Young's Double Slit Experiment (YDSE) Questions in English

(C) For darkness at point $O$,the path difference between the light waves reaching $O$ from $S_1$ and $S_2$ must be an odd multiple of $\frac{\lambda}{2}$.
The path taken by light from $S$ to $O$ via $S_2$ is $SO + OS_2 = D + D = 2D$.
The path taken by light from $S$ to $O$ via $S_1$ is $SS_1 + S_1O$. Here,$SS_1 = \sqrt{D^2 + d^2}$ and $S_1O = \sqrt{D^2 + d^2}$.
Thus,the total path via $S_1$ is $2\sqrt{D^2 + d^2}$.
The path difference $\Delta x = 2\sqrt{D^2 + d^2} - 2D$.
For minimum darkness,$\Delta x = \frac{\lambda}{2}$.
$2\sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2} \implies \sqrt{D^2 + d^2} = D + \frac{\lambda}{4}$.
Squaring both sides: $D^2 + d^2 = D^2 + \frac{\lambda D}{2} + \frac{\lambda^2}{16}$.
Since $d \ll D$,we can neglect the $\lambda^2$ term: $d^2 \approx \frac{\lambda D}{2}$.
Therefore,$d = \sqrt{\frac{\lambda D}{2}}$.

(B) The condition for the angular position of the $n^{th}$ bright fringe in a double-slit experiment is given by $\theta = \frac{n \lambda}{d}$.
Given:
Wavelength $\lambda = 6000 \, Å = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$.
Slit separation $d = 0.1 \, cm = 10^{-3} \, m$.
Order of fringe $n = 10$.
Substituting the values into the formula:
$\theta = \frac{10 \times 6 \times 10^{-7}}{10^{-3}}$
$\theta = 60 \times 10^{-4} \, rad$
$\theta = 6 \times 10^{-3} \, rad$.

(B) Given: Distance between slits $d = 0.1 \ mm = 10^{-4} \ m$,distance of screen $D = 20 \ cm = 0.2 \ m$,and wavelength $\lambda = 5460 \ \mathring{A} = 5460 \times 10^{-10} \ m$.
For the first dark fringe,the path difference is $\Delta x = \lambda / 2$.
The angular position $\theta$ is given by the relation $d \sin \theta = \Delta x = \lambda / 2$.
Since $\theta$ is very small,$\sin \theta \approx \theta$ (in radians).
$\theta = \frac{\lambda}{2d} = \frac{5460 \times 10^{-10}}{2 \times 10^{-4}} = 2730 \times 10^{-6} \ rad$.
To convert radians to degrees,multiply by $(180 / \pi)$:
$\theta_{deg} = 2730 \times 10^{-6} \times \frac{180}{3.14159} \approx 0.1564^\circ$.
Rounding to two decimal places,we get $\theta \approx 0.16^\circ$.

(B) The angular fringe width $\theta$ in a double slit experiment is given by the formula $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the slit separation.
When the apparatus is immersed in a medium with refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
The new angular width $\theta'$ becomes $\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$.
Given $\theta = 0.2^{\circ}$ and $\mu = 4/3$,we have $\theta' = \frac{0.2}{4/3} = 0.2 \times \frac{3}{4} = 0.15^{\circ}$.

(C) The intensity at any point in a double-slit experiment is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.
Given $I = I_{max}/4$,we have $I_{max}/4 = I_{max} \cos^2(\phi/2)$.
This simplifies to $\cos^2(\phi/2) = 1/4$,so $\cos(\phi/2) = 1/2$.
Thus,$\phi/2 = 60^{\circ} = \pi/3$ radians,which means $\phi = 2\pi/3$.
The path difference $\Delta$ is related to the phase difference by $\Delta = (\lambda / 2\pi) \times \phi = (\lambda / 2\pi) \times (2\pi/3) = \lambda/3$.
For small angles,the path difference is also given by $\Delta = d \sin \theta$.
Equating the two expressions for $\Delta$,we get $d \sin \theta = \lambda/3$.
Therefore,$\sin \theta = \lambda/(3d)$,which gives $\theta = sin^{-1}(\lambda/3d)$.

(B) In Young's double-slit experiment,the fringe width $\beta$ is given by the formula:
$\beta = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
From the formula,it is clear that $\beta \propto \frac{1}{d}$.
This relationship represents a rectangular hyperbola,where $\beta$ decreases as $d$ increases.
Therefore,the correct curve is the one showing a rectangular hyperbola.

