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:
$\begin{aligned} & \mathrm{y}=\mathrm{n} .\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right) \\ & \text { for } 8^{\text {th }} \text { fringe } \\ & \mathrm{y}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \mathrm{y}_{\max }=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {min }}} \\ & \mathrm{y}_{\text {min }}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {max }}} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\text {min }}=8 \lambda \mathrm{D}\left[\frac{1}{\mathrm{~d}_{\text {min }}}-\frac{1}{\mathrm{~d}_{\text {max }}}\right] \\ & \lambda=6000 \mathop A\limits^o\\ & \begin{aligned} & \mathrm{D}=1 \mathrm{~m} \\ & \mathrm{~d}_{\max }=0.34 \mathrm{~mm} \\ & \mathrm{~d}_{\min }=0.76 \mathrm{~mm} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\min }=8 \times 6000 \times 10^{-10} \times 1\left[\frac{1}{0.76 \times 10^{-3}}-\frac{1}{0.84 \times 10^{-3}}\right] \\ & \quad=8 \times 6 \times 10^{-4} \times\left[\frac{0.08}{0.76 \times 0.84}\right]=601.5 \mu \mathrm{m} \end{aligned} \end{aligned}$
Explanation:
$\begin{aligned} & \mathrm{y}=\mathrm{n} \cdot \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{n} \cdot \frac{\lambda \cdot \mathrm{d}}{\mathrm{d}^2} \cdot \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \\ & \mathrm{d}=0.8+0.04 \sin \omega \mathrm{t} \\ & \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}=0.04 \omega \cos \omega \mathrm{t} \\ & \text { for } \mathrm{v} \rightarrow \max \Rightarrow \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \\ & \text { For } \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \\ & \cos \omega \mathrm{t}=1 \Rightarrow \sin \omega \mathrm{t}=0 \\ & \Rightarrow\left(\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}\right)_{\max }=0.04 \\ & \Rightarrow \mathrm{d}=0.8 \mathrm{~mm} \\ & \mathrm{v}_{\max }=\frac{8 \times 6000 \times 10^{-10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{-6} \times 10^{-3}}=24 \mu \mathrm{m} / \mathrm{s} \end{aligned}$
Explanation:
To solve this problem, we start by understanding the information given and how the polarization of light affects light intensity.
Initially, we have the ratio of the intensities at points A and B given as:
$ r = \frac{I_A}{I_B} = 2 $
This means:
$ I_A = 2I_B $
Now, we place two polaroids with their pass-axes at an angle of $45^\circ$ between them before point B. When unpolarized light passes through the first polaroid, it gets polarized, and its intensity is reduced to half of its original value:
$ I_{B1} = \frac{1}{2} I_B $
Here, $I_{B1}$ is the intensity of light after passing through the first polaroid.
Next, this polarized light passes through the second polaroid, which is at an angle of $45^\circ$ to the first one. According to Malus's law, the intensity of light after passing through the second polaroid is given by:
$ I_{B2} = I_{B1} \cos^2(45^\circ) $
We know that:
$ \cos(45^\circ) = \frac{1}{\sqrt{2}} $
Thus,
$ I_{B2} = \frac{1}{2} I_B \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} I_B \cdot \frac{1}{2} = \frac{1}{4} I_B $
Now, the new intensity at point B is $I_{B2}$. We need to calculate the new ratio $r'$ of the intensities at points A and B:
$ r' = \frac{I_A}{I_{B2}} $
Substituting the values $I_A = 2I_B$ and $I_{B2} = \frac{1}{4} I_B$, we get:
$ r' = \frac{2 I_B}{\frac{1}{4} I_B} = 2 \times 4 = 8 $
Therefore, the new value of $r$ will be:
$ r' = 8 $
Explanation:
$\mu ({S_2}P) - {S_1}P = m\lambda $
$ \Rightarrow \mu \sqrt {{d^2} + {x^2}} - \sqrt {{d^2} + {x^2}} = m\lambda $
$ \Rightarrow (\mu - 1)\sqrt {{d^2} + {x^2}} = m\lambda $
$ \Rightarrow \left( {{4 \over 3} - 1} \right)\sqrt {{d^2} + {x^2}} = m\lambda $
or, $\sqrt {{d^2} + {x^2}} = 3m\lambda $
Squaring this equation we get,
${x^2} = 9{m^2}{\lambda ^2} - {d^2}$
$ \Rightarrow {p^2} = 9$ or $p = 3$
Consider a system of three connected strings, $S_1, S_2$ and $S_3$ with uniform linear mass densities $\mu$ $\mathrm{kg} / \mathrm{m}, 4 \mu \mathrm{~kg} / \mathrm{m}$ and $16 \mu \mathrm{~kg} / \mathrm{m}$, respectively, as shown in the figure. $S_1$ and $S_2$ are connected at the point $P$, whereas $S_2$ and $S_3$ are connected at the point $Q$, and the other end of $S_3$ is connected to a wall. A wave generator 0 is connected to the free end of $S_1$. The wave from the generator is represented by $y=y_0 \cos (\omega t-k x) \mathrm{cm}$, where $y_0, \omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:
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?
On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct?
Which of the following is(are) true of the intensity pattern on the screen?
In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is $\lambda$. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).
In the Young's double-slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is
Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are $\beta$G, $\beta$R and $\beta$B, respectively. Then,
A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected (R) and transmitted (T) intensities, both as function of $\theta$, are plotted. The correct sketch is
Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S$_1$ and S$_2$. In each of these cases, S$_1$P$_0$ = S$_2$P$_0$, S$_1$P$_1$ $-$ S$_2$P$_1$ = $\lambda/4$ and S$_1$P$_2$ $-$ S$_2$P$_2$ = $\lambda/3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S$_2$. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta$(P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation:
| Column I | Column II | ||
|---|---|---|---|
| (A) |  |
(P) | $\delta ({P_0}) = 0$ |
| (B) | $(\mu-1)t=\lambda/4$ |
(Q) | $\delta ({P_1}) = 0$ |
| (C) | $(\mu-1)t=\lambda/2$ |
(R) | $I({P_1}) = 0$ |
| (D) | $(\mu-1)t=3\lambda/4$ |
(S) | $I({P_0}) > I({P_1})$ |
| (T) | $I({P_2}) > I({P_1})$ |
Speed of the light is