datavorous/entrance-exam-dataset huggingface dataset:
In which one of the following sets of animals to all the four give birth to young ones?

Options:
A kangaroo, hedgehog, dolphin. Loris
B lion, bat, whale, ostrich
C Platypus, penguin, bat, hippopotamus
D shrew, bat, cat, kiwi

The correct answer is:
kangaroo, hedgehog, dolphin. Loris

Solution:
Mammals show viviparity. Kangaroo, hedgehog, dolphin and loris are mammals. These give birth to young ones.
Viviparous Mammals: Kangaroo, Hedgehog, Dolphin, Loris, Lion, Bat, Whale, Hippopotamus, Shrew, Cat
Birds: Ostrich, Penguin, Kiwi
Related Theory
Viviparous animals are present in all groups of vertebrates except birds. Fish, amphibians, reptiles, and mammals all have members that are viviparous even though none of the groups is exclusively viviparous.
Caution
Echidna and Platypus are two egg-laying mammals. Ovoviviparous animals are those wherein the embryos develop in the eggs and remain in the body of the mother until they are capable of hatching.

Topic: Biology, Animal Kingdom, NEET, NEET 2006
In which one of the following, the genus name, its two characters and its phylum are not correctly matched, whereas the remaining three are correct?

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
Molluscs are bilaterally symmetrical, triploblastic, coelomate, soft bodied animals. Their soft body is covered by a calcareous shell and is unsegmented with a distinct head, muscular foot and visceral hump. e.g., Pila (apple snail), Sepia (cuttle fish), Pinctada (pearl oyster), etc.

Topic: Biology, Animal Kingdom, NEET, NEET 2012 (Screening)
In which one of the following, the male and female gametophytes don't have free-living independent existence?

Options:
A Pteris
B Funaria
C Polytrichum
D Cedrus

The correct answer is:
Cedrus

Solution:
Cedrus is a gymnosperm having reduced dependent gametophytes- pollen and endosperm. In gymnosperms the male and the female gametophytes do not have an independent free-living existence and remain within the sporangia and are retained on the sporophytes. Hence, option D is correct.

Topic: Biology, Plant Kingdom, NEET, NEET 2008 (Screening)
In which one of the pairs of ion given, there is an ion that forms a co-ordination compound with
both aqueous sodium hydroxide and ammonia and an other ion that forms a co-ordination
compound only with aqueous sodium hydroxide?

Options:
A \( \mathrm{Zn}^{2+}, \mathrm{Al}^{3+} \)
B \( A l^{3+}, C u^{2+} \)
C \( \mathrm{Pb}^{2+}, \mathrm{Cu}^{2+} \)
D \( C u^{2+}, Z n^{2+} \)

The correct answer is:
\( \mathrm{Zn}^{2+}, \mathrm{Al}^{3+} \)

Solution:
Ion forms a coordination compound with both aqueous sodium hydroxide and ammonia and other ions forms
coordination compound only with aqueous sodium hydroxide. Zinc forms \( \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right] \) coordination compound with
\( \mathrm{NaOH} \)
\( Z n+2 N a O H(a q) \rightarrow N a_{2}\left[Z n(O H)_{4}\right] \)
Zinc forms a colorless tetraammine with the formula
\( \left[Z n\left(N H_{3}\right)_{4}\right]^{2+} \) \( Z n+4 N H_{3}(a q) \rightarrow\left[Z n\left(N H_{3}\right)_{4}\right]^{2+} \)
Aluminum forms complex with only aqueous sodium hydroxide \( \mathrm{N} a_{3}\left[\mathrm{Al}(\mathrm{OH})_{6}\right] \) and not with ammonia.
\( A l+3 N a O H(a q) \rightarrow N a_{3}\left[A l(O H)_{6}\right] \)
and \( \mathrm{Al}+\mathrm{NH}_{3}(a q) \rightarrow \) no reaction

Topic: Chemistry, Structure of Atom, JEE Main
In which phase of the cell cycle, the amount of DNA in a cell remains constant at 4C level, considering that the initial amount of DNA was 2C?

