datavorous/entrance-exam-dataset huggingface dataset:
In yeasts, two organisms fuse together. This type of syngamy is called

Options:
A Isogamy
B Anisogamy
C Oogamy
D Hologamy

The correct answer is:
Hologamy

Solution:
The fusion of gametes that do not differ from each other structurally is called isogamy. The fusion of gametes that differ from each other structurally is called anisogamy. The fusion of a large non-motile male gamete and a small motile female gamete is called oogamy. In yeast, there is no difference between gametes and the somatic cell. Thus, the entire protoplast fuses resulting in hologamy.

Topic: Biology, Reproduction in Organisms, NEET
In Young's double slit experiment a mica plate of refractive index μ is introduced in the path of light coming from one of the slit. If the central bright fringe gets shifted to the point originally occupied by the fourth bright fringe, then the thickness of the mica plate will be (symbols have their usual meaning)

Options:
A 2λ(μ-1)
B 4λ3(μ-1)
C 4Î»(Î¼-1)
D 2λ3(μ-1)

The correct answer is:
4λ(μ-1)

Solution:
Position of fourth bright =4λDd
According to given condition (μ-1)tDd=4λDd
t=4λμ-1

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment, distance between the slits S1 and S2 is d and the distance between slits and screen is D. Then longest wavelength that will be missing on the screen in front of S1 is

Options:
A d2D
B d22D
C Dd2
D None of these

The correct answer is:
d2D

Solution:
Path difference, ∆x=dsin⁡θ≈dtan⁡θ=d22D
For first missing wavelength,
d22D=λ2 or λ=d2D

Topic: Physics, Wave Optics, NEET
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

Options:
A half
B four times
C one-fourth
D double

The correct answer is:
four times

Solution:
β=λDd

β'=λD'd'

D'=2D,d'=d2

β'=λ×2Dd2=4λDd

β'=4β

Fringe width becomes 4 times.

Topic: Physics, Wave Optics, NEET, NEET 2020 (Phase 1)
In Young's double slit experiment if the slit widths are in the ratio 1 : 9. The ratio of the intensity at minima to that at maxima will be

Options:
A 1
B 19
C 14
D 13

The correct answer is:
14

Solution:
amplitude of the superimposing waves are
a1a2=1912=13
Iminima Imaxima=a1-a22a1+a22=14

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment, one of the slit is wider than another, so that amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference ϕ is given by

Options:
A Im94+5cos⁡ϕ
B Im31+2cos2⁡ϕ2
C Im51+4cos2⁡ϕ2
D Im91+8cos2⁡ϕ2

The correct answer is:
Im91+8cos2⁡ϕ2

Solution:
Given, a1=2a2
I1=4I′,Â I2=I′
We know that, Im=I1+I22
=4I′+I′2=9I′
â´I=I1+I2+2I1I2cosâ¡Ï
=4I′+I′+24I′⋅I′cosâ¡Ï
=4I′+I′+4I′cosâ¡Ï
=5I′+4I′cosâ¡Ï
=I′5+4cosâ¡Ï
=Im95+42cos2â¡Ï2-1
=Im95+8cos2â¡Ï2-4
=Im91+8cos2â¡Ï2

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment, slits are separated by 2 mm and the screen is placed at a distance of 1.2 m from the slits. Light consisting of two wavelengths 6500 â« and 5200 Ã are used to obtain interference fringes. Then, the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

Options:
A 0.312 mm
B 0.Â 123Â mm
C 0.213Â mm
D 0.412Â mm

The correct answer is:
0.312 mm

Solution:
y 4 = 4DÎ» d
Separation between the fourth bright fringes
=4DÎ»1d-4DÎ»2d
4DdÎ»1-Î»2=4Ã1.22Ã10-36500-5200Ã10-10
=3120Ã10-7=0.312 mm

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment, the 10 th maximum of wavelength Î»1 is at a distance of y1 from the central maximum. When the wavelength of the source is changed to Î»2, 5th maximum is at a distance of y2 from its central maximum. The ratio y1y2 is