(A) The formula for fringe width $(\beta)$ in $Y.D.S.E$ is given by $\beta = \frac{D \lambda}{d}$.
Given values are:
Distance between slits, $d = 1 \, mm = 1 \times 10^{-3} \, m$.
Distance of screen, $D = 1 \, m$.
Wavelength of light, $\lambda = 500 \, nm = 500 \times 10^{-9} \, m = 5 \times 10^{-7} \, m$.
Substituting these values into the formula:
$\beta = \frac{1 \times 5 \times 10^{-7}}{1 \times 10^{-3}} = 5 \times 10^{-4} \, m$.

(D) The intensity $I$ is directly proportional to the width of the slit $w$,i.e.,$I \propto w$. Also,$I \propto A^2$,where $A$ is the amplitude.
Given $\frac{I_{\max}}{I_{\min}} = \frac{25}{9}$.
We know that $\frac{I_{\max}}{I_{\min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{25}{9}$.
Taking the square root on both sides: $\frac{A_1 + A_2}{A_1 - A_2} = \frac{5}{3}$.
By applying componendo and dividendo: $\frac{A_1}{A_2} = \frac{5+3}{5-3} = \frac{8}{2} = \frac{4}{1}$.
Since the intensity is proportional to the slit width,the ratio of intensities of the two slits is $\frac{I_1}{I_2} = \frac{A_1^2}{A_2^2} = \left(\frac{4}{1}\right)^2 = \frac{16}{1}$.
Therefore,the ratio of the slit widths is $\frac{w_1}{w_2} = \frac{16}{1}$.

(D) The fringe width $\beta$ in Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When a thin glass plate is placed in the path of one of the beams,it introduces an additional path difference,which causes the entire interference pattern to shift on the screen.
However,the fringe width $\beta$ depends only on the wavelength $\lambda$,the distance $D$,and the slit separation $d$. Since none of these parameters are changed by the introduction of the glass plate,the fringe width remains constant.

(D) The condition for the $n^{th}$ minima for wavelength $\lambda$ is given by $y_n = (2n-1) \frac{D\lambda}{2d}$.
For two wavelengths to have overlapping dark fringes,their minima must coincide: $(2n-1) \frac{D\lambda_1}{2d} = (2m-1) \frac{D\lambda_2}{2d}$.
This simplifies to $(2n-1) \lambda_1 = (2m-1) \lambda_2$.
Substituting the values: $(2n-1) 400 = (2m-1) 560$.
$\frac{2n-1}{2m-1} = \frac{560}{400} = \frac{7}{5}$.
Thus,the first overlap occurs at $2n-1 = 7$ (i.e.,$n=4$) and $2m-1 = 5$ (i.e.,$m=3$).
The position of this first overlap is $y_1 = 7 \times \frac{1 \times 400 \times 10^{-6}}{2 \times 0.1 \times 10^{-3}} = 14 \times 10^{-3}\, m = 14\, mm$.
The next overlap occurs at the next odd multiple ratio,which is $\frac{21}{15}$ (where $2n-1=21$ and $2m-1=15$).
The position of this second overlap is $y_2 = 21 \times \frac{1 \times 400 \times 10^{-6}}{2 \times 0.1 \times 10^{-3}} = 42\, mm$.
The distance between two successive regions of complete darkness is $y_2 - y_1 = 42\, mm - 14\, mm = 28\, mm$.

(C) In a Young's double-slit experiment,the path difference between the waves from two slits $S_1$ and $S_2$ to any point $P$ on the screen is given by $\Delta x = S_2P - S_1P = d \sin \theta$.
For a constant path difference,the locus of points $P$ on the screen forms a curve.
The condition for constructive or destructive interference is $\Delta x = n\lambda$ or $\Delta x = (n + 1/2)\lambda$,which implies that the path difference is constant for a given fringe.
The locus of points having a constant difference in distances from two fixed points (the slits) is a hyperbola.
Therefore,the interference fringes formed on the screen are hyperbolic in shape.

(A) The intensity at any point in a Young's double-slit experiment is given by $I = I_{\max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that the intensity is half of the maximum intensity,we have $\frac{I_{\max}}{2} = I_{\max} \cos^2(\frac{\phi}{2})$.
This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which means $\cos(\frac{\phi}{2}) = \frac{1}{\sqrt{2}}$.
Therefore,$\frac{\phi}{2} = \frac{\pi}{4}$,which gives the phase difference $\phi = \frac{\pi}{2}$.
The path difference $\Delta x$ is related to the phase difference by $\Delta x = \frac{\lambda}{2\pi} \phi$.
Substituting $\phi = \frac{\pi}{2}$,we get $\Delta x = \frac{\lambda}{2\pi} \times \frac{\pi}{2} = \frac{\lambda}{4}$.
The distance $x$ from the central point is given by $x = \frac{\Delta x \cdot D}{d}$.
Substituting the value of $\Delta x$,we get $x = \frac{\lambda D}{4d}$.