Options:
A G0 and G1
B G1 and S
C Only G2
D G2 and M

The correct answer is:
Only G2

Solution:
G1 phase corresponds to the interval between mitosis and initiation of DNA replication. During G1 phase, the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time, the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C, then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G1, even after S phase the number of chromosomes remains the same, i.e., 2n. In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm. During the G2 phase, proteins are synthesised in preparation for mitosis, while cell growth continues.

Topic: Biology, Cell Cycle and Cell Division, NEET
In which reaction, hydrogen peroxide neither acts as an oxidising agent nor as a reducing agent?

Options:
A PbS + H 2 O 2 →
B SO 3 - + H 2 O 2 →
C PbO 2 + H 2 O 2 →
D Na 2 CO 3 + H 2 O 2 →

The correct answer is:
Na 2 CO 3 + H 2 O 2 →

Solution:
H2O2 behaves as an acid when reacts with Na2CO3 .
Na 2 CO 3 + H 2 O 2 → Na 2 O 2 + CO 2 + H 2 O

Topic: Chemistry, s Block Elements, JEE Main
In which region of the photodiode does the majority of electron-hole pair generation occur when exposed to light?

Options:
A Depletion region
B Neutral region
C P-region
D N -region

The correct answer is:
Depletion region

Solution:
When exposed to light, the majority of electron-hole pairs in a photodiode are generated in the depletion region, where the electric field separates the carriers, creating a photocurrent.

Topic: Physics, Semiconductors, JEE Main
In which type of inflorescence does the main axis continue to grow?

Options:
A Racemose
B Solitary
C Cymose
D Acropetal

The correct answer is:
Racemose

Solution:
In racemose type of inflorescences the main axis continues to grow.

Topic: Biology, Morphology of Flowering Plants, NEET
In which type of inflorescence does the main axis terminate in a flower?

Options:
A Racemose
B Solitary
C Cymose
D Acropetal

The correct answer is:
Cymose

Solution:
In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth.

Topic: Biology, Morphology of Flowering Plants, NEET
In Whittaker's system of classification, prokaryotes belong to the kingdom

Options:
A Monera
B Protista
C Animalia
D Fungi

The correct answer is:
Monera

Solution:
RH Whittaker (1969) proposed a Five Kingdom Classification. The kingdoms defined by him were named-Monera, Protista, Fungi, Plantae and Animalia. The main criteria for classification used by him include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships. All the prokaryotic organisms were grouped together under kingdom-Monera and unicellular eukaryotic organisms were placed in kingdom Protista. The fungi were placed is a separated kingdom, ie, kingdom-Fungi. The kingdom-Plantae and Animalia, commonly referred to as plant and animal kingdoms, respectively.

Topic: Biology, Biological Classification, JIPMER, JIPMER 2010
In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively:

Options:
A sp3d, trigonal bipyramidal
B d2sp3, octahedral
C dsp2, square planar
D sp3, tetrahedral

The correct answer is:
dsp2, square planar

Solution:
Wilkinson catalyst
[RhClPPh33]
Metal with 4d8 and 5d8 Configuration always forms square planar complexes on account of higher CFSE irrespective of weak field or strong field. The magnetic property is diamagnetic.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (16 Apr Online)
In Williamson synthesis of mixed ether having a primary and a tertiary alkyl group if tertiary halide is used, then :

Options:
A
Rate of reaction will be slow due to slow cleavage of carbon-halogen bond.

B
Alkene will be the main product.

C
Simple ether will form instead of mixed ether.

D
Expected mixed ether will be formed.

The correct answer is:
Alkene will be the main product.

Solution:
The tertiary alkyl halide undergo elimination reaction to give alkenes

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (22 Apr Online)
In Wurtz reaction, alkyl halide reacts with

Options:
A \( \mathrm{Na} \) in water.
B Na in ethereal solution.
C \( \mathrm{Na}-\mathrm{Hg} \) in water.
D Mg in ethereal solution.

The correct answer is:
Na in ethereal solution.

