kRoj.kgSA
(A) 5g/6 (B) 5g/3 (C) g/3 (D*) 17 g/6
Sol. Acceleration of block AB =
3mg 3m m
= 3 g ; acceleration of block CD =
4
2mg 2g
2m m = 3
Acceleration of image in mirror AB
= 2 acceleration of mirror
2. 3g 3
= = g
2
2. 2 g 4g
Acceleration of image in mirror CD =
=
3
4g 3g
17g
Acceleration of the two image w.r.t. each other = 3
= .
2
gy CykWd AB dk Roj.k= 3mg = 3 g; CykWd CD dk Roj.k=
2mg 2g
=
3m m 4
izfrfcEcdkniZ.kABesaRoj.k
= 2niZ.kdkRoj.k
2m m 3
2. 3g 3
= = g
2
2. 2 g 4g
niZ.kCDesaizfrfcEcdkRoj.k=
=
3
4g 3g
17g
blfy,nksuksaizfrfcEcdk,dnwljsdslkis{kRoj.k= 3
= .
2
24.15 The figure shows a parallel slab of refractive index n2 which is surrounded by media of refractive indices n1 and n . Light is incident on the slab at angle of incidence ( 0). The time taken by the ray to cross the slab is ‘t1’ if incidence is from ‘n1’ and it is ‘t2’ if the incidence is from ‘n3’. Then assuming that n2 > n1, n2 > n3 and n3
> n1, then value of t1/t2.
fp=kesa,dlekUrjifV~VdkftldkviorZukadn2gS]nksek/;eksalsf?kjhgqbZgSftlesa,ddkviorZukadn1vkSjnwljsdk n3gSAizdk'k(0) vkirudks.kijifV~Vdkijvkifr gksrkgSAifV~Vdklsxqtjusesaizdk'kfdj.k}kjkfy;kx;k le; ‘t1’gS vxj ‘n1’ls vkifrr gksrh gS vkSj;g ‘t2’gS vxj ;g ‘n3’lsvkifrr gksrhgS rc;g ekfu;s fd n2 >n1,n2 > n3 vkSj n3 > n1, rc t1/t2 dk eku gksxk
(A) = 1 (B) > 1 (C*) < 1 (D) cannotbedecidedKkr ugha fd;k tk ldrkA
Sol. In the first case the distance travelled in the slab < distance travelled in the slab in the 2nd case.
izFkefLFkfresaifV~Vdkesar;dhxbZnwjh V if |u| > |F|
(C*) Vi < VO if |u| > |F| (D) Vi = VO if |u| = |F|
(A*) Vi VO ;fn |u| < |F| (B) Vi > VO ;fn |u| > |F|
(C*) Vi < VO ;fn |u| > |F| (D) Vi = VO ;fn |u| = |F|
Sol. For convex mirror
|m| < 1 for any real object
Now, V
image
= – m2 V
object
| V
image
| < | V
object
| always.
Sol. mÙkyniZ.kdsfy,
|m|< 1okLrfod oLrq ds fy,
vc, VizfrfcEc
2
oLrq
| VizfrfcEc|< | VoLrq| lnSo
24.34 An object AB is placed parallel and close to the optical axis between focus F and centre of curvature C of a converging mirror of focal length f as shown in figure.
,d fcEc AB dks Qksdl nwjh f ds ,d vfHklkjh ds Qksdl F rFkk oØrk dsUnz C ds chp izdkf'kd v{k ds fudV rFkk izdkf'kd v{k ds lekUrj j[kk tkrk gSa] tSlk fd fp=k esa fn[kk;k x;k gSA
(A*) Image of A will be closer than that of B from the mirror.
(B) Image of AB will be parallel to the optical axis.
(C*) Image of AB will be straight line inclined to the optical axis.
(D) Image of AB will not be straight line. (A*)AdkizfrfcEc]BdsizfrfcEcdslkis{kniZ.kdsvf/kdfudVgksxkA (B)ABdkizfrfcEcizdkf'kdv{kdslekUrjgksxkA (C*)ABdkizfrfcEclh/khjs[k esaizdkf'kd&v{kls>qdkgqvkgksxkA (D)ABdkizfrfcEcljyjs[k ughgksxkA
Sol. (Tough) The image of a point closer to the focus will be farther. As the transverse magnification of B will be
more than A, the image of AB will be inclined to the optical axis. Qksdldsvf/kdfudVfcUnqdkizfrfcEcvf/kdnwjgksxkApw¡fdBdkvuqizLFkvko/kZuAlsvf/kdgksxk]blfy;sAB dkizfrfcEcizdkf'kdv{kls>qdkgqvkgksxkA
24.35 Which of the following statements is/are correct about the refraction of light from a plane surface when light ray is incident in denser medium. [C is critical angle] tcfdj.kl?kuek/;eesaizos'kdjrhgSrksizdk'kfdj.kdslerylrglsviorZudsfy,dkSulk@dkSulsdFkulR;
gSA [C ØkfUrd dks.k gS]
(A*) The maximum angle of deviation during refraction is – C, it will be at angle of incidence C.
2
(B*) The maximum angle of deviation for all angle of incidences is – 2C, when angle of incidence is slightly greater than C.
(C*) If angle of incidence is less than C then deviation increases if angle of incidence is also increased. (D*) If angle of incidence is greater than C then angle of deviation decreases if angle of incidence is increased. (A*)viorZudsnkSjkuvf/kdrefopyudks.k –Cgksxk,;gvkirudks.kCijgksxkA
(B*) lHkhvkirudks.k sadsfy,vf/kdrefopyudks.k–2CgS]tcvkirudks.kClsFk sM+klkT;knkgSA (C*) ;fnvkirudks.kClsdegksrksvkirudks.kc<+kusijfopyudks.kc<+sxkA
(D*) ;fnvkirudks.kClsT;knkgksrksvkirudks.kc<+kusijfopyudks.k?kVsxkA
Sol.
24.36 The angle of deviation () vs angle of incidence (i) is plotted for a prism. Pick up the correct statements.
,dfçTedsfy,fopyudks.k"" rFk vkirudks.k"i"dschpxzkQcuk;kx;kgSAlghdFkuksadkspqfu;s&
(A*) The angle of prism is 60°
fçTedks.k60°gSA
(B*) The refractive index of the prism is n =
fçTedk viorZukadn= 3 gSA
(C*) For deviation to be 65° the angle of incidence i1 = 55° 65° fopyu ds fy, vkiru dks.k i1 =55° gSA
(D) The curve of '' vs 'i' is parabolic ''dk'i' ds lkFk oØ ijoy; gSA
Sol. [Moderate] = i + e – A (for minimum derivation i = e)
minimum deviation = 2i – A
= i + e – A(U;wure fopyu ds fy, i = e)
U;wure fopyu= 2i – A
60 = 2 × 60 – A A = 60°
A m 60 60
sin sin
2
n = A
sin 2
2
= sin 60 =
2
1 = i1
+ e – A
65° = i1 + 70° – 60° or i1 = 55°
versus i curve is not parabolic oØijoy;ughagSA
24.37 A luminous point object is placed at O, whose image is formed at as shown in figure. Line AB is the optical axis. Which of the following statement is/are correct ?
(A*) If a lens is used to obtain the image, then it must be a converging lens and its optical centre will be the intersection point of line AB and O.
