A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ________ $\times 10^{-5}$ m.
Explanation:
The wavelength of first light is $\lambda_1=650 \mathrm{~nm}=650 \times 10^{-9} \mathrm{~m}$
The wavelength of second light is $\lambda_2=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m}$
Slit separation is $\mathrm{d}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
Screen distance is $\mathrm{D}=1.2 \mathrm{~m}$
For bright fringes of different wavelengths to coincide at the same distance $(y)$ from the central maximum, their positions must be equal :
$ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{\mathrm{n}_2 \lambda_2 \mathrm{D}}{\mathrm{~d}} $
$ \mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2 $
Where $n_1$ and $n_2$ are the orders of the bright fringes for $\lambda_1$ and $\lambda_2$ respectively.
$ \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_2}{\lambda_1}=\frac{550}{650}=\frac{11}{13} $
Since we are looking for the least distance, we take the smallest integers that satisfy this ratio $\mathrm{n}_1=11$ and $\mathrm{n}_2=13$.
This means the $11^{\text {th }}$ bright fringe of 650 nm coincides with the $13^{\text {th }}$ bright fringe of 550 nm .
Using $\mathrm{n}_1=11$ and $\lambda_1=650 \mathrm{~nm}$ :
$\Rightarrow $ $ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}} $
$\Rightarrow $ $ \mathrm{y}=\frac{11 \times\left(650 \times 10^{-9}\right) \times 1.2}{2 \times 10^{-3}} \mathrm{~m}=429 \times 10^{-5} \mathrm{~m} $
The least distance from the central maximum where the bright fringes coincide is $429 \times 10^{-5} \mathrm{~m}$.
Hence, the correct answer is 429 .
In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $ 2 : 1 $ and $1: 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ( $D_1 / D_2$ ) is $\_\_\_\_$。
Explanation:
In two Young's Double Slit Experiments (YDSE) the fringe widths are equal ( $\beta_1=\beta_2$ ).
Ratio of slit separations (d) is $\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{2}{1}=2$
Ratio of wavelengths ( $\lambda$ ) is $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$
The formula for fringe width ( $\beta$ ) in a YDSE setup is:
$ \beta=\frac{\lambda D}{d} $
where:
$\lambda$ is the wavelength of light.
D is the distance between the slits and the screen.
d is the separation between the slits.
$ \beta_1=\beta_2 \Rightarrow \frac{\lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\lambda_2 \mathrm{D}_2}{\mathrm{~d}_2} $
$ \Rightarrow \frac{\mathrm{D}_1}{\mathrm{D}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \times\left(\frac{\mathrm{d}_1}{\mathrm{~d}_2}\right) $
Since $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$, so, $\frac{\lambda_2}{\lambda_1}=2$
Putting the values into the equation :
$ \frac{\mathrm{D}_1}{\mathrm{D}_2}=(2) \times(2)=4 $
The corresponding ratio of the distances between the slits and the respective screens is $4: 1$.
Hence, the correct answer is $\mathbf{4}$.
In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness $20 \mu \mathrm{~m}$ is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen.
The refractive index of transparent sheet is given by $\frac{\alpha}{10}$, where $\alpha$ is $\_\_\_\_$.
Explanation:
When a transparent sheet of thickness t and refractive index $\mu$ is placed in front of one slit, the optical path length of the light from that slit increases.
Extra path length introduced is $\Delta \mathrm{x}=(\mu-1) \mathrm{t} \ldots$ (i)
In YDSE, the path length at a point is given as $\Delta \mathrm{x}=\mathrm{d} \sin \theta \Rightarrow \sin \theta=\frac{\Delta \mathrm{x}}{\mathrm{d}} \ldots$ (ii)
For small angle, $\sin \theta \approx \tan \theta \approx \theta$ (in radians).
Also, $\tan \theta=\frac{\Delta y}{D} \ldots$ (iii)
From (i), (ii) and (iii);
$ \frac{\Delta y}{D}=\frac{\Delta x}{d} $
$\Rightarrow \Delta y=\frac{D \Delta x}{d}=\frac{D(\mu-1) t}{d} $
$\Rightarrow $ $\Delta y=\frac{D(\mu-1) t}{d} \Rightarrow(\mu-1)=\frac{\Delta y \cdot d}{D \cdot t}$
$ \Rightarrow \mu=\frac{\Delta y \cdot d}{D \cdot t}+1 $
Where :
$\mathrm{D}=$ distance between the slits and the screen $=1 \mathrm{~m}$.
$\mathrm{d}=$ distance between the two slits $=0.4 \mathrm{~mm}=0.4 \times 10^{-3} \mathrm{~m}$.