Options:
A 2Î»1Î»2
B 2 λ 2 λ 1
C Î»12Î»2
D Î»22Î»1

The correct answer is:
2λ1λ2

Solution:
Position of fringe from central maxima

y1=nÎ»1Dd

Given, n=10

y1=10Î»1Dd â¦â¦â¦.(i)

For second source

y2=5Î»2Dd ........... (ii)
y1y2=10Î»1Dd5Î»2Dd

y1y2=2Î»1Î»2

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment, the light emitted from source has Î»=6.5Ã10-7 m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth bright fringe will be

Options:
A 3.2 mm
B 1.63 mm
C 0.585 mm
D 2.31 mm

The correct answer is:
1.63 mm

Solution:
x5=nλDd=5×6.5×10-7×110-3=32.5×10-4 m
x3=2n-1λ2Dd=5×6.3×10-7×12×10-3=16.25×10-4 m
∴x5-x3≈1.63 mm

Topic: Physics, Wave Optics, NEET
In Young's double slit experiment the separation d 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 0.21o (with same λ and D) the separation between the slits needs to be changed to

Options:
A 2.1 mm
B 1.9 mm
C 1.8 mm
D 1.7 mm

The correct answer is:
1.9 mm

Solution:
Angular fringe width, Î¸=Î»d, Î¸1Î¸2=d2d1
0.200.21=d22âd2=0.200.21Ã2=1.9 mm

Topic: Physics, Wave Optics, NEET, NEET 2018
In Young's double slit experiment, the slits are $2 \mathrm{~mm}$ apart and are illuminated by photons of two wavelengths $\lambda_1=12000 Å \quad$ and $\lambda_2=10000 Å$. At what minimum distance from the common central bright fringe on the screen $2 \mathrm{~m}$ from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options:
A $8 \mathrm{~mm}$
B $6 \mathrm{~mm}$
C $4 \mathrm{~mm}$
D $3 \mathrm{~mm}$

The correct answer is:
$6 \mathrm{~mm}$

Solution:
Given $\lambda_1=12000 Å$ and $\lambda_2=10000 Å$, $D=2 \mathrm{~cm}$ and $d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~cm}$.
We have $\frac{\lambda_1}{\lambda_2}=\frac{h_2}{h_1}$
$$
=\frac{12000}{10000}=\frac{6}{5}
$$
as $x=\frac{u_1 \lambda_1 D}{d}=\frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}}$
$=5 \times 1.2 \times 10^4 \times 10^{-10} \times 10^3=6 \mathrm{~mm}$

Topic: Physics, Wave Optics, NEET, NEET 2013 (All India)
In Young's double slits experiment, if the separation between the slits is halved, and the distance between the slits and the screen is doubled, then the fringe width compared to the original one will be

Options:
A Unchanged
B Halved
C Doubled
D Quadrupled

The correct answer is:
Quadrupled

Solution:
The fringe width,
Î²=Î»Dd
Where, Î»= wavelength of light
d= distance of slit
D=Â distance of screen from slit.
According to question,
d′=d2
D=2D
Î²′=?
Î²′=Î»D′d′
=Î»2Dd/2=4Î²
So, the fringe width will be quadrupled.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when the light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by

Options:
A 18
B 24
C 30
D 36

The correct answer is:
18

Solution:
In case of overlapping of fringes.
nÎ²=constent
n1 Î»1=n2 Î»2
â´n2=n1 Î»1Î»2=12Ã600400=18

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, a mica slab of thickness t and refractive index is introduced in the ray from the first source S1. By how much distance, fringes pattern will be displaced?

Options:
A d&ApplyFunction; D μ - 1 t
B D d&ApplyFunction; μ - 1 t
C d&ApplyFunction; μ - 1 D
D D d&ApplyFunction; μ - 1

The correct answer is:
D d&ApplyFunction; μ - 1 t

Solution:
For a path difference (μ - 1)t, the shift is
x = μ - 1 t D d .