Solution:
In Wurtz reaction, alkyl halide reacts with Na in the presence of dry ether solution. If water is used instead of ether, it will react with Na metal.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In $X e O_3$ and $X e F_6$ the oxidation state of $X e$ is

Options:
A $+4$
B $+6$
C $+1$
D $+3$

The correct answer is:
$+6$

Solution:
The oxidation state of $\mathrm{Xe}$ in both $\mathrm{XeO}_3$ and $\mathrm{XeF}_6$ is +6

Topic: Chemistry, Redox Reactions, JEE Main
In XeOF4 how many angles are of 90° ?

Options:

The correct answer is:
0

Solution:
In XeOF4 no angle is of 90° due to lp-bp repulsion.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In XeF 2 ,  XeF 4 ,  XeF 6 the number of lone pairs on Xe are respectively

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

The correct answer is:
3, 2, 1

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In $\mathrm{XeF}_2, \mathrm{XeF}_4, \mathrm{XeF}_6$ the numebr of lone pairs of $\mathrm{Xe}$ are respectively

Options:
A
2, 3, 1

B
1, 2, 3

C
4, 1, 2

D
3, 2, 1

The correct answer is:
3, 2, 1

Solution:
$\begin{array}{lll}\mathrm{XeF}_2 & \mathrm{sp}^3 \mathrm{~d} & 3 \text { lone pairs } \\ \mathrm{XeF}_4 & \mathrm{sp}^3 \mathrm{~d}^2 & 2 \text { lone pairs } \\ \mathrm{XeF}_6 & \mathrm{sp}^3 \mathrm{~d}^3 & \text { 1 lone pair }\end{array}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In $\mathrm{XeF}_2, \mathrm{XeF}_4, \mathrm{XeF}_6$ the numebr of lone pairs of $\mathrm{Xe}$ are respectively

Options:
A
2, 3, 1

B
1, 2, 3

C
4, 1, 2

D
3, 2, 1

The correct answer is:
3, 2, 1

Solution:
$\begin{array}{lll}\mathrm{XeF}_2 & \mathrm{sp}^3 \mathrm{~d} & 3 \text { lone pairs } \\ \mathrm{XeF}_4 & \mathrm{sp}^3 \mathrm{~d}^2 & 2 \text { lone pairs } \\ \mathrm{XeF}_6 & \mathrm{sp}^3 \mathrm{~d}^3 & \text { 1 lone pair }\end{array}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2002
In $\mathrm{XeO}_3 \mathrm{~F}_2$, the number of bond pair(s), $\pi$-bond(s) and lone pair(s) on Xe atom respectively are:

Options:
A
5,3,0

B
$5,2,0$

C
4,2,2

D
4, 4, 0

The correct answer is:
5,3,0

Solution:
Structure of $\mathrm{XeO}_3 \mathrm{~F}_2$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
In YDSE a parallel beam of incident light consists of two wavelengths Î»1=4000 Aâ andÎ»2=5600 Aâ. The minimum distance y on the screen, measured from the central axis, where the bright fringe due to two wavelengths coincide is n Î» 1 D d . Find n.

Options:

The correct answer is:
7

Solution:
n 1 X ¯ 1 = n 2 X ¯ 2
n 1 D λ 1 d = n 2 D λ 2 d
n 1 λ 1 = n 1 λ 2
λ 1 λ 2 = n 2 n 1 = 4000 5600 = 40 58 = 5 7
∴ n 2 n 1 = 5 7 ⇒ X 1 X 2 = 7 5
∴ y = 7 X ¯ 1 = 7 D λ 1 d

Topic: Physics, Wave Optics, JEE Main
In YDSE apparatus as shown in figure wavelength of light used in Î». The screen is moved away from the source with a constant speed V. Initial distance between screen and plane of slits was D. Suppose P is the point where 5th order maxima was lying at t=0, then after how much time third-order maxima will lie at this point P on that screen?