(B) If a lens is used to obtain the image, then it must be a diverging lens and its optical centre will be the intersection point of line AB and O.
(C*) If a mirror is used to obtain the image then the mirror must be concave and object and image subtend equal angles at the pole of the mirror.
(D*) is a real Image.
24.37 ,dizfrfnfIr'khyfcUnqoLrqOijj[khxbZgSAftldkizfrfcEcijfp=k uqlkjcurkgSAjs[k ABizdkf'kdv{kgSaAfuEu esalsdkSulk@dkSulsdFkulR;gSa& (A*);fnçfrfcEcdsfy,ySUldkç;ksxfd;kx;kgSrks;gvfHklkjhySUlgksxk rFk bldkçdkf'kddsUnzjs[k ABvkSjOdsdVkufcUnqijgksxkA (B);fnçfrfcEcizkIrdjusdsfy,ySUlç;ksxfd;kx;kgSrks;gvilkjhgksxkrFk bldkçdkf'kddsUnzjs[k ABvkSj
OdsdVkufcUnqijgksxkA
(C*) ;fnçfrfcEcizkIrdjusdsfy,niZ.kdkmi;ksxfd;kx;kgSrks;gvorygksxkrFk oLrqvkSjçfrfcEcniZ.kds
/kzqoijlekudks.kcukrsgSaA
(D*) ,dokLrfodizfrfcEcgSA
Sol. (A)
(C)
(D) Imageisinverted izfrfcEc mYVk gksrk gSIt shouldberealvr% ;g okLrfod gksxk
SECTION - III : ASSERTION AND REASON TYPE
[k.M- III: dFku vkSj dkj.k izdkj
24.38 STATEMENT-1 : A beam of white light enters the curved surface of a semicircular piece of glass along the normal. The incoming beam is moved clockwise (so that the angle increases), such that the beam always enters along the normal to the curved side. Just before the refracted beam disappears, it
becomes predominantly red.
STATEMENT-2 : The index of refraction for light at the red end of the visible spectrum is more than at the violet end.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
oDrO;-1: 'osrizdk'kdkiqatv)Zo`ÙkkdkjdkapdsVqdM+sdhoØlrgdsvfHkyEcdsvuqfn'kvkifrrgksrkgSA vkifrriqatnf{k.k orZblizdkj?kwerkgSfd(rkfddks.kc<+rkgS)iqatges'k oØlrgdsvfHkyEcdsvuqfn'k izos'kdjrhgSAviofrZriqatdsvn`';gksusdsrqjarigys;geq[;r%ykyjaxdhizrhrgksrhgSA
oDrO;-2: n`';LisDVªeesaizdk'kdsfy,ykyfljsdkviorZukadcSaxuhfljslsvf/kdgksrkgSA
(A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C*) oDrO;-1lR; gS] oDrO;-2vlR; gS ;
(D) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. The index of refraction for light at the red end of the visible spectrum is lesser than at the violet end.Hence statement -2 is false n`';LisDVªeesaizdk'kdsfy,ykyfljsdkviorZukadcSaxuhfljslsdegksrkgSAvr%oDrO;-2xyrgSA
24.39 STATEMENT-1 : There exists two angles of incidence for the same magnitude of deviation (except minimum
deviation) by a prism kept in air.
STATEMENT-2 : In a prism kept in air, a ray is incident on first surface and emerges out of second surface. Now if another ray is incident on second surface (of prism) along the previous emergent ray, then this ray emerges out of first surface along the previous incident ray. This principle is called principle of reversibility of light.
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
oDrO;-1: gokesaj[ksgq;s,dfizTeds}kjk¼U;wurefopyudsvykok½lekuifjek.kdsfopyudsfy,nksvkiru dks.kgksrsgSA
oDrO;-2: gokesaj[ksgq;s,dfizTeesa]igyhlrgij,dfdj.kvkifrrgksrhgSrFk nwljhlrglsckgjfudy tkrhgSAvc;fndksbZnwljhfdj.knwljhlrg¼fizTedh½ijfiNyhfuxZrfdj.kdsvuqfn'kvkifr gksrhgS]rks
;g fdj.k igyh lrg ls fiNyh vkifrr fdj.k ds vuqfn'k fuxZr gksrh gSA bl fl)kUr dks izdk'k dh mRØe.kh;rk (reversibility) dk fl)kUr dgrs gSA
(A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. Statement-2 is correct explanation of statement-1.
oDrO;-2, oDrO;-1 dklgh Li"Vhdj.k gSA
24.40 STATEMENT-1 : A ray is incident from outside on a glass sphere surrounded by air as shown. This ray may suffer total internal reflection at second interface.
STATEMENT 2 : For a ray going from denser to rarer medium, the ray may suffer total internal reflection.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True.
oDrO;-1:,dfdj.kdk¡pdsxksysijfp=k uqlkjckgjlsvkifrrgksrhgSrFkkxksysdsckgjgokgSAblfdj.kdk nwljsvUrZlrgijiw.kZvkUrfjdijkorZugksldrkgSA
oDrO;-2 : ,d fdj.k dk l?ku ek/;e ls fojy ek/;e esa tkus ij] fdj.k dk iw.kZ vkUrfjd ijkorZu gks ldrk gSA
(A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. From symmetry the ray shall not suffer TIR at second interface, because the angle of incidence at first
interface equals to angle of emergence at second interface. Hence statement 1 is false
lefer klsfdj.kdknwljhvUrlZrgijiw.kZvkUrfjdijkorZu(TIR)ughagksxk]D;ksafdizFkelrgijvkirudks.k
dk ekunwljh vUrlZrgij fuxZrdks.k ds cjkcjgksxk vr%dFku1vlR;gSA
24.41 STATEMENT-1 : Keeping a point object fixed, if a plane mirror is moved, the image will also move.
STATEMENT-2 : In case of a plane mirror, distance of object and its image is equal from any point on the
mirror.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True
oDrO;-1: ,dfcUnqoroLrqdksfLFkjtM+or(fixed)j[krsgq,;fn,dleryniZ.kdkspyk;ktkrkgSrksizfrfcEcHkh ges'k xfrdjsxkA
oDrO;-2: leryniZ.kdsfy,]niZ.kijfLFkrfdlhHkhfcUnqlsoLrqdhnwjhomlhfcUnqlsizfrfcEcdhnwjhcjkcj gksrhgSA
(A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D*) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. If the mirror is shifted parallel to itself such that the velocity of the mirror is parallel to its surface, the image
shall not shift. Hence statement 1 is false.
;fnniZ.kdksLoa;dhlrgdslekUrjblizdkjpyk;ktkrkgSfdniZ.kdkosxbldhlrgdslekUrjgS]rksizfrfcEc dkfoLFkkiu'kwU;gksxkAoDrO;-1vlR;gS]oDrO;-2lR;gSA
SECTION - IV : COMPREHENSION TYPE
[k.M- IV:cks/ku izdkj
Comprehension # 1
The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.
A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.
A person suffering for eye defects uses spectacles (Eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.
The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with
100
sign. For example power of lens required is +3D (converging lens of focal length 3 cm) then number of lens will be +3.
