2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
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.
Correct Answer: 450
Explanation:
$y = {d \over 2}$,
$\therefore$ $\Delta x = y{d \over D}$
$ \Rightarrow {{{d^2}} \over {2D}} = {\lambda \over 2}$
$ \Rightarrow \lambda = {{{{(0.6 \times {{10}^{ - 3}})}^2}} \over {0.8}}$
= 450 nm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Morning Shift
Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ___________ $\times$ 10$-$5 m.
Correct Answer: 3
Explanation:
Condition for diffraction maximum is $a \sin \theta=(2 n+1) \frac{\pi}{2}$
For first Maxima, $n=1, a \sin \theta=\frac{3 \lambda}{2}$
Since $\theta$ is very less, so
$ \begin{aligned} & \sin \theta \approx \tan \theta \\\\ & \therefore a \tan \theta=\frac{3 \lambda}{2} \end{aligned} $
$ \begin{aligned} a\left(\frac{y}{D}\right) =\frac{3 \lambda}{2} \\\\ \Rightarrow y =\frac{3 \lambda \mathrm{D}}{2 a} \end{aligned} $
For the two cases given
$ \begin{aligned} y_2-y_1 & =\frac{3 \mathrm{D}}{2 a}\left(\lambda_2-\lambda_1\right) \\\\ & =\frac{3}{2} \times \frac{2}{0 \cdot 5 \times 10^{-3}}(655-650) \times 10^{-9} \\\\ & =6 \times 10^3 \times 5 \times 10^{-9} \\\\ & =3 \times 10^{-5} \mathrm{~m} \end{aligned} $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?
A.
I0
B.
2 I0
C.
5 I0
D.
4 I0
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
In Young's double slit experiment, if the source of light changes from orange to blue then :
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 increases.
D.
the intensity of the minima will increase.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d0 + a0 sin$\omega$t; where d0, $\omega$ and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :
d(t) = d0 + a0 sin$\omega$t; where d0, $\omega$ and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :
A.
${{2\lambda D({d_0})} \over {(d_0^2 - a_0^2)}}$
B.
${{2\lambda D{a_0}} \over {(d_0^2 - a_0^2)}}$
C.
${{\lambda D} \over {d_0^2}}{a_0}$
D.
${{\lambda D} \over {{d_0} + {a_0}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 $\mathop A\limits^o $ is observed at 5896 $\mathop A\limits^o $ ?
A.
306 km/sec
B.
322 km/sec
C.
296 km/sec
D.
336 km/sec
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
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 $\mathop A\limits^o $ is :-
A.
1178 $\times$ 10$-$9 m
B.
1178 $\times$ 10$-$6 m
C.
1178 $\times$ 10$-$12 m
D.
5890 $\times$ 10$-$7 m
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :
A.
0.50 mm
B.
0.25 mm
C.
1 mm
D.
0.75 mm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1 $\mu$m. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that:
A.
its size increases, but intensity decreases
B.
its size increases, and intensity increases
C.
its size decreases, but intensity increases
D.
its sizes decreases, and intensity decreases
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio ${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}$ will be :
A.
${{2\sqrt {2x} } \over {2x + 1}}$
B.
${{2\sqrt {2x} } \over {x + 1}}$
C.
${{\sqrt {2x} } \over {x + 1}}$
D.
${{\sqrt {2x} } \over {2x + 1}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
If the source of light used in a Young's double slit experiment is changed from red to violet :
A.
the fringes will become brighter.
B.
the intensity of minima will increase.
C.
consecutive fringe lines will come closer.
D.
the central bright fringe will become a dark fringe.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
A.
4 : 1
B.
2 : 1
C.
1 : 4
D.
3 : 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x : 4 where x is ____________.
Correct Answer: 1
Explanation:
Given, b1 = 3b2
Here, b1 = width of the one of the two slit and b2 = width of the other slit.
As we know that,
Intensity, I $\propto$ (Amplitude)2
$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$
${I_1} = 9{I_2}$
As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,
${{{I_{\min }}} \over {{I_{\max }}}} = {\left( {{{\sqrt {{I_1}} - \sqrt {{I_2}} } \over {\sqrt {{I_1}} + \sqrt {{I_2}} }}} \right)^2}$
Substituting the values in the above equations, we get
${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$
${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$
Comparing with, ${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$
The value of x = 1.
Here, b1 = width of the one of the two slit and b2 = width of the other slit.
