datavorous/entrance-exam-dataset huggingface dataset:
In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If $I_m$ be the maximum intensity, the resultant intensity I when they interfere at phase difference $\phi$ is given by

Options:
A
$\frac{I_m}{9}(4+5 \cos \phi)$

B
$\frac{\mathrm{l}_{\mathrm{m}}}{3}\left(1+2 \cos ^2 \frac{\phi}{2}\right)$

C
$\frac{I_m}{5}\left(1+4 \cos ^2 \frac{\phi}{2}\right)$

D
$\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$

The correct answer is:
$\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$

Solution:
Let $A_1=A_0, A_2=2 A_0$
If amplitude of resultant wave is $A$ then
$\mathrm{A}^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2 \cos \phi$
For maximum intensity,
$\mathrm{A}_{\max }^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2$
$\therefore\frac{A^2}{A_{\max }^2}=\frac{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}{A_1^2+A_2^2+2 A_1 A_2}$
$=\frac{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right) \cos \phi}{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right)}$
$\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{m}}}=\frac{5+4 \cos \phi}{9}=\frac{1+8 \cos ^2(\phi / 2)}{9}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (Offline)
In young's double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate μ=1.5 of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be

Options:
A 3
B 2
C 1.5
D 0.5

The correct answer is:
2

Solution:
Since the intensity at the central maxima remains unchanged, the change in the path difference should be an integral multiple of the wavelength. Therefore,

Δx=μ-1t=nλ⇒1.5-1xλ=nλ⇒x=n0.5=2n

For n=1

x=2

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (28 Jun Shift 2)
In Young's double slit experiment, point $A$ on the screen has a path difference of $\lambda$ and point $B$ on the screen has a path difference of $\frac{\lambda}{4}$. What is the ratio of the intensities at point $A$ to $B$ ?

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

The correct answer is:
2 : 1

Solution:
In YDSE, let intensities of two identical coherent sources $S_1$ and $S_2$ is $I_0$. Then, resultant intensity,
$$
I=2 I_0 \cos ^2 \frac{\phi}{2}
$$
where, $\phi=$ phase difference.
When path difference is $\lambda$, then
$$
\phi_1=\frac{2 \pi}{\lambda} \text { (path difference) and } \phi_1=\frac{2 \pi}{\lambda} \lambda=2 \pi
$$
When path difference is $\frac{\lambda}{4}$, then
$$
\begin{aligned}
\phi_2 & =\frac{2 \pi}{\lambda} \frac{\lambda}{4}=\frac{\pi}{2} \\
\therefore \quad \frac{I_1}{I_2} & =\frac{2 I_0 \cos ^2 \frac{\phi_1}{2}}{2 I_0 \cos ^2 \frac{\phi_2}{2}}=\frac{\cos ^2\left(\frac{2 \pi}{2}\right)}{\cos ^2\left(\frac{\pi}{4}\right)}=\frac{\cos ^2(\pi)}{\cos ^2 \frac{\pi}{4}} \\
& =\frac{1}{(1 / 2)}=\frac{2}{1}
\end{aligned}
$$
Hence, $I_1: I_2=2: 1$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, the $6^{\text {th }}$ maximum with wavelength ${ }^{\prime} \lambda_{1}{ }^{\prime}$ is at a distance ${ }^{\prime} \mathrm{d}_{1}{ }^{\prime}$ from the central maximum and the $4^{\text {th }}$ maximum with wavelength $\lambda_{2}$
is at distance $\mathrm{d}_{2}$. Then $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}$ is

Options:
A $\frac{2}{3} \frac{\lambda_{1}}{\lambda_{2}}$
B $\frac{3}{2} \frac{\lambda_{1}}{\lambda_{2}}$
C $\frac{2}{3} \frac{\lambda_{2}}{\lambda_{1}}$
D $\frac{3}{2} \frac{\lambda_{2}}{\lambda_{1}}$

The correct answer is:
$\frac{3}{2} \frac{\lambda_{1}}{\lambda_{2}}$

Solution:
$\mathrm{d}_{1}=\frac{6 \lambda_{1} \mathrm{D}}{\mathrm{d}} \quad$ and $\quad \mathrm{d}_{2}=\frac{4 \lambda_{2} \mathrm{D}}{\mathrm{d}}$
$\therefore \frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{3}{2} \frac{\lambda_{1}}{\lambda_{2}}$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, the distance between sources is $1 \mathrm{~mm}$ and distance between the screen and source is $1 \mathrm{~m}$. If the fringe width on the screen is $0.06 \mathrm{~cm}$, then $\lambda=$