For all the calculations required you can use the lens formula and lens maker's formula. Assume that
the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.
us=kdhlhy;jh(ciliary)ekalis'khus=kdsysaldhoØrkfu;af=krdjrhgSrFk QyLo:iizHk ohQksdlnwjhifjofrZr dhtkldrhgSAtcek¡lis'khiw.kZJkarfLFkfresagksrhgS]rksQksdlnwjhegRregksrhgSAtcek¡lis'khfod`r;k f[kphagksrhgSa]rksysaldhoØrkc<+tkrhgS¼vFk Zr~oØrkf=kT;k?kVtkrhgS½rFk Qksdlnwjh?kVtkrhgSALi"V n`';dsfy,izfrfcEcjsfVukijgksukpkfg,ALi"Vn`';dsfy,izfrfcEcnwjhfu;rgksrhgSrFk ;gjsfVukrFk us=k ysaldhnwjhdscjkcjgksrhgSAiw.kZfodflrO;fDrdsfy,;gnwjhyxHkx2.5cm gksrhgSA
,dO;fDr]us=klscgqrnwjhijj[khoLrqvksadhLi"Vn`f"VlSU)kfrd:ilsj[krkgSA,dO;fDrdhLi"Vn`f"V dsfy,fudVrenwjhU;wurelEHk O;QksdlnwjhlslEcfU/krgSAblfLFkfresalhy;jhek¡lis'khlclsT;knkfod`r
;kf[kphgksrhgSA,d vkSlriw.kZfodflrO;fDrdsfy, oLrqdh fudVrenwjh25cm gksuhpkfg,A n`f"Vnks"klsihfM+rO;fDrp'eksa(us=kdkWap)dkiz;ksxdjrkgSAp'eksadsysUldkiz;ksxoLrqvksadkizfrfEc Li"V n`f"VdhijklesacukusesagksrkgSAp'eksadsysUllscukizfrfEc us=kysUldsfy,oLrqgksrkgSrFk ftldkizfrfEc jsfVukijcurkgSA
n`f"Vnks"kfuokj.kdsiz;qDrp'esadsysUldkØaekdvko';dysUldh{kerklsfu/k ZfjrgksrkgSrFk p'esadsysUl dk Øekad ysUl dh {kerk ds fpUà lfgr lkaf[;deku ds cjkcj gksrk gSA mnkgj.k Lo:i;fn vko';d ysUl dh
100
{kerk +3DgSA ( 3 cm Qksdlnwjh dkvfHklkjh ysUl) rcysUl dkØekad +3gksxkA
lHkhvko';dx.kukvksadsfy,vkiysallw=krFk ysalesadjlw=kdkiz;ksxdjldrsgSA;gekfu;sfdus=kysUl leryksRry ysUl gSA us=k ysUl rFkk p'es ds ysUl ds chp dh nwjh dks ux.; ekusaA
24.42 Minimum focal length of eye lens of a normal person is
lkekU;O;fDrdsus=kysUldhU;wureQksdlnwjhgSA
Sol.
25
(A) 25 cm (B) 2.5 cm (C) 9
1 1 1
cm (D*) 25 cm
11
v u
Here v = 2.5 (Distance of Retina as position of image is fixed)
;gk¡v =2.5(çfrfcEcdhfLFkfrvpy gSvr%jsfVukdhnwjh) u = – x
1 1 1
2.5 x
For f
min
: x is minimum
1
min
1 1
2.5 25
1 1 1
f ds fy, : x U;wure gSA
1 1 1
2.5 x
min
min
2.5 25
24.43 Maximum focal length of eye lens of normal person is
lkekU;O;fDrdsus=kysUldhegRreQksdlnwjhgSA
25
(A) 25 cm (B*) 2.5 cm (C) 9
cm (D) 25 cm
11
Sol. For f
max
: x is maximum
1
max
1 1
2.5
fmax
ds fy, : x vf/kdre gSA
1
max
1 1
2.5
24.44 A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be.
,dfudVn`f"VO;fDr100cm nwjhrddhoLrqvksadksLi"V:ilsns[kldrkgSrFkkmllsvf/kdnwjhdhughaA blnks"kfuokj.kdsfy,vko';dp'esadsysUldkuEcjgksxkA
(A) +1 (B*) – 1 (C) + 3 (D) – 3
Sol. For near sighted man lens should make the image of the object with in 100 cm range For lens u = – v = – 100
1
lens
1 1
100
Sol. ,d fudVn`f"V nks"khO;fDr dsfy,ySUldksoLrqdkçfrfcEc100cm dhijklesacukukpkfg,A ySUl ds fy, u = – v = – 100
1
lens
1 1
100
Comprehension # 2
Chromatic Aberration
The image of a white object in white light formed by a lens is usually coloured and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure.
As a result of this, in case of convergent lens if a screen is placed at FV centre of the image will be violet and focused while sides are red and blurred. While at FR, reverse is the case, i.e., centre will be red and focused while sides violet and blurred. The difference between fV and fR is a measure of the longitudinal chromatic aberration (L.C.A), i.e.,
L.C.A. = fR – fV = – df with df = fV – fR (1)
However, as for a single lens,
1 ( 1) 1 1
f R R
............(2)
1 2
df d 1 1
f 2
1 2
...........(3)
Dividing Eqn. (3) by (2) ;
df d
d
= dispersive power (4)
f ( 1)
( 1)
And hence, from Eqns. (1) and (4),
L.C.A. = –df = f
Now, as for a single lens neither f nor can be zero, we cannot have a single lens free from chromatic aberration.
Condition of Achromatism :
In case of two thin lenses in contact
1 1 1
F f1 f2
i.e.,
dF
= df1 df2
F2 1 2
The combination will be free from chromatic aberration if dF = 0
i.e.,
df1 df2 = 0
2 2
1 2
which with the help of Eqn. (4) reduces to
1f1 2f2 = 0
i.e.,
2
1
1 2
2
2
= 0 ...........(5)
f1 f2
This condition is called condition of achromatism (for two thin lenses in contact) and the lens combination which satisfies this condition is called achromatic lens, From this condition, i.e., from Eqn. (5) it is clear that in case of achromatic doublet :
(1) The two lenses must be of different materials.
Since, if
= , 1 1 0
1 2 f1 f2
i.e.,
1 = 0
F
or F =
i.e., combination will not behave as a lens, but as a plane glass plate.
(2) As
and
are positive quantities, for Eqn. (5) to hold, f1
and f2
must be of opposite nature, i.e., if
one of the lenses is converging the other must be diverging.
