Chapter # 22
Photometry
SOLVED EXAMPLES
1.
Find the luminous flux of a 10 W source of 600 nm. The relative luminously at 600 nm is 0.6.
Sol.
The luminous flux of a 1 W source of 555 nm = 685 lumen. Thus, the luminous flux of a 10 W source of 555 nm = 6850 lumen. The luminous flux of a 10 W source of 600 nm is, therefore, 0.6 × 6850 lumen = 4110 lumen.
QUESTIONS
FOR
SHORT
ANSWER
1.
What is the luminous flux of a source emitting radio waves?
2.
The luminous flux of a 1 W sodium vapour lamp is more than that of a 10 kV source of ultraviolet radiation. Comment.
3.
Light is incident normally on a small plane surface. If the surface is rotated by an angle of 30° about the incident light, does the illuminance of the surface increase, decrease or remain same? Does your answer change if the light did not fall normally on the surface?
4.
A bulb is hanging over a table. At which portion of the table is illuminance maximum? If a plane mirror is placed above the bulb facing the table, will the illuminance on the table increase?
5.
The sun is less bright at morning and evening as compared to at noon although its distance from the observer is almost the same. Why ?
6.
Why is the luminous efficiency small for a filament bulb as compared to a mercury vapour lamp?
7.
The yellow colour has a greater luminous efficiency as compared to the other colours. Can we increase the illuminating power of a white source by putting a yellow plastic paper around this source?
Objective - I 1.
The one parameter that determines the brightness of a light source sensed by an eye is (A) energy of light entering the eye per second (B) wavelength of the light (C) total radiant flux entering the eye (D*) total luminous flux entering the eye. vk¡[k }kjk lalwfpr dh x;h izdk'k lzkrs dh rhozrk dks fu/kkZfjr djus okyh ,d jkf'k gS (A) vk¡[k esa izfr lsd.M izos'k djus okyh izdk'k ÅtkZ (B) izdk'k dh rjaxnS/;Z (C) vk¡[k esa izo's k djus okyk dqy fofdj.k ¶yDl (D*) vk¡[k esa izo's k djus okyk dqy izrhiu ¶yDl
2.
Three light sources A, B and C emit equal amount of radiant energy per unit time. The wavelengths emitted by the three sources are 450 nm, 555 nm and 700 nm respectively. The brightness sensed by an eye for the sources are XA, XB and Xc respectively. Then, rhu izdk'k lzkrs A, B vkSj C ,dkad le; esa leku ÅtkZ fofdfjr dj jgs gSAa rhuksa lzkrs ksa ls mRlftZr rjaxnS/;Z Øe'k% 450 nm, 555 nm o 700 nm gSA vk¡[k }kjk vuqHko dh x;h lzkr s dh ped Øe'k% XA, XB gS Xc gS] rks (A) XA > XB, XC > XB (B) XA > XB, XB > XC (C*) XB > XA, XB > XC (D) XB > XA, XC > XB
3.
As the wavelength is increased from violet to red, the luminosity. (A) continuously increases (B) continuously decreases (C*) increases then decreases (D) decreases then increases tSl&s tSls izdk'k dh rjaxnS/;Z cSaxuh ls yky dh vksj c<+rh gS] iznhIrrk (A) lrr~ :i ls c<+rh gSA (B) lrr~ :i ls de gksrh gSA (C*) c<+rh gS rRi'pkr~ de gksrh gSA (D) de gksrh gS rRi'pkr~ c<+rh gSA
4.
An electric bulb is hanging over a table at a height of 1 m above it. The illuminance on the table directly below the bulb is 40 lum. The illuminance at a point on the table 1 m away from the first point will be about ,d Vscy ls 1 m Å¡pkbZ ij ,d fo|qr cYc yVdk gqvk gSA blds Bhd uhps Vscy ij cYc dh iznhfIr 40 yDl gSA Vscy manishkumarphysics.in
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Chapter # 22
Photometry
ij izFke fcUnq ls 1 eh- nwj fLFkr fdlh fcUnq ij iznhfIr yxHkx gS (A) 10 yDl (B*) 14 yDl (C) 20 yDl
(D) 28 yDl
5.
Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by ,d insZ ij fcUnq lzkrs ls izdk'k vkifrr gks jgk gSA ;fn insZ rFkk lzksr ds e/; dh nwjh 1% c<+k nh tk;s] iznhfIr esa deh gksxh - ¼yxHkx½ (A) 0.5% (B) 1% (C*) 2% (D) 4%
6.
A battery-operated torch is adjusted to send an almost parallel beam of light. It produces an illuminance of 40 lux when the light falls on a wall 2 m away. The illuminance produced when it falls on a wall 4 m away is close to ,d cSVjh pkfyr VkpZ dks bl izdkj leaftr fd;k tkrk gS fd blls yxHkx lekukarj izdk'k iqat fudkyrk gSA 2 m nwj fLFkr nhokj ij bldk izdk'k fxjus ij 40 lux iznhiu mRiUu gksrk gSA tc 4 m nwj fLFkr nhokj ij bldk izdk'k fxjsxk rks mRiUu iznhiu yxHkx gS (A*) 40 lux yDl (B) 20 lux yDl (C) 10 lux yDl (D) 5 lux yDl
7.
The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to ,d yEcs csyukdkj izdk'k lzksr }kjk vYi nwjh r nwjh mRiUu rhozrk lekuqikrh gksrh gS -
1 (A) 2 r
1 (B) 3 r
(C*)
1 r
(D) none of these buesa
ls dksbZ ugha
8.
A photographic plate placed at a distance of 5 cm from a weak point source is exposed for 3 s. If the plate is kept at a distance of 10 cm from the source, the time needed for the same exposure is ,d {kh.k fcUnq lzkrs ls 5 lseh nwjh j[kh gqbZ QksVksxkz fQd IysV 3 lsd.M ds fy;s mn~Hkkflr dh x;h gSA ;fn IysV lzkrs ls 10 lseh nwj j[kh nh tk;s] rks leku mn~Hkklu ds fy;s vko';d le; gksxk (A) 3 s lsd.M (B*) 12 s lsd.M (C) 24 s lsd.M (D) 48 s lsd.M
9.
A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12 s to get a good exposure. If the source is rotated by 60° about one of its diameters, the time needed to get the same exposure will be
,d o`Ùkkdkj pdrh ds :i esa izdk'k ds folfjr lzkrs ds Bhd lkeus ,d QksVkxzkfQd IysV j[k nh x;h gSA bldks mn~Hkkflr gksus esa 12 ls- yxrs gSAa ;fn lzkrs dks blds fdlh O;kl ds ifjr% 60° dks.k ls ?kqek fn;k tk;s] leku mn~Hkklu ds fy;s vko';d le; gS (A) 6 s lsd.M (B) 12 s lsd.M (C*) 24 s lsd.M (D) 48 s lsd.M 10.
A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at this portion varies with its distance r from the source as
izdk'k dk ,dfcUnq lzkrs ] ,d lery Vscy ds lekukarj ljy js[kk esa xfr'khy gSA lzkrs dh xeu js[kk ds Bhd uhps Vscy dk ,d NksVs ls Hkkx ij fopkj dhft;s] bl Hkkx ij iznhfIr lzkrs ls nwjh r ij fuEukuqlkj fuHkZj djrh gS (A)
11.
1 r
1 (B) 2 r
1 (C*) 3 r
1 (D) 4 r
Figure shows a glowing mercury tube. the intensities at point A, B and C are related as fp=k esa ,d izdk'k mRiUu djrh gqbZ edZjh V~;cw iznf'kZr dh x;h gSA fcUnqvksa A, B rFkk C ij rhozrkvksa
(A) B > C > A
(B) A > C > B
(C) B = C > A
manishkumarphysics.in
esa lEcU/k gS -
(D*) B = C < A
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Chapter # 22
1.
Photometry
Objective - II
The brightness producing capacity of a source (A) does not depend on its power (C*) depends on its power lzksr dh ped mRiUu djus dh {kerk (A) bldh 'kfDr ij fuHkZj ugha djrh gSA (C*) bldh 'kfDr ij fuHkZj djrh gSA
(B) does not depend on the wavelength emitted (D*) depends on the wavelength emitted (B) mRlftZr rjaxnS/;Z ij fuHkZj ugha djrh (D*) mRlftZr rjaxnS/;Z ij fuHkZj djrh gSA
gSA
2.