Sheet thickness is $\mathrm{t}=20 \mu \mathrm{~m}=20 \times 10^{-6} \mathrm{~m}$
Fringe shift is $\Delta y=20 \mathrm{~mm}=20 \times 10^{-3} \mathrm{~m}$
Substituting these values,
$ \mu=\frac{\left(20 \times 10^{-3}\right) \times\left(0.4 \times 10^{-3}\right)}{1 \times\left(20 \times 10^{-6}\right)}+1 $
$\Rightarrow $ $\mu=\frac{8 \times 10^{-6}}{20 \times 10^{-6}}+1$
$\Rightarrow $ $\mu=0.4+1=1.4=\frac{14}{10}$
$ \Rightarrow \frac{14}{10}=\frac{\alpha}{10} \Rightarrow \alpha=14 $
Therefore, the value of $\alpha$ is 14 .
In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm . If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is $x \times 10^{-3} \mathrm{~cm}$, where $x$-value is _________ .
Explanation:
Width of 20 maxima of double slit $=$ width of central maxima of single slit
$\begin{aligned} & \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm} \end{aligned}$
Value of $x$ is 15
Answer is 15
Explanation:
Calculate Maximum Intensity ($I_{\text{max}}$):
$ I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 $
Substituting the values, we have:
$ I_{\max} = \left(\sqrt{4I} + \sqrt{9I}\right)^2 $
$ I_{\max} = \left(2\sqrt{I} + 3\sqrt{I}\right)^2 = \left(5\sqrt{I}\right)^2 = 25I $
Calculate Minimum Intensity ($I_{\text{min}}$):
$ I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 $
Substituting the values, we have:
$ I_{\min} = \left(\sqrt{4I} - \sqrt{9I}\right)^2 $
$ I_{\min} = \left(2\sqrt{I} - 3\sqrt{I}\right)^2 = \left(-1\sqrt{I}\right)^2 = I $
Calculate the Difference Between Maximum and Minimum Intensities:
$ I_{\max} - I_{\min} = 25I - I = 24I $
Thus, the value of $x$, which represents the difference between the maximum and minimum intensities, is $24$.
If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ $\mu$m.
Explanation:
$\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \theta_1+\theta_2=30^{\circ} \end{aligned}$
$\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\mathrm{a}}\right)+\sin ^{-1}\left(\frac{3 \lambda}{\mathrm{a}}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{2 \lambda}{\mathrm{a}} \sqrt{1-\left(\frac{3 \lambda}{\mathrm{a}}\right)^2}+\frac{3 \lambda}{\mathrm{a}} \sqrt{1+\left(\frac{2 \lambda}{\mathrm{a}}\right)^2}=\sin \frac{\pi}{6} \end{aligned}$
Here $\lambda=628 \mathrm{~nm}$
After solving
$\mathrm{A}=6.07 \mu \mathrm{~m}$
Approximate Method :
$\begin{aligned} & \theta=\theta_1+\theta_2 \\ & \Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{a}+\frac{3 \lambda}{a} \\ & \Rightarrow \frac{\pi}{6}=\frac{5}{a}(628 \mathrm{~nm}) \\ & \Rightarrow a=6 \mu \mathrm{~m} \end{aligned}$
Explanation:
Maxima condition
$2 \mu \mathrm{t}=\mathrm{n} \lambda \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{2 \mu}, \frac{2 \lambda}{2 \mu}, \ldots \ldots$
Minima condition $2 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2$
$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots . . \\ & \Delta \mathrm{t}=\frac{2 \lambda}{4 \mu} \end{aligned}$
Rate of evaporation $=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}$
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10 th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be ________ mm .
Explanation:
In Young's double slit experiment, the distance of $n^{th}$ maxima from central maxima is given by,
$Y=\frac{n\lambda D}{d}$
Here, $n,D$ & $d$ are same for both the cases.
So, $Y \propto \lambda $
$ \Rightarrow {{{Y_1}} \over {{Y_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} \Rightarrow {{10} \over {{Y_2}}} = {{600} \over {660}} = {1 \over {11}}$
$ \Rightarrow {Y_2} = 11\,mm$.
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
$\begin{aligned} & \operatorname{asin} \theta=\lambda \\ & \Rightarrow \sin \theta=\frac{\lambda}{a}=\frac{5000 \times 10^{-10}}{1 \times 10^{-6}}=\frac{1}{2} \\ & \Rightarrow \theta=30^{\circ} \end{aligned}$
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 $=\frac{256}{2} \times \frac{1}{4} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{256 \times 3}{2 \times 4 \times 4}=\mathbf{2 4}$
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} $
Explanation:
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.