(3) If the achromatic combination is convergent,
fC < fD
and as
fC C
fD D
, <
i.e., in a convergent achromatic doublet, convex lens has lesser focal length and dispersive power than the divergent one.
vuqPNsn# 2
o.khZ;nks"k
'osroLrqdk'osrizdk'kesaysUl}kjkcukizfrfcEclkekU;r%jaxhurFk QSykgksrkgSAizfrfcEcdsblnks"kdkso.khZ; nks"kdgrsgSrFk ;gySUldhfHkUu&fHkUujaxksadsfy,fHkUu&fHkUuQksdlnwjhlsmRiUugksrkgSAySUldsdkeku cSaxuhjaxdsfy,egRretcfdykyjaxdsfy,U;wuregksrkgSAfp=kkuqlkjcSaxuhjaxySUldsutnhddsfUnzr
gksrk gS tc fd yky jax ySUl ls nwj dsfUnzr gksrk gSA
bldsifj.k eLo:ivfHklkjhySUldhfLFkfresa;fninsZdksFVijj[ksa]rksizfrfcEcdkdsUnzcSaxuh,oadsfUnzrgksxk tcfdfdukjsykyrFk QSysgksaxsAtcfdinsZdhFR fLFkfrmYVhgksxhvFk Zr~dsUnzykyrFk dsfUnzrgksxktcfd fdukjscSaxuhrFk QSysgksaxsAfVrFk fR dschpdhnwjhvuqnS/;Zo.khZ;nks"kdhekigS&
v- o- nks- = fR – fV = – df tgk¡ df = fV – fR (1)
ijUrq] ,d ySUl ds fy, &
1 ( 1) 1 1
f R R
............(2)
1 2
df d 1 1
f 2
1 2
...........(3)
vFk Zr~]leh-(3)dks(2)lsHk xnsusij&
df
f
d
( 1)
d =fo{ksi.k{kerk (4)
( 1)
vr% leh-(1) rFkk(4) ls & v-o-nks"k=–df =f
ijUrq],dySUldsfy,urksfvkSjugh'kwU;gksldrhgS]vr%ge,dySUldko.kZnks"klseqäughaikldrsA
vo.khZ;ifjfLFkfrdsfy,'krsZ%
nksirysySUlksadslEidZfLFkfresa&
1 1 1
F f1 f2
vFkZr
dF
= df1 df2
F2 1 2
;gla;kstuo.khZ;nks"keqägksxk;fndF=0
vFkZr~
df1 df2 = 0
2 2
1 2
lehdj.k (4) dh lgk;rk ls
1f1 2 f2 = 0
vFkZr
2
1
1 2
2
2
= 0 ...........(5)
f1 f2
blfLFkfrdksvo.khZ;fLFkfrdgrsgS¼nksirysySUlksadhlEidZfLFkfr½rFk tksySUlla;kstublfLFkfrdkslUrq"V djrsgS]vo.khZ;ySUldgykrsgSaAbl'krZls]vFkkZrlehdj.k(5)ls]f}vo.khZ;dsfy,;gLi"VgS&
(1) nksuksa ySUlksa dk inkFkZ fHkUu gksuk pkfg,A
pwafd;fn
= , 1 1 0
1 2 f1 f2
vFkZr
1 = 0 ;k F =
vFkkZrla;kstu ySUldh rjg O;ogkj ughadjds lery dk¡p ifêdkdh rjg O;ogkj djsxkA
(2) pwafd1rFk 2/kukRedjkf'k;k¡gSa]leh-(5)ls]f1rFk f2foijhrizd`frdsgksuspkfg,vFk Zr~,dySUlvfHklkjh rFk nwljkvilkjhgksukpkfg,A
(3) ;fnvo.khZ;la;kstuvfHklkjhgS&
fC < fD
vkSjtSls
fC C
fD D
vFk Zr~,dvfHklkjhf)vo.khZ;esa]vfHklkjhySUldhQksdlnwjhrFk fo{ksi.k{kerkvilkjhySUldhrqyukesade gksrhgSA
24.45 Chromatic aberration in the formation of images by a lens arises because :
(A) of non-paraxial rays.
(B) the radii of curvature of the two sides are not same.
(C) of the defect in grinding.
(D*) the focal length varies with wavelength. ySUlksa}kjkcusizfrfcEcksaesaso.khZ;nks"kmRiUugksrkgSD;ksafd& (A)eq[;v{kdsutnhdfdj.ksaugksusdsdkj.k
(B) ySUldsnksuksarjQdhoØrkf=kT;klekuugksusdsdkj.k (C)ySUldsf?klusesanks"kdsdkj.k (D*)Qksdlnwjh]rjaxnS/;ZlsifjofrZrgksrhgSA
Sol. From passage, (D) is correct.
vuqPNsnls(D) lghgSA
24.46 A combination is made of two lenses of focal lengths and in contact ; the dispersive powers of the materials of the lenses are and . The combination is achromatic when :
rFk QksdlnwjhdsnksysUlksadklEidZla;kstucukrsgSa]ySUlksadsinkFksZdhfo{ksi.k{kerkrFk gSAla;kstu vo.khZ;gksxk;fn&
(A) = , = 2 , = 2 (B) = , = 2 , = /2
0 0 0 0
(C) = , = 2 , = –/2 (D*) = , = 2 , = – 2
0 0 0 0
Sol. From points (2) and (3) of passage :
and must be of opposite sign.
Also
< and
< D
which is satisfied only by (D).
Sol. vuqPNsn (2) o(3) ls:
rFk foifjrfpUgdsgksaxsA rFkk C < D rFkkC < D
;gdsoy (D) }kjklUrq"VgSA
24.47 The dispersive power of crown and flint glasses are 0.02 and 0.04 respectively. An achromatic converging lens of focal length 40 cm is made by keeping two lenses, one of crown glass and the other of flint glass, in contact with each other. The focal lengths of the two lenses are :
(A) 21 cm and 40 cm (B*) 21 cm and –40 cm
(C) –21cm and 40 cm (D) 10 cm and –21cm Økmu,oaf¶yUVdk¡pdhfo{ksi.k{kerkØe'k%0.02rFk 0.04gSAØkWÅudkapof¶yaVdkapdsnksysUlksdks,dnwljsdsikl j[kdj40cmQksdlnwjhdk,dvfHklkjhysUlcukrsgSaAnksuksaySUlksadhQksdlnwjhgS%&
(A) 21 cm rFkk 40 cm (B*) 21 cm rFkk–40 cm
(C) –21cm rFkk40 cm (D) 10 cm rFkk –21cm
Sol.
1 2 = 0
f1 f2
1 f1
= 1 ©1
2 f2 2
.................(1)
1 1 1 =
1
.................(2)
F f1 f2 40
After solving (1) & (2)
(1) o (2) dks gy djus ds i'pkr~ f1 = 20 cm
f2 = – 40 cm.
COMPREHENSION
All objects referred to the subsequent problems lie on the principle axis.
uhps fn, x, iz'u esa lHkh fcEc eq[; v{k ij j[ks gq, gSA
24.48 If light is incident on surface 1 from left, the image formed after the first refraction is definitely :
vxjlrg1ijck;havksjlsizdk'kvkifr gksrkgS]igysviorZudsckncukizfrfcEcfuf'pr:ilsgksxk&
(A) Real for a real object (B*) Virtual for a real object
(C) Real for a virtual object (D) Virtual for a virtual object (A)okLrfodfcEcdsfy, okLrfod (B*)okLrfodfcEcdsfy,vkHkklh (C)vkHk lhfcEcdsfy,okLrfod (D)vkHk lhfcEcdsfy,vkHk lh
2 1 2 1
Sol.
v u R
( – ) is +ve and R is – ve if u is –ve, v will always be –ve
(2–1)/kukRedgSvkSjR,_.k RedgS];fnu_.k RedgSrksvges'k –vegksxk
i.e. for real object image is always virtual.
okLrfodfcEcdsfy;sizfrfcEcges'k vkHklhgksxkA
24.49 In above question if the object is real, then the final image formed after two refractions :
mijksäiz'uesavxjfcEcokLrfodgSrksnksviorZudsi'pkr~vfUreizfrfcEc&
(A) may be real (B) may be virtual (C) must be virtual (D*) both A and B (A)okLrfodgksldrkgS (B)vkHk lhgksldrkgS
(C)vkHklhghgksxk (D*)ArFk Bnksuksa
24.50 If light is incident on surface 2 from right then which of the following is true for image formed after a single
refraction.
vxjizdk'knk;hvksjlslrg2ijvkifr gksrkgSrks,dviorZudsi'pkr~fuEuesalsdkSulklR;gksxk&
(A) Real object will result in a real image (B) Virtual object will result in a virtual image
(C) Real object will result in a virtual image (D*) Virtual object will result in a Real image (A)okLrfodfcEcdkokLrfodizfrfcEccusxk (B)vkHk lhfcEcdkvkHk lhizfrfcEccusxk (C)okLrfodfcEcdkvkHk lhizfrfcEccusxk (D*)vkHk lhfcEcdkokLrfodizfrfcEccusxkA
48. to 50.
