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Physics 28 Years Past year papers for NEET/AIPMT Class 12

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Q.1. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μwater = 4/3)    (2019)
A: 0.266°
B: 0.15°
C: 0.05°
D: 0.1°
Ans: 
B
Solution:
In air angular fringe width
 Notes | EduRev
Angular fringe width in water
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 Notes | EduRev

Q.2. Unpolarised light is incident from air on a plane surface of a material of refractive index 'μ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?    (2018)
A: Reflected light is polarised with its electric vector parallel to the plane of incidence

B: Reflected light is polarised with its electric vector perpendicular to the plane of incidence
C: Notes | EduRev
D: Notes | EduRev
Ans: B
Solution:

 Notes | EduRev

Q.3. In Young's double slit experiment the separation between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to :-    (2018)
A: 1.8 mm
B: 1.9 mm
C: 2.1 mm
D: 1.7 mm

Ans: B
Solution:
 Notes | EduRev

Q.4. Young's double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly :-    (2017)
A: 1.59
B: 1.69
C: 1.78
D: 1.25
Ans: 
C
Solution:

Fringe Width in a medium -
 Notes | EduRev
- where in
Since wavelength in a medium become
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Fringe width in air  Notes | EduRev
Fringe width in any other medium  Notes | EduRev
According to question
 Notes | EduRev

Q.5. Two Polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarised light I0 is incident on P1 . A third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45° with that of P1. The intensity of transmitted light through P2 is :-    (2017)
A:

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B:

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C:

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D:

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Ans: 
B
Solution:

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Q.6. In a diffraction pattern due to a single slit of width 'a' the first minimum is observed at an angle 30º when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of :    (2016)
A: sin-1 (3/4)
B: sin-1 (1/4)
C: sin-1 (2/3)
D: sin-1 (3/2)
Ans: 
A
Solution:

Given that, first minimum is observed at an angle of 30º in a diffraction pattern due to a single slit of width a.
i.e., n = 1, θ = 30º
According to the Bragg's law of diffraction,
 Notes | EduRev
For 1st secondary maxima,
 Notes | EduRev
Putting the value of a from Eqn. (i) to Eqn. (ii), we get
 Notes | EduRev

Q.7. The intensity at the maximum in Young's double slit experiment is I0. Distance between two slits is d = 5λ, where λ is the wavelength of light used is the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?    (2016)
A: I0/2
B: I0
C: I0/4
D: 3I0/4
Ans: 
A
Solution:

Given,
Maximum intensity = Io
Distance between the slits, d = 5λ
Distance of screen from the slit, D = 10d
Using the formula,
Path difference,  Notes | EduRev
Here,
 Notes | EduRev
D = 10d = 50λ
So,
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Corresponding phase difference will be
 Notes | EduRev
 Notes | EduRev
 Notes | EduRev
 Notes | EduRev

Q.8. A radiation of energy = ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light) :    (2015)
A: E/C2
B: E/C
C: 2E/C
D: 2E/C2
Ans: 
C
Solution:

The radiation energy is given by
 Notes | EduRev
Initial momentum of the radiation is
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The reflected momentum is
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So, the change in momentum of light is
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Thus, the momentum transferred to the surface is
 Notes | EduRev

Q.9. In a double slit experiment, the 2 slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?    (2015)
A: 0.02 mm
B: 0.2 mm
C: 0.1 mm
D: 0.5 mm
Ans: 
B
Solution:

In a double slit experiment, the two slits are 1 mm apart.
d = 1 mm = 10-3 m.
The screen is placed at a distance D = 1 m away. Monochromatic light of wave length
λ = 500 nm = 5 x 10-7 m is used.
The distance between two successive maxima or two successive minima is
 Notes | EduRev

 Notes | EduRev

Q.10. For a parallel beam of monochromatic light of wavelength ‘λ’, diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be :    (2015)
A: 2Da/λ
B: 2Dλ/a
C: Dλ/a
D: Da/λ
Ans: B
Solution:

For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width 'a' is of the order of the wavelength we have

 Notes | EduRev

where θ is the angle subtended by the first minima and the central maxima at the slit.

 Notes | EduRev
If x is the width of the central maxima, we have

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where D is the distance of the screen from the slit

Q.11. A beam of light of λ = 600 nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between first dark fringes on either side of the central bright fringe is :    (2014)
A: 2.4 cm
B: 2.4 mm
C: 1.2 cm
D: 1.2 mm
Ans: 
B
Solution:

 Notes | EduRev

From the above figure,
 Notes | EduRev
For small θ and when θ is counted in rad, tan ≈ θ

So,
 Notes | EduRev

Q.12. In the Young’s double−slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be :    (2014)
A: K/2
B: Zero
C: K
D: K/4
Ans: A
Solution:

I = 4Io cos2(S/2)
4I0 = K
∵ δ = 2π if path different = λ
Phase difference when path difference = Δ/4 is equal to   Notes | EduRev

 Notes | EduRev

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