As we know that,
Intensity, I $\propto$ (Amplitude)2
$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$
${I_1} = 9{I_2}$
As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,
${{{I_{\min }}} \over {{I_{\max }}}} = {\left( {{{\sqrt {{I_1}} - \sqrt {{I_2}} } \over {\sqrt {{I_1}} + \sqrt {{I_2}} }}} \right)^2}$
Substituting the values in the above equations, we get
${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$
${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$
Comparing with, ${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$
The value of x = 1.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4 cm. The frequency of light used is ______________ $\times$ 1014 Hz.
Correct Answer: 5
Explanation:
8$\beta$ = 2.4 cm
${{8\lambda \Delta } \over d}$ = 2.4 cm
${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$
f = 5 $\times$ 1014 Hz
${{8\lambda \Delta } \over d}$ = 2.4 cm
${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$
f = 5 $\times$ 1014 Hz
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
A source of light is placed in front of a screen. Intensity of light on the screen is I. Two Polaroids P1 and P2 are so placed in between the source of light and screen that the intensity of light on screen is I/2. P2 should be rotated by an angle of (degrees) so that the intensity of light on the screen becomes ${{3I} \over 8}$.
Correct Answer: 30
Explanation:
$I = {{{I_0}} \over 2}{\cos ^2}\phi $
${{{I_0}} \over 2}{\cos ^2}\phi = {{3I} \over 8}$
${\cos ^2}\phi = {3 \over 4}$
${\cos ^2}\phi = {{\sqrt 3 } \over 2}$
$ \Rightarrow \phi = 30$
${{{I_0}} \over 2}{\cos ^2}\phi = {{3I} \over 8}$
${\cos ^2}\phi = {3 \over 4}$
${\cos ^2}\phi = {{\sqrt 3 } \over 2}$
$ \Rightarrow \phi = 30$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. the difference in wavelengths of red and violet light is ................ nm.
Correct Answer: 300
Explanation:
Position of bright fringe y = n${{D\lambda } \over d}$
y1 of red = ${{D{\lambda _r}} \over d}$ = 3.5 mm
$\lambda$r = 3.5 $\times$ 10$-$3 ${d \over D}$
Similarly, $\lambda$v = 2 $\times$ 10$-$3 ${d \over D}$
${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$
= 3 $\times$ 10$-$7 = 300 nm
y1 of red = ${{D{\lambda _r}} \over d}$ = 3.5 mm
$\lambda$r = 3.5 $\times$ 10$-$3 ${d \over D}$
Similarly, $\lambda$v = 2 $\times$ 10$-$3 ${d \over D}$
${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$
= 3 $\times$ 10$-$7 = 300 nm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is ___________ mm. [Refractive index of air = 1.0003, wavelength of yellow light in vacuum = 6000 $\mathop A\limits^o $]
Correct Answer: 2
Explanation:
Thickness t = n$\lambda$
So, n $\lambda$vac = (n + 1) $\lambda$air
n $\lambda$ = (n + 1) ${\lambda \over {{\mu _{air}}}}$
n = ${1 \over {{\mu _{air}} - 1}} = {{{{10}^4}} \over 3}$
t = n$\lambda$
$ = {{{{10}^4}} \over 3} \times 6000\mathop A\limits^o $
= 2 mm
So, n $\lambda$vac = (n + 1) $\lambda$air
n $\lambda$ = (n + 1) ${\lambda \over {{\mu _{air}}}}$
n = ${1 \over {{\mu _{air}} - 1}} = {{{{10}^4}} \over 3}$
t = n$\lambda$
$ = {{{{10}^4}} \over 3} \times 6000\mathop A\limits^o $
= 2 mm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
A galaxy is moving away from the earth at a speed of 286 kms$-$1. The shift in the wavelength of a redline at 630 nm is x $\times$ 10$-$10 m. The value of x, to the nearest integer, is ____________. [Take the value of speed of light c, as 3 $\times$ 108 ms$-$1]
Correct Answer: 6
Explanation:
From Doppler effect, we know
$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$
here, $v = 286 \times {10^3}$ m/s
$c = 3 \times {10^8}$ m/s
$ \therefore $ $\lambda ' - \lambda = {{\lambda v} \over c}$
$ = {{630 \times {{10}^{ - 9}} \times 286 \times {{10}^3}} \over {3 \times {{10}^8}}}$
= 6 $\times$ 10$-$10 m
$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$
here, $v = 286 \times {10^3}$ m/s
$c = 3 \times {10^8}$ m/s
$ \therefore $ $\lambda ' - \lambda = {{\lambda v} \over c}$
$ = {{630 \times {{10}^{ - 9}} \times 286 \times {{10}^3}} \over {3 \times {{10}^8}}}$
= 6 $\times$ 10$-$10 m
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.
