A beam of unpolarised light of intensity $I_0$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$. The intensity of emergent light is:
The diffraction pattern of a light of wavelength $400 \mathrm{~nm}$ diffracting from a slit of width $0.2 \mathrm{~mm}$ is focused on the focal plane of a convex lens of focal length $100 \mathrm{~cm}$. The width of the $1^{\text {st }}$ secondary maxima will be :
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 $7 \lambda / 4$. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
Monochromatic light of wavelength $500 \mathrm{~nm}$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $=1.5$), the central maximum is shifted to a position previously occupied by the $4^{\text {th }}$ bright fringe. The thickness of the glass-plate is __________ $\mu \mathrm{m}$.
Explanation:
To solve this problem, we need to understand how the interference pattern shifts due to the introduction of a thin glass plate in Young's double slit experiment. This shift occurs because the light passing through the glass experiences a different optical path length compared to the light passing through the other slit without glass.
### Step-by-Step Analysis:
- Calculate the Path Difference Due to the Glass Plate:
- The path difference is influenced by the optical thickness of the glass, which is the product of the physical thickness $ t $ of the glass and the refractive index $ n $ minus the path it would have in air (i.e., $ n \times t - t $).
- The effective additional path difference in the medium of the glass plate is $ (n-1) \times t $.
- Determine the Shift in Fringes:
- The shift of the central maximum to the position previously occupied by the fourth bright fringe indicates that the optical path difference created by the glass plate corresponds to four fringe spacings.
- Each fringe width corresponds to a change in path difference of one wavelength ($ \lambda $).
- Calculating the Thickness $ t $ of the Glass Plate:
- Since the central maximum shifts by four fringes, the path difference $ \Delta $ caused by the glass plate must be equal to $ 4 \lambda $.
- Therefore, $ (n-1) \times t = 4 \lambda $.
- Insert Values and Solve for $ t $:
- Given:
- $ n = 1.5 $ (refractive index of the glass)
- $ \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} $ (wavelength of the light)
- $ (1.5 - 1) \times t = 4 \times 500 \times 10^{-9} \, \text{m} $
- $ 0.5 \times t = 2000 \times 10^{-9} \, \text{m} $
- $ t = \frac{2000 \times 10^{-9} \, \text{m}}{0.5} $
- $ t = 4000 \times 10^{-9} \, \text{m} $
Conversion to Micrometers:
- $ 4000 \times 10^{-9} \, \text{m} $ equals $ 4000 \, \text{nm} $
- Since $ 1 \, \mu\text{m} = 1000 \, \text{nm} $,
- $ t = 4 \, \mu\text{m} $.
Conclusion:
The thickness of the glass plate required to shift the central maximum to the position previously occupied by the fourth bright fringe is $ \mathbf{4 \, \mu\text{m}} $.
In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, the minimum distance of the point from the central maximum is _________ $\mu \mathrm{m}$. (Given : $\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$)
Explanation:
In a Young's double slit experiment, the intensity at a point can be expressed as a function of the phase difference between the light arriving from the two slits. The intensity at any point on the screen is given by:
$I = I_{\text{max}} \cos^2 \left( \frac{\delta}{2} \right)$
Where:
- $I_{\text{max}}$ is the maximum intensity.
- $\delta$ is the phase difference between the light waves from the two slits.
Given the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, we can write:
$\frac{I}{I_{\text{max}}} = \frac{1}{4}$
Substituting this into the intensity equation:
$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} \right)$
Taking the square root of both sides, we get:
$\cos \left( \frac{\delta}{2} \right) = \frac{1}{2}$
The possible solutions for $\delta$ are:
$\frac{\delta}{2} = \frac{\pi}{3}$ or $\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}\right)$ which gives $\delta = \frac{2\pi}{3}$ or $\delta = \frac{4\pi}{3}$.
Considering the smallest phase difference, $\delta = \frac{2\pi}{3}$, we use the relation for the phase difference due to path difference:
$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$
Thus, substituting the value of $\delta$, we have:
$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$
Solving for $\Delta x$, we get:
$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$
The minimum distance of the point from the central maximum on the screen can be found using the interference equation:
$y = \frac{\Delta x \cdot D}{d}$
Substituting the known values:
$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$
Simplifying this expression:
$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$
Hence, the minimum distance of the point from the central maximum is:
$200 \, \mu \mathrm{m}$
Two slits are $1 \mathrm{~mm}$ apart and the screen is located $1 \mathrm{~m}$ away from the slits. A light of wavelength $500 \mathrm{~nm}$ is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is __________ $\times 10^{-4} \mathrm{~m}$.