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, d=0.1 mm, D=20 cm, and Î»=5460 â«, the angular position of first dark fringe will be

Options:
A 0.08o
B 0.24o
C 0.32o
D 0.16o

The correct answer is:
0.16o

Solution:
Position of dark fringe y=2n-1Î»2ÃDd=Î»2ÃDd
â´ Also Î¸=yD
=Î»2d=2730Ã10-6Ã180Ï=0.16o

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment. If the distance between the two slits is equal to the wavelength of used light. Then the maximum number of bright fringes obtained on the screen will be

Options:
A Infinite
B 3
C 7
D 5

The correct answer is:
3

Solution:
We know
dsinâ¡Î¸=Â± m Î»
But d=Î»
So sinâ¡Î¸=Â± m
For max. â sinâ¡Î¸=1
m= Â± 1
So maximum number of bright fringes = 3
Central maxima + either side of central maxima

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, if yellow light is replaced by blue light, the interference fringes become

Options:
A Darker
B Brighter
C Wider
D Narrower

The correct answer is:
Narrower

Solution:
The wavelength Î» of yellow light is greater than blue light.
We know that,
Fringe width w=DÎ»d
wâÎ»
Here, Î» decreases. Hence, the interference fringes become narrower.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, if yellow light is replaced by blue light, the interference fringes becomes

Options:
A Darker
B Brighter
C Wider
D Narrower

The correct answer is:
Narrower

Solution:
The wavelength Î» of yellow light is greater than blue light.
We know that,
Fringe width w=DÎ»d
wâÎ»
Here, Î» decreases. Hence, the interference fringes become narrower.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, in an interference pattern, the second minimum is observed exactly in front of one slit. The distance between the slits is d and the distance between source and screen is D. The wavelength of the light source used is

Options:
A d2D
B d22D
C d23D
D d24D

The correct answer is:
d23D

Solution:
Distance of second minima from central point is y=d2.

We know that minima is given by,

dsinθ= n-12λdyD= 2-12λ

d 2=Dd2×2-12λ
∴d2=Dd×32λ
∴λ=d23D

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment intensity at a point is 14th of the maximum intensity. The angular position of this point is (separation between slits is d and also two slits are similar)

Options:
A sin-1⁡λd
B sin-1⁡λ2d
C sin-1⁡λ3d
D sin-1⁡λ4d

The correct answer is:
sin-1⁡λ3d

Solution:
If a is the amplitude of the wave then
I max 4 = a 2 = a 2 + a 2 +2aa cosϕ
or cosϕ=− 1 2
or ϕ= 2π 3
Corresponding path difference,
Δx= ϕ×λ 2π = ( 2π 3 )×λ 2π = λ 3
So dsinθ= λ 3
Or θ= sin −1 ( λ 3d ).

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference ϕ?

Options:
A Im91-8cos2⁡ϕ2
B Im91+8cos2⁡ϕ2
C Im91-8cos2⁡ϕ
D Im91-sin2⁡ϕ2

The correct answer is:
Im91+8cos2⁡ϕ2

Solution:
Let intensity of two sources in YDSE be I1 and I2 . Assume phase difference between the sources is ϕ.
Resultant intensity in YDSE is given by
I=I1+I2+2I1I2cos⁡ϕ .....(i)
It is given that, a1=2a2
⇒ a12=4a22
⇒ I1=4I2 ∵I∝a2 .....(ii)
Now from Equations. (i) and (ii), we have
I=4I2+I2+2×2×I2cos⁡ϕ
I=5I2+4I2cos⁡ϕ .....(iii)
Given, Im=Imax
=I1+I22
⇒ Im=2I2+I22
Im=9I2
I2=Im9 ......(iv)
Now from Equations. (iii) and (iv), we get
I=5×Im9+4×Im9 cos ϕ
=4Im9+4Im9 cos ϕ+Im9
=4Im91+cos⁡ϕ+Im9
=4Im92 cos2⁡ϕ2+Im9
∵ 1+cos⁡ϕ=2cos2⁡ϕ2
=Im98cos2⁡ϕ2+1
=Im91+8cos2⁡ϕ2

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, spacing between the two slits is 0.1 mm. If the screen is kept at a distance of 1.0 m from the slits and the wavelength of light is 5000 â«, then the fringe width is

Options:
A 1.0Â cm
B 1.5Â cm
C 0.5 cm
D 2.0Â cm

The correct answer is:
0.5 cm

Solution:
Fringe width is given by
Î²=Î»Dd
Î²=5000Ã10-10Ã10.1Ã10-3=5Ã10-3
Î²=0.5 cm

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the distance between slits and the screen is 1 m and monochromatic light of wavelength 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits. If the angular resolution of the eye is 160Â°, then the value of d0 is close to

Options:
A 1 mm
B 3 mm
C 2 mm
D 4 mm

The correct answer is:
2 mm

Solution:
When the angular fringe width in the interference pattern becomes less than the angular resolving power of the person's eye, the fringes will disappear.