Correct Answer: 600
Explanation:
$\beta$ = 6 mm, d = 1 mm, D = 10 m
$\lambda$ = ?
We know, $\beta = {{\lambda D} \over d}$
6 $\times$ 10$-$3 = ${{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}$
$ \therefore $ $\lambda$ = ${{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}$
$\lambda$ = 600 $\times$ 10$-$9 m
$ \therefore $ $\lambda$ = 600 nm
$\lambda$ = ?
We know, $\beta = {{\lambda D} \over d}$
6 $\times$ 10$-$3 = ${{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}$
$ \therefore $ $\lambda$ = ${{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}$
$\lambda$ = 600 $\times$ 10$-$9 m
$ \therefore $ $\lambda$ = 600 nm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
An unpolarised light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30$^\circ$ in clockwise direction, the intensity of emerging light will be _________ Lumens.
Correct Answer: 75
Explanation:
Given, I0 = 100 lumens
When analyser is rotated through an angle $\theta$, the intensity of light will becomes
I = I0 cos2$\theta$ = 100 $\times$ cos230$^\circ$
= 100 $\times$ ${\left( {{{\sqrt 3 } \over 2}} \right)^2}$ = 75 lumens
When analyser is rotated through an angle $\theta$, the intensity of light will becomes
I = I0 cos2$\theta$ = 100 $\times$ cos230$^\circ$
= 100 $\times$ ${\left( {{{\sqrt 3 } \over 2}} \right)^2}$ = 75 lumens
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
In the figure below, P and Q are two equally intense coherent sources emitting radiation of
wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q
by 90o. A, B and C are three distinct points of observation, each equidistant from the midpoint of
PQ. The intensities of radiation at A, B, C will be in the ratio :
A.
4 : 1 : 0
B.
2 : 1 : 0
C.
0 : 1 : 2
D.
0 : 1 : 4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Two coherent sources of sound, S1 and S2,
produce sound waves of the same wavelength,
$\lambda $ = 1 m, in phase. S1 and S2 are placed 1.5 m
apart (see fig). A listener, located at L, directly
in front of S2 finds that the intensity is at a
minimum
when he is 2 m away from S2. The listener moves away from S1, keeping his distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then, d is :
when he is 2 m away from S2. The listener moves away from S1, keeping his distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then, d is :
A.
12 m
B.
2 m
C.
3 m
D.
5 m
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
A beam of plane polarised light of large
cross-sectional area and uniform intensity
of 3.3 Wm-2 falls normally on a
polariser (cross sectional area 3 $ \times $ 10-4 m2) which
rotates about its axis with an angular speed
of 31.4 rad/s. The energy of light passing through
the polariser per revolution, is close to :
cross-sectional area and uniform intensity
of 3.3 Wm-2 falls normally on a
polariser (cross sectional area 3 $ \times $ 10-4 m2) which
rotates about its axis with an angular speed
of 31.4 rad/s. The energy of light passing through
the polariser per revolution, is close to :
A.
1.0 $ \times $ 10-5 J
B.
1.0 $ \times $ 10-4 J
C.
1.5 $ \times $ 10-4 J
D.
5.0 $ \times $ 10-4 J
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Two light waves having the same wavelength $\lambda $ in vacuum are in phase initially. Then the first wave
travels a path L1
through a medium of refractive index n1
while the second wave travels a path of length
L2
through a medium of refractive index n2
. After this the phase difference between the two waves is :
A.
${{2\pi } \over \lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)$
B.
${{2\pi } \over \lambda }\left( {{n_2}{L_1} - {n_1}{L_2}} \right)$
C.
${{2\pi } \over \lambda }\left( {{{{L_1}} \over {{n_1}}} - {{{L_2}} \over {{n_2}}}} \right)$
D.
${{2\pi } \over \lambda }\left( {{{{L_2}} \over {{n_1}}} - {{{L_1}} \over {{n_2}}}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
In a Young’s double slit experiment, light of
500 nm is used to produce an interference
pattern. When the distance between the slits
is 0.05 mm, the angular width (in degree) of
the fringes formed on the distance screen is
close to
A.
0.17o
B.
1.7o
C.
0.57o
D.
0.07o
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
In a Young’s double slit experiment, 16 fringes
are observed in a certain segment of the
screen when light of a wavelength 700 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 would be
A.
28
B.
24
C.
30
D.
18
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Interference fringes are observed on a screen
by illuminating two thin slits 1 mm apart with a
light source ($\lambda $ = 632.8 nm). The distance
between the screen and the slits is 100 cm. If
a bright fringe is observed on a screen at a
distance of 1.27 mm from the central bright
fringe, then the path difference between the
waves, which are reaching this point from the
slits is close is
A.