Explanation:
$\begin{aligned} & \text { Width central maxima }=10 \text { fringe width }=\frac{2 \lambda \theta}{a}=10 \beta \\ & \frac{2 \lambda \theta}{a}=10 \times \frac{\lambda D}{d} \\ & a=\frac{d}{5}=\frac{10^{-3}}{5}=2 \times 10^{-4} \end{aligned}$
A parallel beam of monochromatic light of wavelength $600 \mathrm{~nm}$ passes through single slit of $0.4 \mathrm{~mm}$ width. Angular divergence corresponding to second order minima would be _________ $\times 10^{-3} \mathrm{~rad}$.
Explanation:
$\begin{aligned} \theta & =(2)\left(\frac{2 \lambda}{a}\right) \\ & =\frac{4 \lambda}{a}=\frac{4 \times 600 \times 10^{-9}}{0.4 \times 10^{-3}} \\ & =6 \times 10^{-3} \end{aligned}$
Two coherent monochromatic light beams of intensities I and $4 \mathrm{~I}$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $x \mathrm{~I}$. The value of $x$ is __________.
Explanation:
When two coherent light beams interfere, the resulting intensity at a point depends on the principle of superposition and is related to their amplitudes. Let the amplitude of the first light beam be A, then its intensity, which is proportional to the square of its amplitude, is given as $I \propto A^2$. Since intensity is directly proportional to the square of amplitude, for the second beam with intensity $4I$, its amplitude would be $2A$, as $4I \propto (2A)^2$.
The maximum intensity ($I_{max}$) occurs when the two beams are in phase and their amplitudes add up constructively, which can be represented as:
$I_{max} \propto (A + 2A)^2 = (3A)^2 = 9A^2$
Given that $I \propto A^2$, substituting this relation to express $I_{max}$ in terms of $I$, we get:
$I_{max} = 9I$
The minimum intensity ($I_{min}$) occurs when the two beams are completely out of phase, leading their amplitudes to subtract destructively, thus
$I_{min} \propto (2A - A)^2 = A^2$
Again, using the fact that $I \propto A^2$, we find that:
$I_{min} = I$
Now, the difference between the maximum and minimum intensities in the resulting beam is:
$xI = I_{max} - I_{min} = 9I - I$
$xI = 8I$
Therefore, the value of $x$ is 8.
In a single slit experiment, a parallel beam of green light of wavelength $550 \mathrm{~nm}$ passes through a slit of width $0.20 \mathrm{~mm}$. The transmitted light is collected on a screen $100 \mathrm{~cm}$ away. The distance of first order minima from the central maximum will be $x \times 10^{-5} \mathrm{~m}$. The value of $x$ is :
Explanation:
$\begin{aligned} y & =\frac{n \lambda D}{a} \\ & =\frac{1 \times\left(550 \times 10^{-9}\right)(1)}{\left(0.2 \times 10^{-3}\right)} \\ & =275 \times 10^{-5} \mathrm{~m} \end{aligned}$
In Young's double slit experiment, carried out with light of wavelength $5000~\mathop A\limits^o$, 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 __________ $\mathrm{mm}$.
Explanation:
$\begin{aligned} x & =\frac{3 \lambda D}{d} \\ & =\frac{3 \times 5000 \times 10^{-10} \times 200 \times 10^{-2}}{0.3 \times 10^{-3}} \\ & =10 \mathrm{~mm} \end{aligned}$
Two wavelengths $\lambda_1$ and $\lambda_2$ are used in Young's double slit experiment. $\lambda_1=450 \mathrm{~nm}$ and $\lambda_2=650 \mathrm{~nm}$. The minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is $n$. The value of $n$ is _______.
Explanation:
In Young's double slit experiment, the condition for constructive interference (bright fringes) is given by:
$d \sin \theta = n \lambda$
where:
- $d$ is the distance between the slits
- $\theta$ is the angle of the fringe relative to the central maximum
- $n$ is the order of the fringe (an integer)
- $\lambda$ is the wavelength of the light
We are given two different wavelengths:
$\lambda_1 = 450 \, \text{nm}$
$\lambda_2 = 650 \, \text{nm}$
For the fringes produced by these two wavelengths to overlap, the path difference must be an integer multiple of both wavelengths. This means:
$d \sin \theta = m \lambda_1 = n \lambda_2$
where $m$ and $n$ are the orders of the fringes for $\lambda_1$ and $\lambda_2$, respectively.