Options:
A 2DV
B 2D3V
C 3D2V
D 3DV

The correct answer is:
2D3V

Solution:
At t=0, Δx=5λ=ydD .....(i)
After t times Δx1=3λ=ydD+vt ......(ii)
Dividing (i) and (ii) we get
53=D+vtD
5D=3D+3vt
2D=3vt ⇒ t=2D3v

Topic: Physics, Wave Optics, JEE Main
In YDSE arrangement as shown in the figure, fringes are seen on screen using monochromatic source S having wavelength 3000 Aâ (in the air). S1 and S2 are two slits separated by d=1 mm and D=1 m . Left of slits S1 and S2 medium of refractive index n1=2 is present and to the right of S1 and S2 medium of n2=32 , is present. A thin slab of thickness 't' is placed in front of S1 . The refractive index of n3 of the slab varies with distance from it's starting face as shown in figure.

In order to get central maxima at the centre of the screen, the thickness of the slab (in Î¼m) required is:

Options:

The correct answer is:
2

Solution:
Path difference,
∆x=n1 SS2+n2S2P-n1SS1+n2S1P-∫0tn3-n2dx
=n1 SS2-SS1+n2 S2P-S1P-∫0tn3dx-n2t
In order to get central maxima at the centre of the screen
o=2×1×10-322×1+0-2t+3t2
0.5t=1 μm
t=2 μm

Topic: Physics, Wave Optics, JEE Main
In YDSE, the slits have different widths. As a result, the amplitude of waves from two slits is A and 2A respectively. If I0 be the maximum intensity of the interference pattern then the intensity of the pattern at a point where the phase difference between waves is Ï is given by I0P5+4cosâ¡Ï . Where P is in integer. Find the value of P?

Options:
A 9
B 3
C 6
D 1

The correct answer is:
9

Solution:
As the amplitudes are A and 2A then the ratio of intensities is 1:4
Imax=I0=I1+I2+2I1×I2=I+4I+2×2I
I0=9I⇒I=I09
Intensity at any point:
I'=I1+I2+2I1×I2cosϕ
I'=I+4I+2I×4Icosϕ
I'=5I+4Icosϕ ; I'=I5+4cosϕ
I'=I095+4cosϕ=I0P5+4cosϕ
P=9

Topic: Physics, Wave Optics, JEE Main
In YDSE, the wavelength used is \( 600 \mathrm{~nm} \). A transparent slice of thickness \( 36 \mu \mathrm{m} \) is placed in the path of one wave. The central fringe shifts to \( 30^{\text {th }} \) bright fringe from the centre, then find the refractive index of the slice.

Options:
A \( 1.5 \)
B \( 1.8 \)
C \( 2.0 \)
D None of these

The correct answer is:
\( 1.5 \)

Solution:
Given that the wavelength, thickness of the slab and the shift of the central fringe by the number of fringe widths are,

λ=600 nm, t=36 μm, n=30.

If the central bright fringe is shifted by 30β, then at the centre point of the screen, 30th bright fringe will have a path difference of 30λ. Here, β= Fringe width.

For a bright fringe at the centre of the screen, path difference is, nλ=μ-1t.

30600×10-9=μ-136×10-6

μ-1=12 ⇒μ=32=1.5

Topic: Physics, Wave Optics, JEE Main
In Yong's double slit experiment, in interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent source is ' $\mathrm{d}$ ' and ' $\mathrm{D}$ ' is the distance between the source and screen. The possible wavelength used are inversely proportional to

Options:
A $\mathrm{D}, 5 \mathrm{D}, 9 \mathrm{D} \ldots$
B $\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D} \ldots$
C $3\mathrm{D}, 4 \mathrm{D}, 5 \mathrm{D} \ldots$
D $3\mathrm{D}, 7 \mathrm{D}, 10 \mathrm{D} \ldots$

The correct answer is:
$\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D} \ldots$

Solution:
If $x$ is the fringe width, then three will be a minimum in front of the slit if
$$
\begin{aligned}
& \frac{\mathrm{d}}{2}=\frac{\mathrm{x}}{2}, \frac{3 \mathrm{x}}{2}, \frac{5 \mathrm{x}}{2}, \ldots \ldots \\
& \text { or } \mathrm{d}=\mathrm{x}, 3 \mathrm{x}, 5 \mathrm{x} \ldots . \\
& \therefore \mathrm{x}=\mathrm{d}, \frac{\mathrm{d}}{3}, \frac{\mathrm{d}}{5}, \ldots \ldots \\
& \because \lambda=\frac{\mathrm{xd}}{\mathrm{D}} \\
& \therefore \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}, \frac{\mathrm{d}^2}{3 \mathrm{D}}, \frac{\mathrm{d}^2}{5 \mathrm{D}}
\end{aligned}
$$
$\therefore \lambda$ is inversely proportional to $\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D} \ldots$