The angular fringe width θ 0 = β D = λ d 0
d 0 = λ θ 0 = 600× 10 −9 ( 1 60 × π 180 )

d0=600×60×180×10−9π m=2 mm

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 43, without disturbing the geometrical arrangement, the new fringe width will be

Options:
A 0.30 mm

B 0.40 mm

C 0.53 mm

D 450 microns

The correct answer is:
0.30 mm

Solution:
β′=βμ=0.4(4/3)=0.3 mm

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the intensities at two points P1 and P2 on the screen are I1 and I2, respectively. If P1 is located at the central bright fringe and P2 is located at a distance equal to a quarter of fringe width from P1, then I1I2 is

Options:
A 2
B 12
C 4
D 16

The correct answer is:
2

Solution:
P1 is at central maxima so I1=4I0
For P2, Ï=2ÏÎ²Î²4=Ï2
I2=I0+I0+2I0I0cosâ¡Ï2=2I0
Hence, I1I2=2

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the intensities at two points P1 and P2 on the screen are I1 and I2 respectively. If P1 is located at the central bright fringe and P2 is located at a distance equal to a quarter of fringe width from P1, then I1I2 is

Options:
A 2
B 12
C 4
D 16

The correct answer is:
2

Solution:
P1 is at central maxima so I1=4I0
For P2, ϕ=2πββ4=π2
I2=I0+I0+2I0I0cos⁡π2=2I0
Hence, I1I2=2

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment the intensity at a point where the path difference is Î» 6 ( Î» being the wavelength of light used) is I. If I 0 denotes the maximum intensity, I I 0 is equal to

Options:
A 3 4
B 1 2
C 3 2
D 1 2

The correct answer is:
3 4

Solution:
In Young's double slit experiment intensity at a point is given by
I = I 0 cos 2 Ï 2
where Ï = phase difference,
I 0 = maximum intensity
or I I 0 = cos 2 Ï 2 .......(i)
Phase difference Ï = 2 Ï Î» Ã path difference
â´ Ï = 2 Ï Î» Ã Î» 6 or Ï = Ï 3 .......(ii)
Substitute equation (ii) in equation (i), we get
I I 0 = cos 2 Ï 6 or I I 0 = 3 4 .

Topic: Physics, Wave Optics, NEET
In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is I,λ being the wavelength of light used. The intensity at a point where the path difference is λ4 will be

Options:
A I4
B I2
C I
D Zero

The correct answer is:
I2

Solution:
For path difference Î», phase difference =2Ï(Ï=2ÏÎ»x=2ÏÎ».Î»=2Ï)
âI=I0+I0+2I0cos2Ï
âI=4I0(â´cos2Ï=1)
For x=Î»4, phase difference =Ï2
â´Iâ²=I1+I2+2I1I2cosÏ2
If I1=I2=I0 then I'=2I0=2.I4=I2

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the number of maxima that are obtained on a screen (including the central maxima), if Î»=2000 Ã and d=7000 Ã is

Options:
A 12
B 7
C 18
D 6

The correct answer is:
7

Solution:
For maxima,
Δ x = d sin θ = n λ
∴ sin θ = n λ d = n × 2 0 0 0 7 0 0 0
Since sin θ cannot be greater than 1.
Hence, n=0, 1, 2, 3 only
∴ 3 maxima on both side of screen and 1 central maxima will be obtained.
∴ Total maxima = 7

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the path difference between two interfering waves at a point on the screen is 13.5 times the wavelength. The point is

Options:
A Bright but not central bright
B Neither bright not dark
C Central bright
D Dark

The correct answer is:
Dark

Solution:
Given,
Path difference Δ=13.5λ ......(i)
We know that, for minima
Path difference Δ=n+12λ ......(ii)
We can write Eq. (i) in the following form
Δ=13+12λ ......(iii)
From Eq. (ii) and (iii), it is clear that there will be minima. So, the point will be dark.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the source is white light. One slit is covered with a red filter and the other with a blue filter. There shall be

Options:
A Alternate red and blue fringes
B Alternate dark and pink fringes
C Alternate dark and yellow fringes
D No interference

The correct answer is:
No interference

Solution:
Red and blue filters produce incoherence resulting into no interference.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the spacing between the slits is 'd' and the wavelength of light used is 6000 Å. If the angular width of a fringe formed on a distant screen is 1o, then calculate 'd'.