Options:
A $6000 Å$
B $4000 Å$
C $1200 Å$
D $2400 Å$

The correct answer is:
$6000 Å$

Solution:
$\beta=\frac{\lambda D}{d} \Rightarrow\left(0.06 \times 10^{-2}\right)=\frac{\lambda \times 1}{1 \times 10^{-3}} \Rightarrow \lambda=6000 Å$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits reduced to half. As a result the fringe width:

Options:
A is halved
B becomes four times
C remains unchanged
D is doubled

The correct answer is:
becomes four times

Solution:
In Young's double slit experiment fringe width
$$
\beta=\frac{\lambda D}{d}
$$
where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow$ $\frac{d}{2}$
Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta$
$\beta^1=4 \beta$ i.e., becomes four times

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits reduced to half. As a result the fringe width:

Options:
A is halved
B becomes four times
C remains unchanged
D is doubled

The correct answer is:
becomes four times

Solution:
In Young's double slit experiment fringe width
$$
\beta=\frac{\lambda D}{d}
$$
where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow$ $\frac{d}{2}$
Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta$
$\beta^1=4 \beta$ i.e., becomes four times

Topic: Physics, Wave Optics, NEET, NEET 2013 (Karnataka)
In Young's double slit experiment, the distance between the slits is $3 \mathrm{~mm}$ and the slits are $2 \mathrm{~m}$ away from the screen. Two interference patterns can be obtained on
the screen due to light of wavelength $480 \mathrm{~nm}$ and $600 \mathrm{~nm}$ respectively. The
separation on the screen between the $5^{\text {th }}$ order bright fringes on the two
interference patterns is

Options:
A $6 \times 10^{-4} \mathrm{~m}$
B $8 \times 10^{-4} \mathrm{~m}$
C $12 \times 10^{-4} \mathrm{~m}$
D $4 \times 10^{-4} \mathrm{~m}$

The correct answer is:
$4 \times 10^{-4} \mathrm{~m}$

Solution:
$\mathrm{d}=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$
$D=1 \mathrm{~m} \quad \lambda_{1}=480 \mathrm{~nm} \quad \lambda_{2}=600 \mathrm{~nm}$
${ }^{1} \mathrm{y}_{5}=5 \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}$
${ }^{2} \mathrm{y}_{5}=5 \frac{\lambda_{2} \mathrm{D}}{\mathrm{d}}$
${ }^{2} y_{5}-{ }^{1} y_{5}=\frac{5 D}{d}(600-480) n m$
$=\frac{5 \times 2}{3 \times 10^{-3}} \times 120$
$=10 \times 10^{-3+1} \times 4=4 \times 10^{5-9}$
$=4 \times 10^{-4}=0.4 \mathrm{~mm}$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, the distance between the two slits is $0.1 \mathrm{~mm}$ and the wavelength of light used is $4 \times 10^{-7} \mathrm{~m}$. If the width of the fringe on the screen is $4 \mathrm{~mm}$, the distance between screen and slit is

Options:
A $0.1 \mathrm{~mm}$
B $1 \mathrm{~cm}$
C $0.1 \mathrm{~cm}$
D $1 \mathrm{~m}$

The correct answer is:
$1 \mathrm{~m}$

Solution:
$\beta=\frac{\lambda D}{d} \Rightarrow\left(4 \times 10^{-3}\right)=\frac{4 \times 10^{-7} \times D}{0.1 \times 10^{-3}} \Rightarrow D=1 \mathrm{~m}$

Topic: Physics, Wave Optics, JEE Main
In young's double slit experiment, the fringe width is 12 mm. If the entire arrangement is placed in water of refractive index 43, then the fringe width becomes (in mm)

Options:
A 16
B 9
C 48
D 12

The correct answer is:
9

Solution:
For a given light wavelength corresponding a medium of refractive index μ

λmed =λvacumm μ.

And we know that fringe width β=λD d

Therefore, βmed =βvacuum μ=1243=9 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
In Young's double slit experiment, the intensity of light at a point on the screen is ${ }^{\prime} \mathrm{K}^{\prime}$ unit for path difference ${ }^{\prime} \lambda^{\prime}$. What would be the intensity at a point if path
difference is $\frac{\lambda^{\prime}}{4}$ ?