Topic: Physics, Wave Optics, JEE Main
In young double slit experiment, the wavelength of red light is 7 8 0 0 Ã and that of blue light is 5200 Ã The value of n for which nth bright band due to red light coincides with (n + 1)th bright band due to blue light, is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
For bands to coincide
n λ 1 D d = n + 1 λ 2 D d
⇒ n 7 8 0 0 = n + 1 5 2 0 0
⇒ 7 8 n = n + 1 5 2
⇒ 2 6 n = 5 2
⇒ n = 2

Topic: Physics, Wave Optics, JEE Main
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Ao is :-

Options:
A 1178 × 10–9 m
B 1178 × 10–6 m
C 1178 × 10–12 m
D 5890 × 10–7 m

The correct answer is:
1178 × 10–6 m

Solution:
β=λDd=5890×10-10×0.50.5×10-3
=589×10-6 m
Distance between first and third bright fringe is 2β=2×589×10-6 m
=1178×10-6 m

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (18 Mar Shift 1)
In Young's double slit experiment, a mica slit of thickness $t$ and refractive index $\mu$ is introduced in the ray from the first source $S_1$. By how much distance the fringes pattern will be displaced

Options:
A $\frac{d}{D}(\mu-1) t$
B $\frac{D}{d}(\mu-1) t$
C $\frac{d}{(\mu-1) D}$
D $\frac{D}{d}(\mu-1)$

The correct answer is:
$\frac{D}{d}(\mu-1) t$

Solution:
$\frac{D}{d}(\mu-1) t$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, amplitude of the coherent sources are in the ratio \( 3: 1 \), then ratio of maximum to minimum intensity of the fringe system will be

Options:
A \( 3: 1 \)
B \( 1: 3 \)
C \( 9: 1 \)
D \( 4: 1 \)

The correct answer is:
\( 4: 1 \)

Solution:
Given ab=31
∴Imax.Imin.=a+b2a-b2=3+123-12=164=41

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, angular width of fringes is $0.20^{\circ}$ for sodium light of wavelength $5890 Å Å$. If complete system is dipped in water, then angular width of fringes becomes

Options:
A $0.11^{\circ}$
B $0.15^{\circ}$
C $0.22^{\circ}$
D $0.30^{\circ}$

The correct answer is:
$0.15^{\circ}$

Solution:
Angular fringe width $\theta=\frac{\lambda}{d} \Rightarrow \theta \propto \lambda$
$\lambda_w=\frac{\lambda_a}{\mu_w}$
So
$\theta_w=\frac{\theta_{\text {air }}}{\mu_{\mathrm{w}}} \quad=\frac{0.20}{\frac{4}{3}}=0.15^{\circ}$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, carried out with light of wavelength $5000 \AA$, the distance between the slits is $0.3 \mathrm{~mm}$ and the screen is at $200 \mathrm{~cm}$ from the slits. The central maximum is at $x=0 \mathrm{~cm}$. The value of $x$ for third maxima is _____$\mathrm{mm}$.

Options:

The correct answer is:
10

Solution:
$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}=\frac{10 \times 10^{-3}}{3} \mathrm{~m}$
For $3^{\text {rd }}$ maxima $y_3=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
In Young's double slit experiment, first slit has width four times the width of the second slit. the ratio of the maximum intensity to the minimum intensity in the interference fringe system is :

Options:
A $2: 1$
B $4: 1$
C $9: 1$
D $8: 1$

The correct answer is:
$9: 1$

Solution:
Let intensity of light coming from each slit of a coherent source is $I$.
As first slit has width 4 times the width of the second slit, so
$I_1=4 I \text { and } I_2=I$
$\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}$
$=\frac{(\sqrt{4 I}+\sqrt{I})^2}{(\sqrt{4 I}-\sqrt{I})^2}=\frac{9}{1}$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?