Options:
A 1 mm
B 0.05 mm
C 0.03 mm
D 0.01 mm

The correct answer is:
0.03 mm

Solution:
Angular fringe width Î¸0=Î²D
Î¸0=Î²D=DÎ»dÃ1D=Î»d
Î¸0=1o=Ï180 rad and
Î»=6Ã10-7 m
d=Î»Î¸0=180ÏÃ6Ã10-7
=3.44Ã10-5 m=0.03 mm

Topic: Physics, Wave Optics, NEET
In Youngâs double-slit experiment, the two slits 0.20 mm apart are illuminated by monochromatic light of wavelength 600 nm. The screen is 1.0 m away from the slits. Find the distance of the second bright fringe.

Options:
A 3 mm
B 6 mm
C 4 mm
D 5 mm

The correct answer is:
6 mm

Solution:
Here d=0.20 mm=20×10-5 m

λ=600 nm=600×10-9 m

D=1 m

For the distance of the second bright fringe

x2=2λDd [∵xn=nλDd]

=2×600×10−9×1.020×10−5

=60×10−4=6×10−3 m=6 mm

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment the two slits are illuminated by the light of wavelength 5890 Aâ and the angular distance between the fringes obtained on the screen is 0.2Â°. If the whole apparatus is immersed in water then the angular fringe width will be, (the refractive index of water is 43)

Options:
A 0.30°
B 0.15°
C 15°
D 30°

The correct answer is:
0.15°

Solution:
wa=λd ⇒ wa∝λ
wawaterwa=λwaterλ=λμwaterλ
wawater=0.2×3 4=0.15°

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, the two slits which are separated by 1.2 mm are illuminated with a monochromatic light of wavelength 6000 Aâ. The interference pattern is observed on a screen placed at a distance of 1 m from the slits. Find the number of bright fringes formed over 1 cm width on the screen.

Options:
A 25
B 12
C 15
D 20

The correct answer is:
20

Solution:
Given: d=1.2 mm, Î»=6000 â«=6Ã10-7 m, D=1 m, x=1 cm=1Ã10-2 m
For nth bright fringe, x=nÎ»Dd
â´1Ã10-2=nÃ6Ã10-7Ã11.2Ã10-3
â´n=1.2Ã10-56Ã10-7=0.2Ã102=20
There are 20 bright fringes formed over 1 cm width on the screen.

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, two wavelengths Î»1=780 nm and Î»2=520 nm are used to obtain interference fringes. If the nth bright band due to Î»1 coincides with n+1th bright band due to Î»2, then the value of n is

Options:
A 4
B 3
C 2
D 6

The correct answer is:
2

Solution:
nDÎ»1d=n+1DÎ»2d
nÎ»1=n+1Î»2â 780n=n+1520
n=520260â n=2

Topic: Physics, Wave Optics, NEET
In Young's double-slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. Then, the intensity of light at a point where path difference is λ3 is

Options:
A K2
B 2K
C 4K
D K4

The correct answer is:
K4

Solution:
Phase difference corresponding to path difference Î» is
ÎÏ=2ÏÎ»ÃÎ»=2Ï
Intensity at the point having this phase difference 2Ï is
I=4I0cos2â¡Ï2
=4I0cosâ¡2Ï22
=4I0-12=4I0
Here, I0 is intensity of individual source.
It is given that
I=4I0=K
âÂ Â I0=K4 ......(i)
Phase difference corresponding to path difference of Î»3 is
ÎÏ′=2ÏÎ»ÃÎ»3=2Ï3
Thus, intensity of light corresponding to this phase difference of 2Ï3 is
I′=4I0Ãcosâ¡2Ï32
=KÃ-122 [from Equation. (i)]
=K4

Topic: Physics, Wave Optics, NEET
In Young's experiment, monochromatic light is used to illuminate the two slits A and B. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of the beam coming from the slit, then which of the following explanation holds true with regards to fringes/fringe width?