Explanation:
${{8\lambda \Delta } \over d}$ = 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
Explanation:
${{{I_0}} \over 2}{\cos ^2}\phi = {{3I} \over 8}$
${\cos ^2}\phi = {3 \over 4}$
${\cos ^2}\phi = {{\sqrt 3 } \over 2}$
$ \Rightarrow \phi = 30$
Explanation:
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
Explanation:
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
Explanation:
$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$
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
Explanation:
$\lambda$ = ?
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
Explanation:
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
Explanation:
$\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
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 ___________.
Explanation:
$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
Explanation:
15 $ \times $ 500 $ \times $ ${D \over \lambda }$ = 10 $ \times $ ${{\lambda D} \over \lambda }$
$ \Rightarrow $ 15 × 50 =$\lambda $
$ \Rightarrow $ $\lambda $ = 750 nm
Explanation:
$\begin{aligned} & x+y=12 \mu m \\ & \frac{4}{12}=\frac{1}{x} \\ & x=3 \mu m \\ & y=6 \mu m\end{aligned}$
$\begin{aligned} & \therefore \Delta=\frac{y d}{D}-\left(\mu_B-1\right) x-\left(\mu_A-1\right) y+\left(\mu_B-1\right) y+\left(\mu_A-1\right) x \\ & \frac{-y d}{D}=-0.5 \times 3-0.7 \times 9+0.5 \times 9+0.7 \times 3 \\ & \frac{-y d}{D}=-0.5 \times 6-0.7 \times 6 \\ & \Rightarrow \frac{-y d}{D}=-1.2 \mu \mathrm{~m} \\ & \Rightarrow y=\frac{1.2 \times D}{d}=\frac{1.2 \times 2}{2 \times 10^{-3}} \times 10^{-6} \\ & =1.2 \mathrm{~mm}\end{aligned}$
Explanation:
Given. $\frac{b d}{d}=m \lambda$
so $\quad b=\frac{m \lambda D}{d}$
Error propagation:
$ \frac{\Delta b}{b}=\frac{\Delta m}{m}+\frac{\Delta \lambda}{\lambda}+\frac{\Delta D}{D}+\frac{\Delta d}{d} $
as $\lambda $ and $ m$ are known precisely so $\Delta \lambda$ and $\Delta$m are zero.
So ${\Delta }{b}=\frac{\Delta D}{D}+\frac{\Delta d}{d}$
$ \begin{aligned} & \Rightarrow {\Delta }{b}=\left(\frac{\Delta D}{D}+\frac{\Delta d}{d}\right)\left(\frac{m \lambda D}{d}\right) \\ & \Rightarrow {\Delta }{b} =\left(\frac{1}{100}+\frac{0.1}{0.5}\right)\left(\frac{3 \times 600 \times 10^{} \times 100}{0.5}\right) \\ &=\left(\frac{1+20}{100}\right)\left(\frac{18}{5} \times \frac{1}{100}\right) \\ &=\frac{21}{100} \times \frac{18}{5} \times \frac{1}{100} \mathrm{~cm} \\ & \end{aligned} $
$ \begin{aligned} &\begin{gathered} \Delta b=\frac{1890}{25000} \times 10 \mu\mathrm{m}=\frac{1890}{25} \mu\mathrm{m} \\ \Rightarrow \Delta b=75.60 \mu\mathrm{m} \end{gathered}\\ &\text { } \end{aligned} $
Explanation:
$\begin{aligned} & \mathrm{y}=\mathrm{n} .\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right) \\ & \text { for } 8^{\text {th }} \text { fringe } \\ & \mathrm{y}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \mathrm{y}_{\max }=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {min }}} \\ & \mathrm{y}_{\text {min }}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {max }}} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\text {min }}=8 \lambda \mathrm{D}\left[\frac{1}{\mathrm{~d}_{\text {min }}}-\frac{1}{\mathrm{~d}_{\text {max }}}\right] \\ & \lambda=6000 \mathop A\limits^o\\ & \begin{aligned} & \mathrm{D}=1 \mathrm{~m} \\ & \mathrm{~d}_{\max }=0.34 \mathrm{~mm} \\ & \mathrm{~d}_{\min }=0.76 \mathrm{~mm} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\min }=8 \times 6000 \times 10^{-10} \times 1\left[\frac{1}{0.76 \times 10^{-3}}-\frac{1}{0.84 \times 10^{-3}}\right] \\ & \quad=8 \times 6 \times 10^{-4} \times\left[\frac{0.08}{0.76 \times 0.84}\right]=601.5 \mu \mathrm{m} \end{aligned} \end{aligned}$