Consider a object on left side of spherical surface seperating two media. If real object is in rarer media i.e., n1 < n2
Then n2 n2 n1 n1
= – ve
v (u) (R)
Hence image shall be virtual for a real object lying on concave side with rarer media. (1)
If real object is in denser media i.e., n1 > n2
n2 (n1 n2 ) n1 = n1 n2 n1
v (u) (R) u R
Image is real if
n1 n2
u
> n1 or u <
R
(n1 n2 )R n1
.... (2)
n1 n2
and image is virtual if u > R
n1
.... (3)
From statements 1, 2 and 3 we can easily conclude the answers.
Sol. 48 to 50
nksek/;eksadsi`Fkddjusokysxksyh;lrgdscka;hvksjfcEcdksysrsgSA;fnfojyek/;eesaokLrfodfcEcgSvFk Zr~
n1 < n2
rks n2 n2 n1 n1
=–ve
v (u) (R)
blfy;s fojy ek/;e esa vory lrg ij fLFkj okLrfod fcEc ds fy, izfrfcEcvkHklhgksxkA (1)
;fnl?kuek/;eesa okLrfodfcEcgS]vFkkZr n1>n2
n2 (n1 n2 ) n1 = n1 n2 n1
v (u) (R) u R
izfrfcEc okLrfod ;fn n1 n2 > n1 or u< (n1 n2 )R
.... (2)
u R n1
n1 n2
rFkk izfrfcEc vkHkklh gS ;fn u > R
1
.... (3)
dFku 1, 2 o 3 ls ge ljyrk ls mÙkjksa dk fu"d"kZ fudky ldrs gSA
SECTION - V : MATRIX - MATCH TYPE
[k.M- V:eSfVªDl&lqesy izdkj
24.51 Four particles are moving with different velocities in front of stationary plane mirror (lying in y-z plane).
At t = 0, velocity of A is → ˆ, velocity of B is → ˆi 3 ˆj, velocity of C is → ˆ 6 ˆj , velocity of D
v A i vB vC 5 i
is → ˆ ˆ. Acceleration of particle A is →
2 ˆi ˆj and acceleration of particle C is → ˆ. The
vD 3 i j a A
aC 2 t j
particle B and D move with uniform velocity (Assume no collision to take place till t = 2 seconds). All quantities are in S.I. Units. Relative velocity of image of object A with respect to object A is denoted by
VA, A. Velocity of images relative to corresponding objects are given in column I and their values are given in column II at t = 2 second. Match column I with corresponding values in column II.
fLFkj lery niZ.k ds lkeus pkj d.k fofHkUu osx ls xfr dj jgs gS
¼niZ.k yz ry esa gS½A t = 0 le; ij A dk osx → ˆi , B dk osx → ˆi 3 ˆj, C dk
osx → 5 ˆi 6 ˆj , D dk osx
→ 3 ˆi ˆj gSA d.k A dk Roj.k → 2 ˆi ˆj vkSj d.k C dk
Roj.k → ˆ gSA d.k B o D ds osx fu;r gS ¼;g ekfu;s fd t = 2 ls0 rd
aC 2 t j
niZ.k ls dksbZ VDDj ugh gksrh gS½A fcEc (oLrq) A ds lkis{k] A ds
izfrfcEc dk lkis{k osx VA, A }kjk n'kkZ;k tkrk gSA lHkh jkf'k;k¡ S.I.
ek=kd esa gSA oLrqvks ds laxr izfrfcEcksa ds lkisf{kd osx dkWye-I esa fn, x, gS rFkk muds eku dkWye-II esa t = 2 ls0 ij fn, x, gSA rks dkWye-I ds laxr dkWye-II dks lqesfyr dhft,A
D A B
y
l er y ni Z. k
Plane mirror
C
x
Column I Column II
dkWyeI dkWyeII
(A) VA, A
(p)
2 ˆi
(B)
VB, B
(q)
6 ˆi
(C) VC, C
(D) VD,D
Ans. (A) s, (B) p, (C) s, (D) q
(r)
(s)
12 ˆi 4 ˆj
10 ˆi
Sol.
→ ˆi → t =
→ ˆ
ˆi (2 ˆi ˆj)(2) ˆ
= 5 ˆi 2 ˆj
v A 5 i 2 j
→ → → ˆ
v A, A = v A v A 10 i
→ ˆ ˆ → ˆ ˆ → ˆ
vB ( i 3 j) , vB i 3 j so vB,B 2 i
For particle C (d.k C ds fy,)
dvy
dt = 2t vy – 6 = t2 vy = 6 + 4 = 10
→ ˆ ˆ →
ˆ ˆ → ˆ
vC 5 i 10 j ,
vC 5 i 10 j so
vC,C 10 i
→ ˆ ˆ →
ˆ ˆ → ˆ
vD 3 i j
, vD 3 i j ,
vD,D 6 i
24.52 An object O (real) is placed at focus of an equi-biconvex lens as shown in figure 1. The refractive index of lens is = 1.5 and the radius of curvature of either surface of lens is R. The lens is surrounded by air. In each statement of column-I some changes are made to situation given above and information regarding final image
formed as a result is given in column-II. The distance between lens and object is unchanged in all statements of column-I. Match the statements in column-I with resulting image in column-II.