Options:
A Reducing the separation between the slits.
B Using blue light instead of green light.
C Using red light instead of green light.
D Moving the screen away from the slits.

The correct answer is:
Using blue light instead of green light.

Solution:
Fringe width $\propto \lambda$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, how many maxima can be obtained on a screen (including central maxima), if d = 5 Î» 2 (where, Î» is the wavelength of light)?

Options:

The correct answer is:
5

Solution:
In YDSE, path difference, Î x = d sin Î¸
For maxima, Î x = n Î» , where n = 0 , Â± 1 , Â± 2 ....
5 Î» 2 sin Î¸ = n Î» (given that d = 5 Î» 2 )
âsinÎ¸= 2n 5 .........(i)
Since, â1â¤sinÎ¸â¤1 From (i) ââ1â¤ 2n/5 â¤1

So, - 5 2 â¤ n â¤ 5 2
So, possible values of n are -2, -1, 0, 1, 2 Thus a total of 5 maximas will be obtained.

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, how many maxima can be obtained on a screen (including central maxima), if d = 5 Î» 2 (where Î» is the wavelength of light)?

Options:
A 5
B 4
C 7
D 1

The correct answer is:
5

Solution:
In YDSE, path difference, Î x = d sin Î¸
For maxima, Î x = n Î» , where n = 0 , Â± 1 , Â± 2 ....
5 Î» 2 sin Î¸ = n Î» (given that d = 5 Î» 2 )

âsinÎ¸= 2n 5 .........(i)
Since, â1â¤sinÎ¸â¤1 Form (i) ââ1â¤2n/5â¤1
So, - 5 2 â¤ n â¤ 5 2
So, possible values of n are -2, -1, 0, 1, 2 . Thus a total of 5 maxima will be obtained.

Topic: Physics, Wave Optics, JEE Main
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 $1 / 9$
C 1,4
D $1 / 3$

The correct answer is:
1,4

Solution:
Amplitude of the superimposing waves are
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}}=\left(\frac{1}{9}\right)^{1 / 2}=\frac{1}{3} \\
\frac{I_{\operatorname{minima}}}{I_{\operatorname{maxima}}}=\frac{\left(a_{1}-a_{2}\right)^{2}}{\left(a_{1}+a_{2}\right)^{2}}=\frac{1}{4}
\end{array}
$$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, if the source of light changes from orange to blue then:

Options:
A the central bright fringe will become a dark fringe.
B the distance between consecutive fringes will decrease.
C the distance between consecutive fringes will increase.
D the intensity of the minima will increase.

The correct answer is:
the distance between consecutive fringes will decrease.

Solution:
Fringe width =λDd.
As λ decreases, fringe width also decreases.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7λ4. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

Options:
A 12
B 34
C 13
D 14

The correct answer is:
12

Solution:
Given the path difference is ∆x=7λ4

Hence, the phase difference can be found as follows:

ϕ=2πλ∆x=2πλ×7λ4=7π2

The formula to calculate the intensity is given by

I=Imaxcos2ϕ2 ...1

From equation (1), it follows that

IImax=cos2ϕ2=cos27π2×2=cos27π4=cos22π-π4=cos2π4=12

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
In Young's double slit experiment, monochromatic light of wavelength 5000 A∘ is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______ ×10-6 m.

Options:

The correct answer is:
125

Solution:
Let intensity of light on screen due to each slit is I0.

So internity at centre of screen is 4I0(as cosϕ=1).

Intensity at distance y from centre,

I=I0+I0+2I0I0cosϕ

As we know, Imax=4I0. Therefore,

⇒Imax2=2I0=2I0+2I0cosϕ

⇒cosϕ=0

⇒ϕ=π2

Hence, kΔx=π2

⇒2πλdsinθ=π2

⇒2λd×yD=12

⇒y=λD4 d=5×10-7×14×10-3

=125×10-6

=125

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (01 Feb Shift 2)