A room it illuminated by an extended source. The illuminance at a particular portion of a wall can be increased by (A*) moving the source (B*) rotating the source (C*) bringing some mirrors in proper positions (D*) changing the colour of the source ,d dejk foLr`r lzkrs ls izdkf'kr gSA fdlh fof'k"V nhokj ij iznhiu c<+k;k tk ldrk gS (A*) lzksr dks xfr djkdj (B*) lzkr s dks /kwf.kZr djkdj (C*) dqN fof'k"V fLFkfr;ksa esa niZ.kksa dks j[kdj (D*) lzksr dk jax ifjofrZr djdsA
3.
Mark the correct options. (A) The luminous efficiency of a monochromatic source is always greater than that of a white light source of same power. (B*) The luminous efficiency of a monochromatic source of wavelength 555 nm is always greater than that of a white light source of same power. (C*) The illuminating power of a monochromatic source of wavelength 555 nm is always greater than that of a white light source of same power. (D) the illuminating power of a monochromatic source is always greater than that of a white light source of same power. lgh dFkuksa dks fpfUgr dhft;s (A) fdlh ,do.khZ izdk'k lzkr s dh iznhiu n{krk] lnSo leku 'kfDr okys 'osr izdk'k lzkrs ls vf/kd gksrh gSA (B*) 555 nm rjaxnS/;Z okys ,do.khZ izdk'k lzkr s dh iznhiu n{krk] lnSo leku 'kfDr okys 'osr izdk'k lzkrs ls vf/kd gksrh
gSA
(C*) 555 nm rjaxnS/;Z
okys ,d o.khZ lzkrs dh iznhiu mRiUu djus dh 'kfDr] lnSo leku 'kfDr okys 'osr izdk'k lzkrs ls
vf/kd gksrh gSA (D) ,d o.khZ izdk'k lzkr s dh iznhiu mRiUu djus dh 'kfDr] lnSo leku 'kfDr okys 'osr izdk'k lzkrs ls vf/kd gksrh gSA 4.
Mark the correct options. (A) luminous flux and radiant flux have same dimensions (B*) luminous flux and luminous intensity have same dimensions (C*) radiant flux and power have same dimensions (D*) relative luminously is a dimensionless quantity lgh dFkuksa dks fpfUgr dhft;s (A) iznhiu ¶yDl rFkk fofdj.k ¶yDl dh foek,¡ leku gksrh gSA (B*) iznhiu ¶yDl rFkk iznhiu rhozrk dh foek,¡ ,d leku gksrh gSA (C*) fofdj.k ¶yDl rFkk 'kfDr dh foek,¡ ,d leku gksrh gSA (D*) vkisf{kd iznhiu ,d foekghu jkf'k gSA
WORKED OUT EXAMPLES 1.
A source emits 12.0 J of light of wavelength 620 nm and 8.0 J of light of wavelength 580 nm per second. The relative luminosity at 620 nm is 35% and that at 580 nm is 80%. Find (a) the total radiant flux, (b) the total luminous flux and (c) the luminous efficiency.
Sol.(a) The total radiant flux = total energy radiated per unit time = 12 J/s + 8 J/s = 20 J/s = 20 W. (b) The luminous flux corresponding to the 12 W of 620 nm radiation is 0.35 × (12 W) × 685 lumen/W = 2877 lumen. Similarly, the luminous flux corresponding to the 8W of 580 radiation is 0.80 × 10(8W) × 685 lumen/W =4384 lumen. The luminous flux of the source is 2877 lumen + 4384 lumen
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Chapter # 22
Photometry = 7261 lumen = 7260 lumen.
Total lu min ous flux Total radiant flux
The luminous efficiency =
2.
A circular area of radius 1.0 cm is placed at a distance of 2.0 m from a point source. The source emits light uniformly in all directions. The line joining the source to the centre of the area is normal to the area. It is found that 2.0 × 10–3 lumen of luminous flux is incident on the area. Calculate the total luminous flux emitted by the source and the luminous intensity of the source along the axis of area.
