33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

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Wave OpticsTelegram @unacademyplusdiscounts93020 . dOn dividing we get,=021 . 2mm\ d = 1.9 mm17. (c) : Position of 8 th bright fringe in medium,Dx = 8λ mdPosition of 5 th dark fringe in air,⎛ 1 ⎞− D⎝⎜5⎠⎟ λair2 45 . λairDx′==dd8λm 4 5λGiven x = x′ ⇒ D . airD=d dλµ m = airλ= 8=m 45 .178 .d18. (b) : Here, d= 5λ, D= 10d,y=2dResultantIntensity at y = Iy =2 , ?dThe path difference between two waves at y = 2d ×dyd∆x= dtanθ= d × =2 5λλ= = =D 10d20 20 42ππCorresponding phase difference, φ = ∆x =λ 2Now, maximum intensity in Young’s double slit experiment,I max = I 1 + I 2 + 2I 1 I 2 or I 0 = 4I( I 1 = I 2 = I)I∴ I = 04Required intensity,IIy = I1 + I2 + I12I = I = 02 cos π 22 219. (c) : As, intensity I ∝ width of slit WAlso, intensity I ∝ square of amplitude AI1W1A1 2 ∴ = =I2W2A2 2ButW1= 1W2 25(given)A1 2 1 A11 1∴ = = =AA2 2 or252 25 52⎛ A ⎞+I + ⎝⎜⎠⎟max ( A1 A2)21 1A2∴ = =Imin( A1−A2)2 2⎛ A1⎞−⎝⎜ 1A ⎠⎟22⎛ ⎞ ⎛+⎝⎜⎠⎟⎝⎜⎞ 21=5 1 6=5 ⎠ ⎟ 36 9= =2 2⎛ 1 ⎞−⎝⎜⎠⎟ ⎛−⎝⎜ ⎞⎠⎟5 1 4 16 4520. (c) : Intensity at any point on the screen is2I = 4I0cos φ2where I 0 is the intensity of either wave and f is the phasedifference between two waves.2πPhasedifference, φ = × PathdifferenceλWhen path difference is l, then2πφ = × λ=2πλ2 ⎛ 2π⎞2∴ I = 4I0cos⎝⎜⎠⎟ = 4I0cos ( π ) = 4I0= K ...(i)2When path difference is λ 4 , then2πλ πφ = × =λ 4 22 ⎛ π ⎞ K∴ I = 4I0cos⎝⎜⎠⎟ = 2I0=[ Using(i) ]4 221. (d) : Let n 1 bright fringe of l 1 coincides with n 2bright fringe of l 2 . Thenn1λ1Dn2λ2D= or n1λ1 = n2λ2d dn1n= λ210000 52 λ= 1 12000= 6Let x be given distance.n1λ1D∴ x =dHere, n 1 = 5, D = 2 m, d = 2 mm = 2 × 10 –3 ml 1 = 12000 Å = 12000 × 10 –10 m = 12 × 10 –7 mx = × × −75 12 10 m × 2 m −3= 6× 10 m=6mm−32×10 m22. (b) : Fringe width, β = λDdwhere D is the distance between slits and screen and d isthe distance between the slits.When D is doubled and d is reduced to half, then fringewidth becomesλ( 2D)4λDβ′ = = = 4β( d / 2)d23. (a)24. (d) : Separations between the slitsd 1 = 16 cm and d 2 = 9 cmActual distance of separationd= d1d2= 16 × 9 = 12 cm25. (d) : In vacuum, l increases very slightly comparedto that in air. As b ∝ l, therefore, width of interferencefringe increases slightly.

Telegram @unacademyplusdiscounts94 NEET-AIPMT Chapterwise Topicwise Solutions Physics26. (a) : As β λ D= and λb < λy d,\ Fringe width b will decrease.27. (b) :D−102x= ( n)λ = 3 × 5000 × 10 ×d−302 . × 10= 1.5 × 10 –2 m = 1.5 cmλD28. (d) : For dark fringe, x= ( 2n−1)2 dλ= = × −3 × × −32xd2 10 0.9 10( 2n−1)D ( 2× 2− 1)× 1l = 0.6 × 10 –6 m = 6 × 10 –5 cmβ 04 .29. (a) : β′ = = = 03 . mmµ 4/330. (c) : Distance of n th maxima x= nλD ∝λdAs l b < l g\ x(blue) < x(green).31. (a) : Given l = 600 nm = 600 × 10 –9 m and D = 2 m9122 . λ 1.22 600 10\ Limit of resolution =D = × × −2= 366 × 10 –9 = 3.66 × 10 –7 rad32. (c) : For telescope, angular magnification = f 0f eDAngular resolution = should be large.122 . λSo, objective lens should have large focal length (f 0 ) andlarge diameter D for large angular magnification andhigh angular resolution.33. (b) : The resolving power of an optical microscope,RP = 2µ sin θλFor wavelength l 1 = 4000 Å, resolving power will beRP 1 = 2 µ sin θ...(i)4000For wavelength l 2 = 6000 Å, resolving power will beRP 2 = 2 µ sin θ...(ii)6000On dividing eqn. (i) by eqn. (ii)RP16000RP= 34000= 2234. (d) : Here, a = 0.02 cm = 2 × 10 –4 ml = 5 × 10 –5 cm = 5 × 10 –7 mD = 60 cm = 0.6 mPosition of first minima on the diffraction pattern,−7Dλ 06 . × 5×10−4y = == 15 × 10 m=015 . cma−42×1035. (b) : For first minimum, the path difference betweenextreme waves,asinq = l1Here θ= 30° ⇒ sinθ=2\ a = 2l ...(i)For first secondary maximum, the path differencebetween extreme waves33asin θ′= λ or ( 2λ)sinθ′ = λ223−1⎛ 3 ⎞or sinθ′ = ∴ θ′ = sin4⎝⎜4 ⎠⎟[Using eqn(i)]36. (c) : For double slit experiment,d = 1 mm = 1 × 10 –3 m, D = 1 m, l = 500 × 10 –9 mλFringe width β = D dWidth of central maxima in a single slit = 2λDaAs per question, width of central maxima of single slitpattern = width of 10 maxima of double slit pattern2λD⎛ λD⎞= 10a ⎝⎜d ⎠⎟ or2da = = × − 32 1010 10= 0.2 × 10 –3 m = 0.2 mm37. (a) : The situation is shown in the figure.In figure A and B represent the edges of the slit AB ofwidth a and C represents the midpoint of the slit.For the first minimum at P,asinq = l...(i)where l is the wavelength of light.The path difference between the wavelets from A to C isa 1 λ∆x = sin θ= ( asin θ)= (using(i))2 2 2The corresponding phase difference Df is∆φ2 π π λ= ∆x2 = × = πλ λ 238. (d) : Here, l = 600 nm = 600 × 10 –9 ma = 1 mm = 10 –3 m, D = 2 mDistance between the first dark fringes on either sideof the central bright fringe is also the width of centralmaximum.