Options:
A zero
B $\frac{K}{4}$
C $\mathrm{K}$
D $\frac{K}{2}$

The correct answer is:
$\frac{K}{2}$

Solution:
When path difference is $\lambda$, phase difference $\phi$ is $2 \pi$ and when path difference is $\frac{\lambda}{4}$, the phase difference is $\frac{\pi}{2}$. If $\mathrm{I}_{0}$ is the intensity of the two waves then the resultant intensity is given by
$$
\begin{aligned}
I &=4 I_{0} \cos ^{2} \frac{\phi}{2} \\
\therefore K &=4 I_{0} \cos ^{2} \frac{2 \pi}{2}=4 I \cos ^{2} \pi=4 I_{0} \quad \because \cos \pi=-1
\end{aligned}
$$
$$
\begin{array}{l}
\text { When } \phi=\frac{\pi}{2} \\
\qquad \begin{aligned}
\mathrm{I}=4 \mathrm{I}_{0} \cos ^{2} \cdot \frac{\pi}{4} & \\
&=4 \mathrm{I}_{0} \frac{1}{2}=\frac{\mathrm{K}}{2} \quad \because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}
\end{aligned}
\end{array}
$$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, the intensity on the screen at a point where path difference \( \lambda \) is \( \mathrm{K} \). What will be the intensity at the point where path difference is \( \frac{\lambda}{4} \) :

Options:
A \( \frac{K}{4} \)
B \( \frac{K}{2} \)
C \( \mathrm{K} \)
D Zero

The correct answer is:
\( \frac{K}{2} \)

Solution:
By using phase difference ϕ=2πλΔ
For path difference λ ,
Phase difference ϕ1=2π and for path difference λ4 , phase difference ϕ2=π2 .
Also by using I=4I0cos2⁡ϕ2
⇒I1I2=cos2⁡ϕ12 cos2⁡ϕ22
⇒KI2=cos2⁡2π2 cos2⁡π2=112
⇒I2=K2

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment, the screen is kept 1.6 m from the slits. The coherent sources are 0.032 cm apart and fringes are observed on the screen. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 1.06 cm from the central fringe. The wavelength of the light used is

Options:
A 530 nm
B 265Â nm
C 1060Â nm
D 132.5Â nm

The correct answer is:
530 nm

Solution:
For maxima Îx=nÎ»
dyD=nÎ»
For n=4
Î»=dy4D
=0.032Ã10-2Ã1.06Ã10-24Ã1.6
=530 nm

Topic: Physics, Wave Optics, JEE Main
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, JEE Main
In Young's double slit experiment, the slit separation is made threefold by keeping remaining setup same, the fringe width becomes

Options:
A $\frac{1}{9}$ times
B $\frac{1}{3}$ times
C 9 times
D 3 times

The correct answer is:
$\frac{1}{3}$ times

Solution:
Fringe width $\beta$, in terms of slit separation $d$ and screen distance $D$ is given by,
$\beta=\frac{\lambda D}{d}$
If $d$ is made three times, fringe width becomes $\frac{1}{3}$ times

Topic: Physics, Wave Optics, MHT CET, MHT CET 2022 (11 Aug Shift 1)
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, JEE Main
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be _____ nm.

Options:

The correct answer is:
450

Solution:
According to the question, the first dark fringe is observed directly opposite to one of the slits. Therefore, half of the fringe width is equal to half of the distance between the slits.

β2=d2⇒β=d=6 mm

⇒λDd=0.6×10-3⇒ λ=0.6×10-3dD⇒ λ=0.6×10-3×0.6×10-380×10-2

⇒λ=450×10-9 m=450 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (27 Jun Shift 1)
In Young's double slit experiment, the two slits are $d$ distance apart. Interference pattern is observed on the screen at a distance $D$ from the slits. Fist dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of the light is.

Options:
A $\frac{D^2}{2 d}$
B $\frac{d^2}{D}$
C $\frac{d^2}{2 D}$
D $\frac{D^2}{d}$