1.27 $\mu $m
B.
2.05 $\mu $m
C.
2.87 nm
D.
2 nm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
In a double slit experiment, at a certain point
on the screen the path difference between the
two interfering waves is ${1 \over 8}$th of a wavelength.
The ratio of the intensity of light at that point
to that at the centre of a bright fringe is :
A.
0.853
B.
0.568
C.
0.672
D.
0.760
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment,
a source of light of wavelengh 589 nm is used and the interference pattern is observed on a
screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :
A.
4.9 mm
B.
5.9 mm
C.
6.9 mm
D.
3.9 mm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Visible light of wavelength 6000 $ \times $ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at $\theta $1, then $\theta $1, is close to :
A.
45o
B.
30o
C.
25o
D.
20o
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be
zero, is :
A.
71.6o
B.
90o
C.
18.4o
D.
45o
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
A Young's double-slit experiment is performed using monochromatic light of wavelength $\lambda $. The
intensity of light at a point on the screen, where the path difference is $\lambda $, is K units. The intensity
of light at a point where the path difference is ${\lambda \over 6}$ is given by ${{nK} \over {12}}$, where n is an integer. The value
of n is __________.
Correct Answer: 9
Explanation:
From 1st case,
$\Delta $$\phi $ = ${{2\pi } \over \lambda } \times \lambda $ = 2$\pi $
$ \therefore $ Inet = 4Icos2 ${{\Delta \phi } \over 2}$ = 4I = K (Given)
From 2nd case,
$\Delta $$\phi $ = ${{2\pi } \over \lambda } \times {\lambda \over 6}$ = ${\pi \over 3}$
$ \therefore $ Inet = 4I cos2 ${{\Delta \phi } \over 2}$
= 4I $ \times $ ${3 \over 4}$ = ${3 \over 4}K$
$ \therefore $ According to question,
${{nK} \over {12}}$ = ${3 \over 4}K$
$ \Rightarrow $ n = 9
$\Delta $$\phi $ = ${{2\pi } \over \lambda } \times \lambda $ = 2$\pi $
$ \therefore $ Inet = 4Icos2 ${{\Delta \phi } \over 2}$ = 4I = K (Given)
From 2nd case,
$\Delta $$\phi $ = ${{2\pi } \over \lambda } \times {\lambda \over 6}$ = ${\pi \over 3}$
$ \therefore $ Inet = 4I cos2 ${{\Delta \phi } \over 2}$
= 4I $ \times $ ${3 \over 4}$ = ${3 \over 4}K$
$ \therefore $ According to question,
${{nK} \over {12}}$ = ${3 \over 4}K$
$ \Rightarrow $ n = 9
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
Orange light of wavelength 6000 $ \times $ 10–10 m illuminates a single
slit of width 0.6 $ \times $ 10–4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is ___________.
slit of width 0.6 $ \times $ 10–4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is ___________.
Correct Answer: 200
Explanation:
For minima
$d\,\sin \theta = n\lambda $
or $\sin \theta = {{n\lambda } \over d}$
$ \because $ maximum value of sin$\theta $ is 1
$ \therefore $ ${{n\lambda } \over d} \le 1$
$n \le {d \over \lambda }$
$n \le {{0.6 \times {{10}^{ - 4}}} \over {6000 \times {{10}^{ - 10}}}}$
$n \le 100$
For both sides 100 + 100 = 200
$d\,\sin \theta = n\lambda $
or $\sin \theta = {{n\lambda } \over d}$
$ \because $ maximum value of sin$\theta $ is 1
$ \therefore $ ${{n\lambda } \over d} \le 1$
$n \le {d \over \lambda }$
$n \le {{0.6 \times {{10}^{ - 4}}} \over {6000 \times {{10}^{ - 10}}}}$
$n \le 100$
For both sides 100 + 100 = 200
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
In a Young's double slit experiment 15 fringes
are observed on a small portion of the screen
when light of wavelength 500 nm is used. Ten
fringes are observed on the same section of
the screen when another light source of
wavelength $\lambda $ is used. Then the value of $\lambda $ is
(in nm) __________.
Correct Answer: 750
Explanation:
The length of the screen used portion for 15
fringes, and also for ten fringes
15 $ \times $ 500 $ \times $ ${D \over \lambda }$ = 10 $ \times $ ${{\lambda D} \over \lambda }$
$ \Rightarrow $ 15 × 50 =$\lambda $
$ \Rightarrow $ $\lambda $ = 750 nm
15 $ \times $ 500 $ \times $ ${D \over \lambda }$ = 10 $ \times $ ${{\lambda D} \over \lambda }$
$ \Rightarrow $ 15 × 50 =$\lambda $
$ \Rightarrow $ $\lambda $ = 750 nm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1.
