In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $2: 1$. The ratio of the maximum to minimum intensity in the interference pattern is:
The ratio of intensities at two points $\mathrm{P}$ and $\mathrm{Q}$ on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are $\pi / 3$ and $\pi / 2$, respectively are
The width of fringe is $2 \mathrm{~mm}$ on the screen in a double slits experiment for the light of wavelength of $400 \mathrm{~nm}$. The width of the fringe for the light of wavelength 600 $\mathrm{nm}$ will be:
'$n$' polarizing sheets are arranged such that each makes an angle $45^{\circ}$ with the preceeding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be $I / 64$. The value of $n$ will be:
Two polaroide $\mathrm{A}$ and $\mathrm{B}$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $\mathrm{C}$ is placed between $\mathrm{A}$ and $\mathrm{B}$ bisecting angle between them. If intensity of unpolarized light is $\mathrm{I}_{0}$ then intensity of transmitted light after passing through polaroid $\mathrm{B}$ will be:
In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800 \mathrm{~nm}$. The line joining $A_{1} P$ is perpendicular to $A_{1} A_{2}$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation 'a' will be:
The distance of screen from slits D = 5 cm
In Young's double slits experiment, the position of 5$\mathrm{^{th}}$ 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 :
Given below are two statements :
Statement I : If the Brewster's angle for the light propagating from air to glass is $\mathrm{\theta_B}$, then the Brewster's angle for the light propagating from glass to air is $\frac{\pi}{2}-\theta_B$
Statement II : The Brewster's angle for the light propagating from glass to air is ${\tan ^{ - 1}}({\mu _\mathrm{g}})$ where $\mathrm{\mu_g}$ is the refractive index of glass.
In the light of the above statements, choose the correct answer from the options given below :
Unpolarised light of intensity 32 Wm$^{-2}$ passes through the combination of three polaroids such that the pass axis of the last polaroid is perpendicular to that of the pass axis of first polaroid. If intensity of emerging light is 3 Wm$^{-2}$, then the angle between pass axis of first two polaroids is ______________ $^\circ$.
Explanation:
When dealing with three polaroids, the intensity after the second polaroid will be given by Malus' law as $I_0 \cos^2 \theta$, where $\theta$ is the angle between the pass axes of the first two polaroids. However, as the third polaroid is orthogonal to the first, no light from the first polaroid passes through, only light from the second polaroid. So the final intensity is also modulated by a $\sin^2 \theta$ term (as the second and third polaroids are orthogonal).
So if we set up the equation for the final intensity $I_{\text{net}}$:
$I_{\text{net}} = I_0 \cos^2 \theta \sin^2 \theta$
And we substitute the given values $I_{\text{net}} = 3 \, \text{W/m}^2$ and $I_0 = \frac{32 \, \text{W/m}^2}{2} = 16 \, \text{W/m}^2$:
$3 = 16 \cos^2 \theta \sin^2 \theta$
This simplifies to:
$\frac{3}{16} = \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2$
Taking the square root of both sides gives:
$\frac{\sqrt{3}}{2} = \left|\sin 2\theta\right|$
The solutions for this are $\theta = 30^\circ$ and $\theta = 60^\circ$.
So, the angle between the pass axes of the first two polaroids is either $30^\circ$ or $60^\circ$.
A beam of light consisting of two wavelengths $7000~\mathop A\limits^o $ and $5500~\mathop A\limits^o $ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is $2.5 \mathrm{~mm}$ and the distance between the plane of slits and the screen is $150 \mathrm{~cm}$. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is $n \times 10^{-5} \mathrm{~m}$. The value of $n$ is __________.
Explanation:
In Young's double slit experiment, we have two slits separated by a distance $d$, and a screen placed at a distance $L$ from the slits. When light with a single wavelength $\lambda$ passes through the slits, an interference pattern is formed on the screen with bright and dark fringes.
In this problem, we have a beam of light consisting of two wavelengths $\lambda_1 = 7000~\mathop A\limits^o$ and $\lambda_2 = 5500~\mathop A\limits^o$. We are given the distance between the slits $d = 2.5 \mathrm{~mm}$ and the distance between the plane of the slits and the screen $L = 150 \mathrm{~cm}$. Our goal is to find the least distance from the central fringe where the bright fringes due to both wavelengths coincide.