The correct answer is:
$\frac{d^2}{D}$

Solution:
Consider the following diagram:
The dark fringe is produced at a point just opposite to the slit $S_1$, i.e.,
$y_1=\frac{d}{2}$
For dark fringe, path difference is odd multiple of half of the wavelength:
$\Delta x=(2 n-1) \frac{\lambda}{2}$
From the figure, relation between fringe location and path difference can be obtained as:
$$
\begin{aligned}
& \tan \theta=\frac{y}{D} \approx \sin \theta=\frac{\Delta x}{d} \\
& \therefore y=\frac{\Delta x D}{a}=(2 n-1)\left(\frac{\lambda}{2}\right) \cdot\left(\frac{D}{a}\right)=(2 n-1) \frac{\lambda D}{2 d}
\end{aligned}
where $n=1,2,3, \ldots \ldots \ldots$
$\begin{aligned} & \Rightarrow \frac{d}{2}=(2 n-1) \frac{\lambda D}{2 d} \\ & \Rightarrow \lambda=\frac{d^2}{(2 n-1) D}\end{aligned}$
For $n=1$,
$$
\Rightarrow \lambda=\frac{d^2}{D}
$$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit experiment when a glass plate of refractive index 1.44 is introduced in the path of one of the interfering beams, the fringes are displaced by a distance ' $y$ '. If this plate is replaced by another plate of same thickness but of refractive index 1.66, the fringes will be displaced by a distance

Options:
A $\frac{3 y}{2}$
B $\frac{2 y}{3}$
C $\frac{5 y}{4}$
D $\frac{4 y}{5}$

The correct answer is:
$\frac{3 y}{2}$

Solution:
As a glass plate is used in one of the paths,
$$
\begin{aligned}
& \mathrm{y}_1=\frac{\beta}{\lambda}(1.44-1) \mathrm{t} \\
& \mathrm{y}_1=0.44 \mathrm{t} \times \frac{\beta}{\lambda}
\end{aligned}
$$
New displacement is:
$$
\begin{aligned}
\mathrm{y}_2 & =\frac{\beta}{\lambda}(1.66-1) \mathrm{t} \\
\mathrm{y}_2 & =0.66 \mathrm{t} \times \frac{\beta}{\lambda} \\
\frac{\mathrm{y}_2}{\mathrm{y}_1} & =\frac{0.66}{0.44} \\
\therefore \quad \mathrm{y}_2 & =\frac{3 \mathrm{y}}{2}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main
In young's double slit experiment with a source of light of wavelength $6320 \hat{A}$, the first maxima will occur when

Options:
A Path difference is $9480 \tilde{A}$
B Phase difference is $\pi$ radian
C Path difference is $6320 Å$
D Phase difference is $\pi$ radian

The correct answers are:
Path difference is $6320 Å$

Solution:
For maxima, path difference $\Delta=n \lambda$
So for $n=1, \Delta=\lambda=6320 Å$

Topic: Physics, Wave Optics, JEE Main
In Young's double slit interference experiment, the slit widths are in the ratio $1: 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

Options:
A
$3: 2$

B
$1: 25$

C
$9: 4$

D
$1: 5$

The correct answer is:
$9: 4$

Solution:
We know that,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{\omega_1}{\omega_2}+1}\right)^2}{\left(\sqrt{\frac{\omega_1}{\omega_2}-1}\right)^2}
$$
$I_{\max }$ and $I_{\min }$ are maximum and minimum intensity
$\omega_1$ and $\omega_2$ are widths of two slits
$$
\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{1}{25}}+1\right)^2}{\left(\sqrt{\frac{1}{25}}-1\right)^2}\left(\frac{\omega_1}{\omega_2}=\frac{1}{25} \text { given }\right)
$$
On solving we get,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9: 4
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (26 May Online)
In Young's double slit interference experiment the wavelength of light used is $6000 Å$. If the path difference between waves reaching a point $P$ on the screen is 1.5 microns, then at that point $P$.

Options:
A Second bright band occurs
B Second dark band occurs
C Third dark band occurs
D Third bright band occurs

The correct answer is:
Third dark band occurs

Solution:
$$
\begin{aligned}
\lambda & =6000 Å, \Delta x=1.5 \mu \\
& =6 \times 10^{-7} \mathrm{~m},=1.5 \times 10^{-6} \mathrm{~m}
\end{aligned}
$$
Path difference for maxima
$$
\begin{aligned}
\Delta x & =n \lambda \\
1.5 \times 10^{-6} & =n \times 6 \times 10^{-7} \\
n & =\frac{1.5 \times 10^{-6}}{6 \times 10^{-7}}=\frac{15}{6}=2.5=\frac{5}{2}
\end{aligned}
$$
Value of $n$ is always an integer for maxima, so point $P$ is not a maxima.
$\therefore$ For maxima
$$
\begin{aligned}
\Delta x & =(2 n+1) \frac{\lambda}{2} \\
1.5 \times 10^{-6} & =(2 n+1) \times \frac{6 \times 10^{-7}}{2}
\end{aligned}
$$
$$
\begin{aligned}
(2 n+1) & =\frac{1.5 \times 10^{-6} \times 2}{6 \times 10^{-7}} \\
2 n & =5-1 \\
2 n & =4 \Rightarrow n=2
\end{aligned}
$$
So, at $P$ third dark band occurs.