Sol.
The solid angle subtended by the area on the point source is =
(1.0 cm)2 (2.0m)
2
=
=
2760 lumen 20 W
(c)
× 10–4 sr.. 4
× 10–4 sr.. 4 The total solid angle at the source is 4., As the source radiates uniformly in all directions, the total luminous intensity = F/. Thus, 2.0 × 10–3 lumen of flux is emitted in
=
2.0 10 3 lumen = 25 cd. 10 4 sr 4
3.
The overall luminous efficiency of a 100 W electric lamp is 25 lumen/W. Assume that light is emitted by the lamp only in the forward half, and is uniformly distributed in all directions in this half. Calculate the luminous flux falling on a plane object of area 1 cm2 placed at a distance of 50 cm from the lamp and perpendicular to the line joining the lamp and the object.
Sol.
The power input to the bulb = 100 W. The luminous flux emitted by the bulb = (25 lumen/W) × 100 W = 2500 lumen. Since light is emitted only in the forward half and is distributed uniformly in this half, the luminous intensity is = F/
=
2500 lumen 2 sr
The solid angle subtended by the object on the lamp is =
1cm 2 (50 cm)
2
=
1 sr.. 25000
The luminous flux emitted in this solid angle is F = 2500 lumen 1 sr = sr 2500 2
=
1 lumen = 0.16 lumen. 2
4.
A point source emitting uniformly in all directions is placed above a table top at a distance of 0.50 m from it. The luminous flux of the source is 1570 lumen. Find the illuminance at a small surface area of the table-top (a) directly below the source and (b) at the distance of 0.80 m from the source.
Sol.
Consider the situation shown in figure. Let A be the point directly below the source S and B be the point at 0.80 m from the source.
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Chapter # 22 Photometry The luminous flux of 1570 lumen is emitted uniformly in the solid angle 4. The luminous intensity of the source in any direction is =
1570 lumen 4 sr
= 125 cd.
The illuminance is E=
cos
r2 At the point A, r = 0.50 m and = 0. Thus, 125 cd
EA =
= 500 lux.
0.25 m 2
At the point B, r = 0.80 m and cos =
SA 0.50 5 = = . SB 0.80 8
(125 cd) Thus,
EB =
5 8
0.64 m 2
= 122 lux.
5.
The luminous intensity of a small plane source of light along the forward normal is 160 candela. Assuming the source to be perfectly diffused, find the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of 60°C with the forward normal.
Sol.
The situation is shown in figure. By Lambert’s cosine law, the intensity in the direction SB is = 0 cos 60°, where 0 = 160 candela is the intensity along the forward normal. 1 Thus, = (160 candela) 2 = 80 candela. The luminous flux emitted in the cone shown in the figure is F = = (80 canedela) (0.02 sr) = 1.6 lumen.
EXERCISE 1.
A source emits 45 joules of energy in 15 s. What is the radiant flux of the source? ,d lzksr 15 lsd.M esa 45 twy ÅtkZ mRlftZr djrk gSA lzksr dk fofdj.k ¶yDl fdruk gSA Ans. 3 W
2.
A photographic plate records sufficiently intense lines when it is exposed for 12 s to a source of 10 W. How long should it be exposed to a 12 W source radiating the light of same colour to get equal intense lines? tc ,d QksVkxzkfQd IysV dks 10 okWV ds lzkrs ls 12 ls ls mn~Hkkflr fd;k tkrk gSA ;g i;kZIr rhozrk dh js[kk,¡ fjdkMZ djrh gSA leku rhozrk dh js[kk,¡ fjdkMZ djrh gSA leku rhozrk dh js[kk,¡ izkIr djus ds fy;s leku jax okys 12 okWV ds izdk'k
lzksr ls fdrus le; rd mn~Hkkflr djuk gksxk\ Ans. 10 s 3.