The pass axis of P2 is inclined at 60o to the pass axis of P3. When a beam of unpolarized light of intensity I0 is
incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (${{{I_0}} \over I}$) equals (nearly) :
A.
10.67
B.
5.33
C.
16.00
D.
1.80
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 $\mathop A\limits^o $
is used, the minimum separation between two points, to be seen as distinct, will be :
A.
0.12 $\mu $m
B.
0.38 $\mu $m
C.
0.24 $\mu $m
D.
0.48 $\mu $m
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
In a double slit experiment, when a thin film of thickness t having refractive index $\mu $. is introduced in front of
one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is
($\lambda $ is the wavelength of the light used) :
A.
${\lambda \over {2\left( {\mu - 1} \right)}}$
B.
${\lambda \over {\left( {2\mu - 1} \right)}}$
C.
${{2\lambda } \over {\left( {\mu - 1} \right)}}$
D.
${\lambda \over {\left( {\mu - 1} \right)}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
In a Young's double slit experiment, the ratio of the slit's width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :
A.
25 : 9
B.
4 : 1
C.
${\left( {\sqrt 3 + 1} \right)^4}:16$
D.
9 : 1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Diameter of the objective lens of a telescope is
250 cm. For light of wavelength 600nm.
coming from a distant object, the limit of
resolution of the telescope is close to :-
A.
3.0 × 10–7 rad
B.
4.5 × 10–7 rad
C.
1.5 × 10–7 rad
D.
2.0 × 10–7 rad
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The figure shows a Young's double slit
experimental setup. It is observed that when a
thin transparent sheet of thickness t and
refractive index μ is put in front of one of the
slits, the central maximum gest shifted by a
distance equal to n fringe widths. If the
wavelength of light used is $\lambda $, t will be :
A.
${{D\lambda } \over {a\left( {\mu - 1} \right)}}$
B.
${{2nD\lambda } \over {a\left( {\mu - 1} \right)}}$
C.
${{2D\lambda } \over {a\left( {\mu - 1} \right)}}$
D.
${{n\lambda } \over {\left( {\mu - 1} \right)}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
Calculate the limit of resolution of a telescope
objective having a diameter of 200 cm, if it has
to detect light of wavelength 500 nm coming
from a star :-
A.
610 × 10–9 radian
B.
457.5 × 10–9 radian
C.
305 × 10–9 radian
D.
152.5 × 10–9 radian
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
In an interference experiment the ratio of amplitudes of coherent waves is ${{{a_1}} \over {{a_2}}} = {1 \over 3}$ . The
ratio of maximum and minimum intensities of
fringes will be :
A.
2
B.
4
C.
18
D.
9
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
In a double-slit experiment, green light (5303$\mathop A\limits^ \circ $) falls on a double slit having a separation of 19.44 $\mu $m and awidht of 4.05 $\mu $m. The number of bright fringes between the first and the second diffraction minima is :
A.
04
B.
05
C.
10
D.
09
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is ${1 \over 8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
A.
0.94
B.
0.85
C.
0.74
D.
0.80
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength $\lambda $ such that the first minima occurs directly in front of the slit (S1) ?
A.
${\lambda \over {2\left( {5 - \sqrt 2 } \right)}}$
B.
${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$
C.
${\lambda \over {\left( {5 - \sqrt 2 } \right)}}$
D.
${\lambda \over {\left( {\sqrt 5 - 2} \right)}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle ${1 \over {40}}$ by using light of wavelength $\lambda $1. When the light of wavelength $\lambda $2 is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda $1 and $\lambda $2 are in visible range (380 nm to 740 nm), their values are -
A.
400 nm, 500 nm
B.
625 nm, 500 nm
C.
380 nm, 500 nm
D.
380 nm, 525 nm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength $\lambda $ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range $-$ 30o$ \le $$\theta $$ \le $30o is :
A.
640
B.
320
C.
321
D.
641
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :
A.
16 : 9
B.
25 : 9
C.
4 : 1
D.
5 : 3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
Consider a tank made of glass(refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index $\mu $. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of $\mu $ is :
A.
$\sqrt {{5 \over 3}} $
B.
${3 \over {\sqrt 5 }}$
C.
${5 \over {\sqrt 3 }}$
D.
${4 \over 3}$
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $\theta $. Then :
A.
cos$\theta $ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$
B.
cos$\theta $ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$
C.
cos$\theta $ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$
D.
cos$\theta $ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$