,d fcEc O(okLrfod)fp=k-1 ds vuqlkj ,d lef}mRry (equi-biconvex) ysUl ds Qksdl ij fLFkr gSA ysUl dk viorZukad=1.5gSrFk ysUldsfdlhHkhlrgdhoØrkf=kT;kRgSAysUlok;qlsf?kjkgqvkgSAdkWye-IdsizR;sd dFkuesaÅijnhxbZfLFkfresadqNifjorZufd;sx;sgSrFk ifj.k eLo:icu sokysvfUreizfrfcEclslEcfU/kr lwpuk;sadkWye-I esanhxbZgSaAdkWye-IdslHkhdFkuksaesaysUlrFk oLrq¼fcEc½dschpnwjhvifjofrZrgSAdkWye- Iesafn;sx;sdFkuksadksdkWye-I esafn;sx;sifj.k ehizfrfcEcksalslqesfyrdhft;sA
Column-I Column-II
(A) If the refractive index of the lens is (p) final image is real doubled (that is, made 2 ) then
(B) If the radius of curvature is doubled (q) final image is virtual (that is, made 2R) then
(C) If a glass slab of refractive index = 1.5 (r) final image becomes smaller in
is introduced between the object size in comparison to size of image
and lens as shown, then before the change was made
(D) If the left side of lens is filled with a medium (s) final image is of same size of object. of refractive index = 1.5 as shown, then (t) final image is of larger size of object
dkWye-I dkWye-II
(A) ;fnysUldkviorZukadnqxqukfd;ktkrkgS (p) vfUreizfrfcEcokLrfodgksrkgSA (vFk Zr~2 fd;ktkrkgS)rc
(B) ;fn oØrk f=kT;k nqxquh dh tkrh gS (q) vfUre izfrfcEc vkHkklh gksrk gSA
(vFkkZr~ 2R dh tkrh gS) rc
(C) ;fn viorZukad =1.5ds dk¡p dh iêh (r) ifjorZudjusdsigysdsizfrfcEc dksfcEcrFk ysUldschp ds vkdkjdh rqyukesa vccuk vfUre
fp=kuqlkjj[k tkrkgS]rks izfrfcEcvkdkjesaNksVkgSA
(D) ;fnysUldscka;hvksjviorZukad=1.5 (s) vfUreizfrfcEc]fcEcdsleku dk,dek/;efp=k uqlkjHkjktkrkgSA vkdkjdkgS
(t) vfUreizfrfcEc]fcEclscM+svkdkjdkgSA
Ans. (A) p,r (B) q, r (C) q, r (D) q, r
Sol. Initially the image is formed at infinity.
(A) As is increased the focal length decreases. Hence the object is at a distance larger than focal length. Therefore final image is real. Also final image becomes smaller is size in comparision to size of image before the change was made.
(B) If the radius of curvature is doubled, the focal length decreases. Hence the object is at a distance lesser than focal length. Therefore final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made.
(C) Due to insertion of slab the effective object for lens shifts right wards. Hence final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made.
(D) The object comes to centre of curvature of right spherical surface as a result. Hence the final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made.
izkjEHkesaizfrfcEcvuUrijcurkgSA
(A) tcdksc<+k;ktkrkgS]rksQksdlnwjh?kVrhgSAblfy;sfcEcQksdlnwjhlsvf/kdnwjhijgksxkAblfy;svfUre izfrfcEcokLrfodgSAlkFkghifjorZudjusdsigysdsizfrfcEcdsvkdkjdhrqyukesavfUreizfrfcEcvkdkjesa NksVk curk gSA
(B) ;fn oØrk f=kT;k nqxquh dh tkrh gS] rks Qksdl nwjh ?kV tkrh gSA blfy;s fcEc Qksdl nwjh ls de nwjh ij gksxkA blfy;s vfUre izfrfcEc vkHkklh gksxkA lkFk gh ifjorZu djus ds igys ds izfrfcEc ds vkdkj dh rqyuk vfUre izfrfcEc vkdkj esa NksVk curk gSA
(C) iV~Vhdsizos'kdsdkj.kysUldsfy,izHkohfcEcnak;hvksjfoLFkfirgkstkrkgSAblfy;svfUreizfrfcEcvkHklhgksxkA lkFkghifjorZudjusdsigysdsizfrfcEcdsvkdkjdhrqyukesavfUreizfrfcEcvkdkjesaNksVkcurkgSA
(D) ifj.k eLo:ifcEcnak;sxksyh;lrgdsoØrkdsUnzijvktkrkgSAblfy;sizfrfcEcvkHk lhgksxkAlkFkgh ifjorZu djusdsigysdsizfrfcEcdsvkdkjdhrqyukesavfUreizfrfcEcvkdkjesaNksVkcurkgSA
24.53 Match the following :
An object O is kept perpendicular to the principal axis of a spherical mirror. Each situation (A,B,C and D) gives object coordinate u in centimeters with sign, the type of mirror, and then the distance (centimeters, without sign) between the focal point and the pole of the mirror. On the right side information regarding the image is given. Correctly match the situations on the left side with the images described on the right side.
Image
(P) Real, Errect, Enlarged
(Q) Virtual, Errect, Diminished (R ) Real Inverted, Enlarged
(S) Virtual, Erect, Enlarged
,doLrqOxksyh;niZ.kdseq[;v{kdsyEcorj[khgSAizR;sdfLFkfr(A,B,CvkSjD)dsfy,oLrqdsfunsZ'k adufpUg dslkFkrFk lseh-esaniZ.kdkvkdkjrFk nwjh(lseh0]fpUgjfgr)QksdlfcUnqrFk niZ.kds/kzqodschpnhxbZgSAnka;h vksjizfrfcEcdhlwpuknhxbZgSAcka;hrjQdhfLFkfr;ksadsfy,nka;hrjQdsizfrfcEcksadkfeykudfj;sA
Ans. (A) r , (B) s (C) q (D) q
Sol. Image by convex mirror is always virtual, erect and diminished.
In case of concave mirror, see using position of object.
mÙkyniZ.kds}kjkcuk;kx;kizfrfcEcges'k vkHklh]lh/k oNksVkgksrkgSAvoryniZ.kdsfy,oLrqdhfLFkfrns[ksaA
24.54 A white light ray is incident on a glass prism, and it create four refracted rays A, B, C and D. Match the refracted rays with the colours given (1 & D are rays due to total internal reflection.): dk¡pdsfizTeij,d'osrizdk'kfdj.kvkifrrgS]rFkk;gpkjviofrZrfdj.ksaA,B,CrFkkDmRiUudjrhgSA viofrZrfdj.k sadksfn;sx;sjaxksalsfeykvks(1rFk Dfdj.ksaiw.kZvkUrfjdijkorZudsdkj.kgSa)
B A
1
C D
incident ray
v ki fr r fd j . k
Ray Colour
(A) A (p) red
(B) B (q) green
(C) C (r) yellow
(D) D (s) blue
fdj.ksa jax
(A) A (p) yky
(B) B (q) gjk
(C) C (r) ihyk
(D) D (s) uhyk
Ans. (A) p, (B) r, (C) q, (D) s
Sol. By snell law Lusy fu;e ls n =
sin i sin r
Since for 1st case angular incidence is same for all ray. So r will be less for red.
izFkefLFkrhdsfy;sdks.kh;vkirulHkhfdj.kdsfy;slekugksxkAvr%ykydsfy;srdegksxkA
24.55 In each situation of column-I, an incident wavefront and its corresponding reflected or refracted wavefront is shown. In column-II the optical instrument used for reflection or refraction is given. Always take the optical instrument to the right of incident wavefront. The incident wavefront is moving towards right. Match each pair of incident and reflected/refracted wavefront in column-I with the correct optical instrument given in column-II.