Using figure, find the relative luminosity of wavelength (a) 480 nm, (b) 520 nm (c) 580 nm and (d) 600 nm. fp=k dk mi;ksx djrs gq,] rjaxnS/;Z (a) 480 nm, (b) 520 nm (c) 580 nm vkSj (d) 600 nm dk vkisf{kd iznhiu Kkr manishkumarphysics.in
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Chapter # 22
Photometry
dhft;sA
Ans. (a) 0.14 4.
(b) 0.68
(c) 0.92
(d) 0.66
The relative luminosity of wavelength 600 nm is 0.6. Find the radiant flux of 600 nm needed to produce the same brightness sensation as produced by 120 W of radiant flux at 555 nm. 600 nm rjaxnS/;Z dh vkisf{kd iznhIrrk 0.6 gSA 600 nm dk fofdj.k ¶yDl Kkr dhft;s tks 555 nm ij 120 okWV okys
lzksr ds leku rhozrk mRiUu djsA Ans. 200 W 5.
The luminous flux of monochromatic source of 1 W is 450 lumen/watt. Find the relative luminosity at the wavelength emitted. 1 okWV ds ,d o.khZ lzksr dk iznhiu ¶yDl 450 Y;weu@okWV gSA mRlftZr rjaxnS/;Z ij vkisf{kd iznhIrrk Kkr dhft;sA Ans. 66%
6.
A source emits light of wavelength 555 nm and 600 nm. The radiant flux of the 555 nm part is 40 W and of the 600 nm part is 30 W. The relative luminosity at 600 nm is 0.6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminous efficiency. ,d lzkrs 555 nm rFkk 600 nm rjaxnS/;Z mRlftZr dj jgk gSA 555 nm dk fofdj.k ¶yDl 40 okWV ,oa 600 nm dk Hkkx 30 okWV gSA 600 nm ij vkisf{kd iznhIrrk 0.6 gSA Kkr dhft;s % (a) dqy fofdfjr ¶yDl (b) dqy iznhIrk ¶yDl , (c) iznhIru n{krk Ans. (a) 70 W (b) 39730 lumen (c) 568 lumen/W
7.
A light source emits monochromatic light of wavelength 555 nm. The source consumes 100 W of electric power and emits 35 W of radiant flux. Calculate the overall luminous efficiency. ,d o.khZ izdk'k lzkrs 555 nm rjaxnS/;Z dk izdk'k mRlftZr dj jgk gSA lzkrs 100 okWV fo|qr 'kfDr O;; djrk gS ,oa 35
okWV fofdj.k ¶yDl mRiUu djrk gSA dqy nhfIr {kerk dh x.kuk dhft;sA Ans. 240 lumen/W 8.
A source emits 31.4 W of radiant flux distributed uniformly in all directions. The luminous efficiency is 60 lumen/watt. What is the luminous intensity of the source? ,d lzkrs leLr fn'kkvksa esa leku :i ls forfjr 31.4 okWV fofdj.k ¶yDl mRlftZr djrk gSA iznhiu n{krk 60 Y;weu@okWV
gSA lzksr dh iznhiu rhozrk fdruh gS\ Ans. 150 cd 9.
A point source emitting 628 lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at (1.0 m, 0, 0) in such a way that the normal to the area makes an angle of 37° with the X-axis. ewy fcUnq ij fLFkr ,d lzksr leLr fn'kkvksa esa ,d leku :i ls 628 Y;weu iznhiu ¶yDl mRlftZr dj jgk gSA (1.0 eh-, 0, 0) ij ,d vYi {ks=kQy] bl izdkj j[kk gqvk gS fd {ks=kQy ij vfHkyEc x-v{k ls 37° dks.k cukrk gSA bl vYi
{ks=kQy] ij iznhIru dh x.kuk dhft;sA Ans. 40 lux 10.