To find the minimum order of fringe $n$ for $\lambda_2$ that coincides with a fringe for $\lambda_1$, we need to find the least common multiple (LCM) of these wavelengths in terms of their smallest integers. This can be formulated as:
$m \lambda_1 = n \lambda_2$
Dividing both sides by $\lambda_1$ and $\lambda_2$, we get:
$\frac{m}{\lambda_2} = \frac{n}{\lambda_1}$
Cross-multiplying, we get:
$m \lambda_1 = n \lambda_2$
Using the given wavelengths:
$m \times 450 = n \times 650$
Simplifying this equation, we get:
$\frac{m}{n} = \frac{650}{450}$
$\frac{m}{n} = \frac{13}{9}$
For the fringes to overlap, $m$ and $n$ must be integers. The smallest integers that satisfy this ratio are:
$m = 13$
$n = 9$
Therefore, the minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is:
$n = 9$
Explanation:
So internity at centre of screen is $4 \mathrm{I}_0$
Intensity at distance y from centre-
$ \begin{aligned} & I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi \\\\ & I_{\max }=4 I_0 \\\\ & \frac{I_{\max }}{2}=2 I_0=2 I_0+2 I_0 \cos \phi \end{aligned} $
$\begin{aligned} & \cos \phi=0 \\\\ & \phi=\frac{\pi}{2} \\\\ & K \Delta \mathrm{x}=\frac{\pi}{2} \\\\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2} \\\\ & \frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}\end{aligned}$
$\begin{aligned} & \mathrm{K} \Delta \mathrm{x}=\frac{\pi}{2} \\\\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2} \\\\ & \frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2} \\\\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} \\\\ & =125 \times 10^{-6} \\\\ & =125\end{aligned}$
Two waves of intensity ratio $1: 9$ cross each other at a point. The resultant intensities at that point, when (a) Waves are incoherent is $I_1$ (b) Waves are coherent is $I_2$ and differ in phase by $60^{\circ}$. If $\frac{I_1}{I_2}=\frac{10}{x}$ then $x=$ _________.
Explanation:
For incoherent wave $\mathrm{I}_1=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{I}_1=\mathrm{I}_0+9 \mathrm{I}_0$
$\mathrm{I}_1=10 \mathrm{I}_0$
For coherent wave $\mathrm{I_2=I_A+I_B+2 \sqrt{I_A I_B} \cos 60^{\circ}}$
$\begin{aligned} & \mathrm{I}_2=\mathrm{I}_0+9 \mathrm{I}_0+2 \sqrt{9 \mathrm{I}_0^2} \cdot \frac{1}{2}=13 \mathrm{I}_0 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{10}{13} \end{aligned}$
In a single slit diffraction pattern, a light of wavelength 6000$\mathop A\limits^o$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 \mathrm{~mm}$ when the screen in placed $50 \mathrm{~cm}$ away from slits. The width of the slit is _________ $\times 10^{-4} \mathrm{~m}$.
Explanation:
For $\mathrm{n}^{\text {th }}$ minima
$\mathrm{b} \sin \theta=\mathrm{n} \lambda$
($\lambda$ is small so $\sin \theta$ is small, hence $\sin \theta \simeq \tan \theta$)
$\mathrm{btan} \theta=\mathrm{n} \lambda$
$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$
$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(\text { Position of } \mathrm{n}^{\mathrm{th}}\right. \text { minima) }$
$\begin{aligned} & \mathrm{B} \rightarrow 1^{\text {st }} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rd}} \text { minima } \\ & \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \\ & \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \\ & 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \\ & \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \\ & \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \\ & x=2 \end{aligned}$
In a double slit experiment shown in figure, when light of wavelength $400 \mathrm{~nm}$ is used, dark fringe is observed at $P$. If $\mathrm{D}=0.2 \mathrm{~m}$, the minimum distance between the slits $S_1$ and $S_2$ is _________ $\mathrm{mm}$.
Explanation:
Path difference for minima at $\mathrm{P}$
$\begin{aligned} & 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \\ & \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \\ & \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \\ & \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\ & \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \\ & \therefore \mathrm{d}=20 \times 10^{-5} \\ & \Rightarrow \mathrm{d}=0.20 \mathrm{~mm} \end{aligned}$
A parallel beam of monochromatic light of wavelength 5000 $\mathop A\limits^o$ is incident normally on a single narrow slit of width $0.001 \mathrm{~mm}$. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of _________ (degree).
Explanation:
For first minima
$\begin{aligned} & \operatorname{asin} \theta=\lambda \\ & \Rightarrow \sin \theta=\frac{\lambda}{a}=\frac{5000 \times 10^{-10}}{1 \times 10^{-6}}=\frac{1}{2} \\ & \Rightarrow \theta=30^{\circ} \end{aligned}$
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 $=\frac{256}{2} \times \frac{1}{4} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{256 \times 3}{2 \times 4 \times 4}=\mathbf{2 4}$
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 $\alpha = 4$