First, let's convert the wavelengths and distances to meters:
$\lambda_1 = 7 \times 10^{-7} \mathrm{~m}$ $\lambda_2 = 5.5 \times 10^{-7} \mathrm{~m}$ $d = 2.5 \times 10^{-3} \mathrm{~m}$ $L = 1.5 \mathrm{~m}$
The fringe width $\beta$ for a single wavelength is given by:
$\beta = \frac{\lambda D}{d}$
Now, let's consider the condition for the bright fringes due to both wavelengths to coincide. Let the $n^{\text{th}}$ bright fringe of $\lambda_1$ match with the $m^{\text{th}}$ bright fringe of $\lambda_2$. In this case, we have:
$n \beta_1 = m \beta_2$
Substituting the expression for fringe width, we get:
$n \frac{\lambda_1 L}{d} = m \frac{\lambda_2 L}{d}$
Simplifying, we get:
$\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{11}{14}$
The least values of $n$ and $m$ that satisfy this condition are $n = 11$ and $m = 14$.
Now, let's find the position $y$ of the coincident bright fringe on the screen:
$y = n \beta_1 = n \frac{\lambda_1 L}{d} = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m}}$
$y = k \times 10^{-5} \mathrm{~m}$
Calculating the value of $k$:
$k = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m} \times 10^{-5} \mathrm{~m}} = 462$
So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $462 \times 10^{-5} \mathrm{~m}$.
As shown in the figure, in Young's double slit experiment, a thin plate of thickness $t=10 \mu \mathrm{m}$ and refractive index $\mu=1.2$ is inserted infront of slit $S_{1}$. The experiment is conducted in air $(\mu=1)$ and uses a monochromatic light of wavelength $\lambda=500 \mathrm{~nm}$. Due to the insertion of the plate, central maxima is shifted by a distance of $x \beta_{0} . \beta_{0}$ is the fringe-width befor the insertion of the plate. The value of the $x$ is _____________.
Explanation:
$ \begin{aligned} & \mu=1.2 \\\\ & \lambda=500 \times 10^{-9} \mathrm{~m} \end{aligned} $
When the glass slab inserted infront of one slit then the shift of central fringe is obtained by
$ \mathrm{t}=\frac{\mathrm{n} \lambda}{(\mu-1)} $
$ \Rightarrow \quad 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{(1.2-1)} $
$ \Rightarrow $ $ 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{0.2} $
$ \Rightarrow $ $ \mathrm{n}=4 $
Explanation:
$\omega_{1}=16 \mathrm{~mm} $ and $ \omega_{2}=12 \mathrm{~mm}$
So $\operatorname{LCM}\left(\omega_{1}, \omega_{2}\right)=48 \mathrm{~mm}$
So at $48 \mathrm{~mm}$ distance both bright fringes will be found.
In a Young's double slit experiment, the intensities at two points, for the path differences $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda$ being the wavelength of light used) are $I_{1}$ and $I_{2}$ respectively. If $I_{0}$ denotes the intensity produced by each one of the individual slits, then $\frac{I_{1}+I_{2}}{I_{0}}=$ __________.
Explanation:
$I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)$
so ${I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)$
${I_1} = 2{I_0}$
& ${I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)$
${I_2} = {I_0}$
So ${{{I_1} + {I_2}} \over {{I_0}}} = 3$
In Young's double slit experiment, two slits $S_{1}$ and $S_{2}$ are '$d$' distance apart and the separation from slits to screen is $\mathrm{D}$ (as shown in figure). Now if two transparent slabs of equal thickness $0.1 \mathrm{~mm}$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $(\lambda=4000$ $\mathop A\limits^o $) from $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ respectively. The central bright fringe spot will shift by ___________ number of fringes.
Explanation:
Path difference introduced by two slabs $=(\mu_2-\mu_1)t$
$\Rightarrow$ Number of shifts $ = {{({\mu _2} - {\mu _1})t} \over \lambda }$
$ = {{0.04 \times 0.1\,\mathrm{mm}} \over {4000\,\mathop A\limits^o }}$
$ = {{4 \times {{10}^{ - 2}} \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 7}}}} = 10$
Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2.8 (medium $-1$) and 6.8 (medium $-2$), respectively. To satisfy the condition, so that the reflected and refracted rays are perpendicular to each other, the angle of incidence should be ${\tan ^{ - 1}}{\left( {1 + {{10} \over \theta }} \right)^{{1 \over 2}}}$ the value of $\theta$ is __________.