The illuminance of a small area changes from 900 lumen/m2 to 400 lumen/m2 when it is shifted along its normal by 10 cm. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position. tc ,d vYi {ks=kQy dks bl ij vfHkyEc dh vuqfn'k 10 lseh foLFkkfir fd;k tkrk gS bldh nhfIr 900 Y;weu@eh2 ls 400
Y;weu@eh2 rd ifjofrZr gks tkrh gSA ;g ekurs gq, fd bldks vfHkyEc ij j[ks gq, ,d fcUnq lzkrs ls iznhIr fd;k tk jgk gS] lzksr rFkk {ks=kQy dh ewy fLFkfr esa nwjh Kkr dhft;sA Ans. 20 cm manishkumarphysics.in
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Chapter # 22 Photometry 11. A point source emitting light uniformly in all directions is placed 60 cm above a table - top. The illuminance at a point on the table-top, directly below the source, is 15 lux. Find the illuminance at a point on the table top 80 cm away from the first point. ,d Vscy ds ry ls 60 lseh Åij fLFkr fcUnq lzksr leLr fn'kkvksa esa leku :i ls izdk'k mRlftZr dj jgk gSA lzksr ds Bhd uhps Vscy ij fLFkr fcUnq ij nhfIr 15 yDl gSA Vscy ij izFke fcUnq ls 80 lseh nwj fLFkr fcUnq ij nhfIr Kkr dhft;sA Ans. 3.24 lux 12.
Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by a angle of 60°, by what fraction will the illuminance change? vkifrr izdk'k ds yEcor~ j[ks gq, ,d vYi {ks=kQy ij] ,d fcUnq lzkrs ls izdk'k ds ifjr% 60° dks.k ls ?kqek fn;k tk;s]
iznhfIr fdl xq.kkad ls ifjofrZr gks tk;sxh\ Ans. it will not change 13.
A student is studying a book placed near the edge of a circular table of radius R. A point source of light is suspended directly above the centre of the table. What should be the height of the source above the table so as to produce maximum illuminance at the position of the book? ,d fo|kFkhZ R f=kT;k dh o`Ùkkdkj Vsfcy ds fdukjs ij ,d iqLrd j[kdj i<+ jgk gSA Vscy ds dsUnz ds Bhd Åij izdk'k
dk ,d fcUnq lzksr yVdk;k x;k gSA izdk'k lzkrs dh Vscy ls fdruh Å¡pkbZ j[kh tk;s fd ;g Vscy ij iqLrd dh fLFkfr ij vf/kdre nhfIr ij iqLrd dh fLFkfr ij vf/kdre nhfIr mRiUu djs\ Ans. R/2 14.
Figure shows a small diffused plane source S placed over a horizontal table top at a distance of 2.4 m with its plane parallel to the table-top. The illuminance at the point A directly below the source is 25 lux. Find the illuminance at a point B of the table at a distance of 1.8 m from A. fp=k esa iznf'kZr fd;k x;k gS fd ,d NksVk folfjr lzkrs S, ,d {ksfrt Vscy VkWi ls 2.4 eh Åij] Vscy ds VkWi ds lekukarj j[kk gqvk gSA Bhd uhps fLFkr fcUnq A ij nhfIr 25 yDl gSA Vsfcy ij A ls 1.8 eh nwj fLFkr fcUnq B ij nhfIr Kkr dhft;sA
Ans. 6.1 lux 15.
An electric lamp and a candle produce illuminance at a photometer screen when they are placed at 80 cm and 20 cm from the screen respectively. The lamp is now covered with a thin paper which transmits 49% of the luminous flux. By what distance should the lamp be moved to balance the intensities at the screen again? ,d fo|qr ySEa i rFkk ,d eksecÙkh tc ,d QksVksehVj ds insZ ls Øe'k% 80 lseh rFkk 20 lseh nwj fLFkr gks] rks insZ ij ,d leku nhfIr mRiUu djrs gSAa vc ySEa i dks ,d irys dkxt ls
nsrk gSA ySia dks fdruh nwjh rd f[kldk;k tk;s fd insZ ij rhozrk,sa iqu% ,d leku gks tk;s\ Ans. 24 cm 16.
Two light sources of intensities 8 cd and 12 cd are placed on the same side of a photometer screen at a distance of 40 cm from it. Where should a 80 cd source be placed to balance the illuminance? 8 cd o 12 cd rhozrk okys nks izdk'k lzkr s ] ,d QksVksehVj ds insZ ls 40 lseh nwj ,d gh vksj j[ks gq, gS 80 cd rhozrk dk
lzksr dgk¡ j[kk tk;s fd iznhfIr larqfyr gks tk;s\ Ans. 80 cm.
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