Options:
A fringes will disappear
B fringe width will increase
C fringe width will increases
D there will be no change in the fringe width but fringe pattern will shift

The correct answer is:
there will be no change in the fringe width but fringe pattern will shift

Solution:
In the presence of thin glass plate, the fringe width pattern shifts, but no change in fringe width. It is because of the uniform path difference that gets introduced by adding a glass plate.

Topic: Physics, Wave Optics, NEET
In Young's experiment, the ratio of maximum and minimum intensities in the fringe system is 9:1 . The ratio of amplitudes of coherent sources is

Options:
A 9:1
B 3:1
C 2:1
D 1:1

The correct answer is:
2:1

Solution:
ImaxImin=a1+a22a1-a22
91=a1+a2a1-a22
a1+a2a1-a2=31
By solving this:
a1+a2=3a1-3a2
4a2=2a1
a1a2=21

Topic: Physics, Wave Optics, NEET
In Young's experiment, the third bright band for the wavelength of light source A having wavelength 6000 Ã coincides with the fourth bright band for source B in the same arrangement. Wavelength of light emitted by source B is

Options:
A 6289 Ã
B 4500 Ã
C 2250 Ã
D 6000 Ã

The correct answer is:
4500 Ã

Solution:
As given, DdÃ3Ã6Ã10-7
=DdÃ4ÃÎ»2

âÎ»2=3Ã6Ã10-74

=4.5Ã10-7 m
=4500 Ã

Topic: Physics, Wave Optics, NEET
In Young's experiment, the wavelength of red light is 7.8Ã10-8 cm and that of blue light is 5.2Ã10-8 cm . Value of n for which (n+1)th blue bright fringe coincides with nth red fringe is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
As we know that n1 λ1=n2 λ2
∴n2n1=λ1λ2=7.8×10-85.2×10-8=32
Also n2-n1=1
∴ n1=2

Topic: Physics, Wave Optics, NEET
In your opinion, which is the most effective way to conserve the plant diversity of an area?

Options:
A by tissue culture method
B by creating biosphere reserve
C by creating botanical garden
D by developing seed bank.

The correct answer is:
by creating biosphere reserve

Solution:
Biosphere reserve is a in situ method of conservation, while rest options are ex situ method. Biosphere reserves are the protected areas meant for the conservation of plants and animals. It also restores the traditional life of the tribals living in that vicinity. They conserve the biodiversity of that area.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2004
In Z - scheme, Z - shape is formed when

Options:
A carriers are placed uphill
B carriers are placed downhill
C carriers are placed in sequence on a redox potential scale
D carriers are placed in increasing order of energy

The correct answer is:
carriers are placed in sequence on a redox potential scale

Solution:
In photosystem II the reaction centre chlorophyll a absorbs 680 nm wavelength of red light causing electrons to become excited and jump into an orbit farther from the atomic nucleus. These electrons are picked up by an electron acceptor which passes them to an electrons transportsystem consisting of cytochromes. This movement of electrons is downhill, in terms of an oxidation-reduction or redox potential scale. The electrons are not used up as they pass through the electron transport chain, but are passed on to the pigments of photosystem PS I. Simultaneously, electrons in the reaction centre of PS I are also excited when they receive red light of wavelength 700 nm and are transferred to another accepter molecule that has a greater redox potential. These electrons then are moved downhill again, this time to a molecule of energy-rich NADP+. The addition of these electrons reduces NADP+ to NADPH + H+ . This whole scheme of transfer of electrons, starting from the PS II, uphill to the acceptor, down the electron transport chain to PS I, excitation of electrons, transfer to another acceptor, and finally downhill to NADP+ reducing it to NADPH + H+ is called the Z scheme, due to its characteristic shape. This shape is formed when all the carriers are placed in a sequence on a redox potential scale.

Topic: Biology, Photosynthesis in Higher Plants, NEET