(Given for dielectric media, $\mu_r=1$)
Explanation:
We know that
$\tan {\theta _0} = {{{\mu _2}} \over {{\mu _1}}}$
$\tan {\theta _0} = \sqrt {{{6.8} \over {2.8}}} = \sqrt {{{17} \over 7}} $
${\theta _0} = {\tan ^{ - 1}}\sqrt {1 + {{10} \over 7}} \Rightarrow \theta = 7$
As shown in the figure, three identical polaroids P$_1$, P$_2$ and P$_3$ are placed one after another. The pass axis of P$_2$ and P$_3$ are inclined at angle of 60$^\circ$ and 90$^\circ$ with respect to axis of P$_1$. The source S has an intensity of 256 $\frac{W}{m^2}$. The intensity of light at point O is ____________ $\frac{W}{m^2}$.
Explanation:
Intensity out $=\frac{256}{2} \times \frac{1}{4} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{256 \times 3}{2 \times 4 \times 4}=\mathbf{2 4}$
A point source of light is placed at the focus of a concave mirror. Consider only paraxial rays. The shapes of the wavefronts of incident and reflected lights respectively are
spherical, spherical
spherical, planar
spherical, cylindrical
planar, spherical
Two slits separated by a distance of 1 mm are illuminated with light of wavelength $6.5 \times 10^{-7} \mathrm{~m}$. The interference fringes are observed on a screen placed at 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to
0.655 mm
1.625 mm
3.125 mm
4.785 mm
Two slits are made one millimetre apart and the screen is placed one metre away from the slits. The fringe width when light of wavelength 500 nm is used is
5 m
0.5 mm
0.5 m
5 cm
If the slit width is 2 mm and wavelength of light used is $4000\mathop {\rm{A}}\limits^{\rm{o}}$, then Fresnel distance is nearly
An unpolarised light beam of intensity $2 I_{0}$ is passed through a polaroid P and then through another polaroid Q which is oriented in such a way that its passing axis makes an angle of $30^{\circ}$ relative to that of P. The intensity of the emergent light is
Two coherent sources of light interfere. The intensity ratio of two sources is $1: 4$. For this interference pattern if the value of $\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$ is equal to $\frac{2 \alpha+1}{\beta+3}$, then $\frac{\alpha}{\beta}$ will be :
In Young's double slit experiment, the fringe width is $12 \mathrm{~mm}$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$, then the fringe width becomes (in mm):
Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)
Using Young's double slit experiment, a monochromatic light of wavelength 5000 $\mathop A\limits^o $ produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 $\mathop A\limits^o $ is used and the separation between the slits is doubled, then the new fringe width will be :
In Young's double slit experiment performed using a monochromatic light of wavelength $\lambda$, when a glass plate ($\mu$ = 1.5) of thickness x$\lambda$ 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 :
For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is :
The interference pattern is obtained with two coherent light sources of intensity ratio 4 : 1. And the ratio ${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}}$ is ${5 \over x}$. Then, the value of x will be equal to :
A light whose electric field vectors are completely removed by using a good polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism.
The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi$/2 at point P and $\pi$ at point Q. Then the difference between the resultant intensities at P and Q will be :
Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be :
Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is $\pi$. The difference of resultant intensities, at the point A and B, will be _________ I.
Explanation:
${I_A} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} = 25I$
${I_B} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} = I$
So, ${I_A} - {I_B} = 24I$
In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is __________ nm.
Explanation:
$\lambda = 560 \times {10^{ - 9}}$
${B_1} = 7.2 \times {10^{ - 3}}$
${B_2} = 8.1 \times {10^{ - 3}}$
${{{B_1}} \over {{B_2}}} = {{{\lambda _1}} \over {{\lambda _2}}}$
$ \Rightarrow {\lambda _2} = {{560 \times {{10}^{ - 9}} \times 8.1 \times {{10}^{ - 3}}} \over {7.2 \times {{10}^{ - 3}}}}$
$ = 6.3 \times {10^{ - 7}}$ m
$ = 630$ nm
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are $\pi / 2$ and $\pi / 3$ at points $\mathrm{A}$ and $\mathrm{B}$ respectively. The difference between the resultant intensities at the two points is $x I$. The value of $x$ will be ________.