Wave Optics

Telegram @unacademyplusdiscounts

93

020 . d

On dividing we get,

=

021 . 2mm

\ d = 1.9 mm

17. (c) : Position of 8 th bright fringe in medium,

D

x = 8λ m

d

Position of 5 th dark fringe in air,

⎛ 1 ⎞

− D

⎝

⎜5

⎠

⎟ λair

2 45 . λairD

x′=

=

d

d

8λm 4 5λ

Given x = x′ ⇒ D . airD

=

d d

λ

µ m = air

λ

= 8

=

m 45 .

178 .

d

18. (b) : Here, d= 5λ, D= 10d,

y=

2

d

ResultantIntensity at y = Iy =

2 , ?

d

The path difference between two waves at y = 2

d ×

d

y

d

∆x= dtanθ

= d × =

2 5λ

λ

= = =

D 10d

20 20 4

2π

π

Corresponding phase difference, φ = ∆x =

λ 2

Now, maximum intensity in Young’s double slit experiment,

I max = I 1 + I 2 + 2I 1 I 2 or I 0 = 4I( I 1 = I 2 = I)

I

∴ I = 0

4

Required intensity,

I

Iy = I1 + I2 + I12

I = I = 0

2 cos π 2

2 2

19. (c) : As, intensity I ∝ width of slit W

Also, intensity I ∝ square of amplitude A

I1

W1

A1 2 ∴ = =

I2

W2

A2 2

But

W1

= 1

W2 25

(given)

A1 2 1 A1

1 1

∴ = = =

A

A

2 2 or

25

2 25 5

2

⎛ A ⎞

+

I + ⎝

⎜

⎠

⎟

max ( A1 A2)

2

1 1

A2

∴ = =

Imin

( A1−

A2)

2 2

⎛ A1

⎞

−

⎝

⎜ 1

A ⎠

⎟

2

2

⎛ ⎞ ⎛

+

⎝

⎜

⎠

⎟

⎝

⎜

⎞ 2

1

=

5 1 6

=

5 ⎠ ⎟ 36 9

= =

2 2

⎛ 1 ⎞

−

⎝

⎜

⎠

⎟ ⎛

−

⎝

⎜ ⎞

⎠

⎟

5 1 4 16 4

5

20. (c) : Intensity at any point on the screen is

2

I = 4I0

cos φ

2

where I 0 is the intensity of either wave and f is the phase

difference between two waves.

2π

Phasedifference, φ = × Pathdifference

λ

When path difference is l, then

2π

φ = × λ=

2π

λ

2 ⎛ 2π

⎞

2

∴ I = 4I0

cos

⎝

⎜

⎠

⎟ = 4I0

cos ( π ) = 4I0

= K ...(i)

2

When path difference is λ 4 , then

2π

λ π

φ = × =

λ 4 2

2 ⎛ π ⎞ K

∴ I = 4I0

cos

⎝

⎜

⎠

⎟ = 2I0

=

[ Using(i) ]

4 2

21. (d) : Let n 1 bright fringe of l 1 coincides with n 2

bright fringe of l 2 . Then

n1λ1D

n2λ2D

= or n1λ1 = n2λ2

d d

n1

n

= λ2

10000 5

2 λ

= 1 12000

= 6

Let x be given distance.

n1λ

1D

∴ x =

d

Here, n 1 = 5, D = 2 m, d = 2 mm = 2 × 10 –3 m

l 1 = 12000 Å = 12000 × 10 –10 m = 12 × 10 –7 m

x = × × −7

5 12 10 m × 2 m −3

= 6× 10 m=

6mm

−3

2×

10 m

22. (b) : Fringe width, β = λD

d

where D is the distance between slits and screen and d is

the distance between the slits.

When D is doubled and d is reduced to half, then fringe

width becomes

λ( 2D)

4λD

β′ = = = 4β

( d / 2)

d

23. (a)

24. (d) : Separations between the slits

d 1 = 16 cm and d 2 = 9 cm

Actual distance of separation

d= d1d

2= 16 × 9 = 12 cm

25. (d) : In vacuum, l increases very slightly compared

to that in air. As b ∝ l, therefore, width of interference

fringe increases slightly.

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