Topic: Physics, Wave Optics, JEE Main
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:

Options:
A 60 μm
B 48 μm
C 12 μm
D 36 μm

The correct answer is:
60 μm

Solution:
Given here,

D=1 m, λ=600×10-9 m and n=5

Distance of nth bright fringe is given by, yn=nλDd, where, d is separation between the slits.

⇒5×600×10-9×1d=5×10-2

⇒d=5×600×10-9×15×10-2=60×10-6 m

⇒d=60 μm.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
In Young's double-slit experiment, 12 fringes are obtained on a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed on the same segment of the screen is given by

Options:
A 18
B 24
C 30
D 36

The correct answer is:
18

Solution:
If on a certain segment of the screen, n1 fringes of wavelength λ1 are present and n2 fringes of wavelength λ2 are present, then

n1β1=n2β2

⇒n1λ1=n2λ2
∴n2=n1 λ1λ2=12×600400=18

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment, both the slits produce equal intensities on a screen. A 100% transparent thin film of refractive index Î¼=1.5 is kept in front of one of the slits, due to which the intensity at the point O on the screen becomes 75% of its initial value. If the wavelength of monochromatic light is 720 nm, then what is the minimum thickness (in nm) of the film?

Options:

The correct answer is:
240

Solution:
The path difference at the point O is

Δx=μ-1t

I=4l0cos2⁡ϕ2

0.754I0=4I0cos2⁡ϕ2

⇒cosϕ2=±32

⇒ϕmin=60°

ϕmin=2πλΔx=2πλμ-1t=π3

t=λ6μ-1=λ061.5-1=240 nm

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment intensity at a point is (3/4)th of the maximum intensity. The possible angular position of this point is

Options:
A sin -1 λ 3d
B sin -1 λ 2d
C sin-1Î»6d
D sin -1 λ 4d

The correct answer is:
sin-1λ6d

Solution:
I = I max cos 2 ϕ 2
∴ 3I max 4 = I max cos 2 ϕ 2
cosϕ2=32
or, ϕ2=π6
∴ ϕ=π3=2πλΔx .....(i)
where, Δx=d sinθ
Substituting in Eq.(i) we get
sinθ=λ6d
or, θ=sin-1λ6d

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment interference bands are produced on the screen placed at 1.5 m from two slits 0.15 mm apart and illuminated by the light of wavelength 6000 â«. If the screen is now taken away from the slits by 50 cm then find the change in fringe width (in mm)?

Options:

The correct answer is:
2

Solution:
D2-D1=50 cm ; D1=1.5 m=150 cm
Î»=6000 â«=6000Ã10-8 cm ; d=0.15 mm
Change in fringe width âÎ²=Î²2-Î²1
âÎ²=D2-D1Î»d=50Ã6000Ã10-80.15Ã10-1
âÎ²=20Ã10-410-3=2Ã10-1 cm=2mm
âÎ²=2 mm

Topic: Physics, Wave Optics, JEE Main
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, JEE Main
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, JEE Main, JEE Main 2016 (09 Apr Online)
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, JEE Main
In Young's double-slit experiment, the distance between slits is d = 0.25 cm and the distance of the screen D = 120 cm from the slits. If the wavelength of light used is Î»=6000 AÂ° and I0 is the intensity of central maximum, then the minimum distance of the point from the centre, where the intensity is I02 is kÃ10-5 m. What is the value of k?

Options:

The correct answer is:
7.2

Solution:
I=I02=I0cos2ϕ2
cos ϕ 2 = 1 2
ϕ 2 = π 4
ϕ=π2
2 π λ y D d = π 2
y = λ D 4 d = 6 0 0 0 × 1 0 - 10 × 1.2 4 × 0.25 × 1 0 - 2
=7 · 2 × 1 0 - 5 m

Topic: Physics, Wave Optics, JEE Main
In Youngâs double-slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is :

Options:
A 3
B 6
C 24
D 12

The correct answer is:
12

Solution:
d = separation between 2 slits
a = slit width
β y = fringe width in YDSE & β D = width of central maxima
βy=λDd=λD6.1aβD=2λDa
β D =2×6.1  β y
=12.2  β y

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (19 Apr Online)