Explanation:
${I_{{R_1}}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
${I_A} = I + 4I + 2\sqrt {I.4I} \cos 90^\circ $
$ = 5I$
${I_B} = I + 4I + 2\sqrt {I.4I} \cos 60^\circ $
$ = 7I$
${I_B} - {I_A} = 2I$
In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 $\times$ 10$-$2 m towards the slits, the change in fringe width is 3 $\times$ 10$-$3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ____________ nm.
Explanation:
Fringe width $\beta = {{\lambda D} \over d}$
$ \Rightarrow \left| {d\beta } \right| = {\lambda \over d}\left| {d(D)} \right|$
$ \Rightarrow 3 \times {10^{ - 3}}\,cm = {\lambda \over {1\,mm}}\left( {5 \times {{10}^{ - 2}}\,m} \right)$
$ \Rightarrow \lambda = {{3 \times {{10}^{ - 8}}} \over {5 \times {{10}^{ - 2}}}}\,m$
$ \Rightarrow \lambda = 600\,nm$
In a Young's double slit experiment, an angular width of the fringe is 0.35$^\circ$ on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is ${1 \over \alpha }$. The value of $\alpha$ is ___________.
Explanation:
Angular fringe width $\theta = {\lambda \over D}$
So ${{{\theta _1}} \over {{\lambda _1}}} = {{{\theta _2}} \over {{\lambda _2}}}$
${\theta _2} = {{0.35^\circ } \over {450\,nm}} \times {{450\,nm} \over {7/5}} = 0.25^\circ = {1 \over 4}$
So $\alpha = 4$
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.
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
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.
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} $
A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index $n_{2}$. The other slit is at the interface of this medium with another medium 1 of refractive index $n_{1}\left(\neq n_{2}\right)$. The line joining the slits is perpendicular to the interface and the distance between the slits is $d$. The slit widths are much smaller than $d$. A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle $\theta$ from the line joining them, so that $\theta$ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.
Which of the following statement(s) is(are) correct?
In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become $\_\_\_\_$ times the original value
2
4
8
16
In an interference pattern of Young's double slit experiment, at a point we observe the 12 th order maximum for a monochromatic light source with wavelength $6000 \mathop {\rm{A}}\limits^{\rm{o}} $. What order will be visible here, if the source is replaced by a light of wavelength $4800 \mathop {\rm{A}}\limits^{\rm{o}} $ ?
15
10
8
18
In a double slit experiment performed in air, the angular width of a fringe is found to be $0.15^{\circ}$ on a screen placed 80 cm away. The wavelength of light is used 490 nm . The angular width of the fringe, if the entire apparatus is immersed in a medium of refractive index $\frac{5}{3}$ is
$0.09^{\circ}$
$0.7^{\circ}$
$0.9^{\circ}$
$0.11^{\circ}$
In Young's double slit experiment for what order does the wavelength of red light $(\lambda=780 \mathrm{~nm})$ coincide with $(n+1)$ th order of blue light $(\lambda=520 \mathrm{~nm})$ ?
1
2
3
4
The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen 1 m away the wavelength of light used is 600 nm . The change in angular width of the fringe, if the entire measurement system is immersed in water is [use refractive index of water as $4 / 3$ ]
$0.05^{\circ}$
$0.10^{\circ}$
$0.15^{\circ}$
$0.20^{\circ}$
A Young's double slit experiment apparatus has slits separated by 0.2 mm and a screen 60 cm away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $11 / 9$ and the slits are illuminated with green light ( $\lambda=550$ nm in vacuum). Find the fringe width of the pattern formed on the screen.
0.95 mm
1.25 mm
1.35 mm
1.45 mm
Young's double slit experiment is conducted with monochromatic light of wavelength 5000$\mathop A\limits^o $, with slit separation of 3 mm and observer at 20 cm away from the slits. If a 1 mm transparent plate is placed infront of one of the slits, the fringes shift by 6 mm . The refractive index of the transparent plate is