dkWye-I dh izR;sd fLFkfr esa] ,d vkifrr rjaxkxz rFkk mlds laxr ijkofrZr
;k viofrZr rjaxkxz (wavefront) fn[kk;k x;k gSA dkWye-II esa izdk'kh; midj.k tks fd ijkorZu ;k viorZu ds fy, gS og fn;k gSA ges'kk vkifrr rjaxkxz ds nkfguh rjQ gh izdk'kh; midj.k gSA vkifrr rjaxkxz nkfguh rjQ xfr dj jgrk gS rks dkWye-I esa fn;s x;s vkifrr rFkk ijkofrZr@viofrZr rjaxkxz ds fy, dkWye-II esa fn;s lgh izdk'kh; midj.k dks lqesfyr dfj;sA
Column-I Column-II
Incident Reflected/Refracted Optical instrument used wavefront Wavefront
dkWye-I dkWye-II
vkifrr ijkofrZr@viofrZr iz;qDrizdk'kh;midj.k rjaxkxz rjaxkxz
(A) (p)
(B) (q)
(C) (r)
(D) (s)
Ans. (A) p, r (B) q, s (C) q,r (D) p, s
Sol. (A) For plane incident wave fronts a divergent refracted beam can be formed only by diverging action and convergent reflected beam can be formed by converging action. Hence (A) p, r
(B) For plane incident wave fronts a convergent refracted beam can be formed only by converging action and divergent reflected beam can be formed by diverging action. Hence (B) q, s
(C) For a incident divergent beam a parallel reflected or refracted beam can be formed only by converging
action.Hence (C) q, r
(D) For a incident convergent beam a parallel reflected of refracted beam can be formed only by diverging action.Hence (D) q, r
(A) ,d lery vkifrr rjaxkxz ds fy, ,d vilkjh viofrZr fdj.k dsoy vilkjh fØ;k ds dkj.k gh cu ldrh gS rFkk vfHklkjh ijkofrZr fdj.k dsoy vfHklkjh fØ;k ds dkj.k gh cu ldrh gSA vr% (A) p, r
(B) ,dleryvkifrrrjaxkxzdsfy,vfHklkjhviofrZrfdj.kdsoyvfHklkjhfØ;kdsdkj.kghculdrhgSrFkk vilkjh ijkofrZr fdj.k dsoy vilkjh fØ;k ls gh cu ldrh gS vr% (B) q, s
(C) ,dvkifrr vilkjhfdj.k dsfy, lekUrjijkofrZr ;kviofrZr fdj.kdsoy vfHklkjhfØ;k dsdkj.k lEHko gS vr% (C) q, r
(D) ,dvfHklkjh vkifrrfdj.k dsfy, lekUrjijkofrZr ;kviofrZr fdj.k dsoy vilkjhfØ;k ds dkj.k ghcu ldrh gS vr% (D) q, r
24.56 Column-I gives certain situations regarding a point object and its image formed by an optical instrument. The possible optical instruments are concave and convex mirrors or lenses as given in Column-II. Same side of principal axis means both image and object should either be above the principal axis or both should be below the principal axis as shown in figure. Same side of optical instrument means both image and object should be either left of the optical instrument or both should be on right of the optical instrument as shown in figure. Match the statements in column-I with the corresponding statements in column- .
dkWye- esa ,d fcUnq fcEc rFkk ,d izdkf'kd ;a=k }kjk cus blds izfrfcEc ls lEcfU/kr dqN fuf'pr fLFkfr;k¡ nh xbZ gSaA lEHko izdkf'kd ;U=k vory rFkk mÙky niZ.k ;k ysUl gSa tks fd dkWye- esa fn;s x;s gSaA eq[; v{k ds
,d gh vksj gksus dk vFkZ gS fd fcEc rFkk izfrfcEc nksuksa ;k rks eq[; v{k ds Åij gksaxs ;k fQj nksuksa eq[; v{k ds uhps gksaxs tSlk fd fp=k esa fn[kk;k x;k gSA izdkf'kd ;U=k ds ,d gh vksj gksus dk vFkZ gS fd fcEc rFkk izfrfcEc nksuksa ;k rks izdkf'kd ;U=k ds cka;h vksj gksaxs
;k fQj nksuksa izdkf'kd ;U=k ds nka;h vksj gksaxs tSlk fd fp=k esa fn[kk;k x;k gSA dkWye- esa fn;s x;s dFkuksa dks dkWye- esa fn;s x;s laxr dFkuksa ls lqesfyr dhft,A
Column I Column II
(A) If a point object and its image are on same side of (p) Concave mirror principal axis and opposite sides of the optical
instrument then the optical instrument is
(B) If a point object and its image are on opposite side (q) Convex mirror of principal axis and same sides of the optical
instrument then the optical instrument is
(C) If a point object and its image are on same side (r) Concave lens of principal axis and same sides of the optical
instrument then the optical instrument is
(D) If a point object and its image are on opposite side (s) Convex lens of principal axis and opposite sides of the optical
instrument then the optical instrument is
dkWyeI dkWye II
(A) ;fnfcUnqfcEcrFk bldkizfrfcEceq[;v{kds,dghvksjgksa (p) vory niZ.k rFkk izdkf'kd ;U=k ds foijhr vksj gksa rks izdkf'kd ;U=k gS
(B) ;fn fcUnq fcEc rFkk bldk izfrfcEc eq[; v{k ds foijhr vksj gks
(q) mÙky niZ.k
rFkk izdkf'kd ;U=k ds ,d gh vksj gks] rks izdkf'kd ;U=k gS
(C) ;fnfcUnqfcEcrFk bldkizfrfcEceq[;v{kds,dghvksjgks (r) vory ysUl rFk izdkf'kd;U=kdsHkh,dghvksjgksrksizdkf'kd;U=kgS
(D) ;fn fcUnq fcEc rFkk bldk izfrfcEc eq[; v{k ds foijhr vksj gks
(s) mÙky ysUl
rFkk izdkf'kd ;U=k ds Hkh foijhr vksj gks rks izdkf'kd ;U=k gS
Ans. (A) p, q (B) p, q (C) r, s (D) r, s
Sol. For a spherical mirror, line joining object and its image crosses principal axis at centre of curvature. The line joining object and image inverted about principal axis cuts the principal axis at the pole. Then from figure below.
xksyh;niZ.kdsfy,]fcEcrFk bldsizfrfcEcdkstksM+usokyhjs[k eq[;v{kdksoØrkdsUnzijdkVrhgSAfcEcrFk izfrfcEc dkstksM+usokyhjs[k eq[;v{klsmYVhgkstkrhgSrFk eq[;v{kdks/kzqoijdkVrhgSAuhpsfn;sx;sfp=kls
We can conclude
(A) If object and image are on same side of principal axis, they are on opposite side of mirror.
(B) If object and image are on opposite side of principal axis, they are on same side of mirror. For a lens, the line joining object and image cuts the principal axis at optical centre.
Then from figures below.
gefu"d"kZfudkyldrsgSA
(A);fnfcEcrFk izfrfcEceq[;v{kds,dghvksjgSrksosniZ.kdsfoijhrvksjgksaxsA (B);fnfcEcrFk izfrfcEceq[;v{kdsfoijhrvksjgS]rksosniZ.kds,dghvksjgksaxsA
ysUldsfy,]fcEcrFk izfrfcEcdkstksM+usokyhjs[k eq[;v{kdksizdkf'kddsUnzijdkVrhgSrksuhpsdsfp=ksals]
We can conclude
(C) If object and image are on same side of principal axis, they are also on same side of lens.
(D) If object and image are on opposite side of principal axis. They are also on opposite side of lens.
gefu"d"kZfudkyldrsgSA
(C) ;fnfcEcrFk izfrfcEceq[;v{kds,dghvksjgSa]rksosysUldsHkh,dghvksjgksaxsA
(D) ;fn fcEc rFkk izfrfcEc eq[; v{k ds foijhr vksj gS] rks os ysUl ds Hkh foijhr vksj gksaxsA
SECTION - VI : INTEGER TYPE
[k.M-VI:iw.kkZadizdkj
SHORT SUBJECTIVE
y?kqÙkjkRediz'u
24.57 As shown in the figure, an object O is at the position ( 10, 2) with respect to the origin P. The concave mirror M1 has radius of curvature 30 cm. A plane mirror M2 is kept at a distance 40 cm infront of the concave mirror. Considering first reflection on the concave mirror M1 and second on the plane mirror M2. if x co-ordinate is -x0 of the second image w.r.t. the origin P.then x0 is
fp=kesn'k Z;svuqlkj,doLrqOewyfcUnqPdslkis{kfLFkfr(10,2)ijj[krsgSAvoryniZ.kM1dhoØrkf=kT;k 30cm gSA,dleryniZ.kM2dksvoryniZ.kdslkeus40cm nwjhijj[krsgSA;gekursgSfdigykijkorZuvory niZ.kM1onwljkM2 lsgksrkgSAnwljsçfrfcEcdsPdslkis{kx funsZ'k ad-x0 gSrksxgksxk
Ans. 40
Sol. For m1, u = 10, f = 15, h = 2.
Using mirror formula
1 1
= 1 1 1 = 1
v u f
v 10
15
1
v =
1 1
10 15
3 2
= 30
= v = 30 cm &
h v
2 = h2
h1
= 6 cm
6
40
The image formed by the plane mirror is at 70 below the principal axis & 70 + 6 30 = 46 of the concave mirror. ∴ coordinates of I2 w.r.t. P = (− 46, − 70) Ans. ]
gy m1 ds fy, = 10, f = 15, h = 2.
niZ.k lw=k dks yxkusij
1 1
= 1 1 1 = 1
v u f
v 10
15
1
v =
1 1
10 15
3 2
= 30
= v = 30 cm &
h v
2 = h2
h1
= 6 cm
6
40
lery niZ.k}kjk cukbZxbZizfrfcEceq[;v{kls70cm uhpsgS vkSj70+ 6 30= 46cm voryniZ.kls
∴ I2 ds P ds lkis{k funsZ'kkad= (− 46, − 70) Ans. ]
24.58 A ray of light travelling in air is incident at angle of incident 30° on one surface of a slab in which refractive index varies with y. The light travels along the curve y = 4x2 (y and x are in metre) in the slab. if the refractive index of the slab at y = 1/2 m in the slab is 15/x0.then x0 is
,d izdk'k dh fdj.k ,d ifV~Vdk dh ,d lrg ij 30° dks.k ij vkifrr gksrh gS ftlesa fdj.k dk viorZukad y ds lkFk cnyrk gSA izdk'k y = 4x2 oØ ds vuqfn'k ifV~Vdk esa pyrk gS (y vkSj x ehVj esa gSa)l y = 1/2m ij ifV~Vdk esa] ifV~Vdk dk viorZukad dk eku 15/x0 gks rks x0 gksxkA
Ans. 10
Sol. By snell’s law. Lusydsfu;els
1 × sin 30º = ................. = = n sini
where n is R. at y and i is angle of incidence at y.
tgk¡n,yijviorZukadgSvkSji,yijvkirudks.kgSa
dy
tan (90 – i) = dx = 8x = 4
cot i = 4 = 4 = 2
sin i = 1 3
n =
sin 30 sini
1/ 2
= 1/ 3
= 1.5 Ans.
LONG SUBJECTIVE
fucU/k Rediz'u
24.59 In the figure shown L is a converging lens of focal length 10cm and M is a concave mirror of radius of curvature 21cm. A point object O is placed in front of the lens at a distance 15cm. AB and CD are optical axes of the lens and mirror respectively. if the distance of the final image formed by this system from the optical
centre of the lens is 6 cm. then x is The distance between CD & AB is 1 cm.
n'k Z;sx;sfp=kesaL,10cmQksdlnwjhokyk,dvfHklkjhySalgSrFk M]21cm.oØrkf=kT;kdk,dvoryniZ.k gSAySsaldslkeus15cmdhnwjhij,dfcUnqor~oLrw Oj[khgq;hgSAABoCDØe'k%ySaloniZ.kdsizdkf'kd v{k gaSA bl fudk; }kjk cuus okys vfUre izfrfcEc dh ySal ds izdkf'kd
dsUnz ls nwjh
gSA
Ans. 26
Sol.
6 cm gks rks x Kkr dhft;s A AB o CD ds chp nwjh 1 cm
I1 is the image of object O formed by the lens.
ySal}kjkcukfcEcOdkizfrfcEcI1gS
1 1 1
v1 – u1 = f
u1 = – 15 f1 = 10
Solving we get
gy djus ij
v1 = 30 cm
I1 acts as source for mirror
I1 niZ.kdsfy,,dlzksrdhrjgdkedjrkgS
u2 = –(45 – v1) = – 15 cm
I2 is the image formed by the mirror I2niZ.k}kjkcuk;kx;kizfrfcEcgS
1
v 2 =
1
fm –
1
u2 = –
1 – 1 v
10 15 2
= – 30 cm
v 2
The height of I2 above principal axis of lens is = u2 × 1 + 1 = 3cm
v 2
I2 dh ySal ds eq[; v{k ls Å¡pkbZ gS = u2 ×1+1=3cm
I2 acts a source for lens u3 = – (45 – v2) = – 15 cm
I2 ySaldsfy,,dlzksrdhrjgdk;ZdjrkgS u3 = – (45 – v2) = – 15 cm
Hence the lens forms an image I3 at a distance v3 = 30 cm to the left of lens and the image of
v 3
I3, u3 × 3 = 6cm below the principal exis of lens.
v 2
The height of I2 above principal axis of lens is = u2
× 1 + 1 = 3cm
blfy, ySal v3 = 30 cm dh nwjh ij ySal ds ck;ha vksj I3 izfrfcEc cukrk gS
vkSj
v3
u3 × 3 = 6 cm ySal ds eq[; v{k ds uhps curk gSA
ysUl ds eq[; v{k ds Åij I dh ÅWpkbZ = v 2 × 1+1= 3cm
2
required distance =
vko';d nwjh =
u2
= 6 cm
= 6 cm
24.60 A stationary observer O looking at a fish F (in water of, = 4/3) through a converging lens of focal length 90.0 cm. The lens is allowed to fall freely from a height 62.0 cm with its axis vertical. The fish and the observer are on the principal axis of the lens. The fish moves up with constant velocity 100 cm/s. Initially it was at a depth of 44.0 cm. if the velocity with which the fish appears to move to the observer at t = 0.2 sec.is x/4 cms-1.then x is (g = 10 m/s2)
[M.Bank_G.O._7.39]
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Ans. 91
Sol. At t = 0.2 sec , velocity of lens
V𝑙
= gt = 2m/s (downwards)
1 3 11
for lens the fish appears to approach with a speed of 2 +
= m/s
4
Lens t = 0.2s
V𝑙
at a distance of 42
24 = 60 cm. 42 cm
4
3 4
t = 0.2s
n = 3
V 24 cm
F
image of fish from lens , V =
60 90
60 90
= –180 cm.
v 2 du
– 180 2 11 99
Velocity of image w.r.t. lens V =
2 =
=
m/s
i dt
60 4 4
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