(D) The formula for fringe width in $YDSE$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the slits.
From the formula,it is clear that fringe width $\beta$ is inversely proportional to the slit separation $d$,i.e.,$\beta \propto \frac{1}{d}$.
If the distance between the slits is increased $4$ times,the new distance $d' = 4d$.
Therefore,the new fringe width $\beta' = \frac{\lambda D}{d'} = \frac{\lambda D}{4d} = \frac{\beta}{4}$.
Thus,the fringe width decreases $4$ times.

(C) The interference pattern disappears when the angular separation between the fringes becomes less than the angular resolution of the human eye.
The angular fringe width is given by $\beta_{\theta} = \frac{\lambda}{d}$.
The angular resolution of the eye is $\Delta\theta = \frac{1}{60}^o = \frac{1}{60} \times \frac{\pi}{180} \text{ radians}$.
For the pattern to disappear,the angular fringe width must be equal to the angular resolution of the eye: $\frac{\lambda}{d_0} = \Delta\theta$.
Substituting the values: $d_0 = \frac{\lambda}{\Delta\theta} = \frac{600 \times 10^{-9} \text{ m}}{\frac{1}{60} \times \frac{\pi}{180} \text{ rad}}$.
$d_0 = \frac{600 \times 10^{-9} \times 60 \times 180}{\pi} \approx \frac{6.48 \times 10^{-3}}{3.14} \approx 2.06 \times 10^{-3} \text{ m}$.
Thus,$d_0 \approx 2 \text{ mm}$.

(C) The intensity at any point in a Young's double-slit experiment is given by $I = 4I_0 \cos^2(\frac{\Delta \phi}{2})$,where $I_0$ is the intensity of each individual slit and $\Delta \phi$ is the phase difference.
At the central maximum,the intensity is $I_{max} = 4I_0$.
We want to find the position $y$ where the intensity $I$ falls to half of the maximum intensity,i.e.,$I = \frac{I_{max}}{2} = 2I_0$.
Substituting this into the intensity formula: $2I_0 = 4I_0 \cos^2(\frac{\Delta \phi}{2}) \implies \cos^2(\frac{\Delta \phi}{2}) = \frac{1}{2} \implies \cos(\frac{\Delta \phi}{2}) = \frac{1}{\sqrt{2}}$.
This implies $\frac{\Delta \phi}{2} = \frac{\pi}{4}$,so the phase difference is $\Delta \phi = \frac{\pi}{2}$.
Since the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$,we have $\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives the path difference $\Delta x = \frac{\lambda}{4}$.
For small angles,the path difference is $\Delta x = \frac{dy}{D}$.
Equating the two: $\frac{dy}{D} = \frac{\lambda}{4} \implies y = \frac{\lambda D}{4d}$.
Since the fringe width is $\beta = \frac{\lambda D}{d}$,we substitute this to get $y = \frac{\beta}{4}$.

(B) Given the thickness of the mica sheet $t = 1.8 \times 10^{-6} \ m$,refractive index $\mu = 1.6$,and the shift $y = (\mu - 1)t \frac{D}{d}$.
The fringe width in the first case is $\beta = \frac{\lambda D}{d}$.
The fringe width in the third case is $\beta' = \frac{\lambda (2D)}{d} = 2\beta$.
Given the condition $y = \beta'$,we have $(\mu - 1)t \frac{D}{d} = 2 \frac{\lambda D}{d}$.
Simplifying,we get $(\mu - 1)t = 2\lambda$.
Substituting the values: $(1.6 - 1) \times 1.8 \times 10^{-6} = 2\lambda$.
$0.6 \times 1.8 \times 10^{-6} = 2\lambda$.
$1.08 \times 10^{-6} = 2\lambda$.
$\lambda = 0.54 \times 10^{-6} \ m = 540 \ nm$.

(C) The angular width of the central maximum in a single slit diffraction pattern is given by $\theta = \frac{2 \lambda}{b}$,where $b$ is the slit width.
The angular separation between consecutive interference maxima in a Young's double slit experiment is $\Delta \theta = \frac{\lambda}{d}$,where $d$ is the distance between the slits.
The number of interference maxima $n$ that fit within the central diffraction maximum is given by the ratio of the angular width of the central diffraction maximum to the angular separation of the interference fringes:
$n = \frac{2 \lambda / b}{\lambda / d} = \frac{2d}{b}$.
Given that $d = 6.1b$,we substitute this into the equation:
$n = \frac{2 \times (6.1b)}{b} = 12.2$.
Since the number of maxima must be an integer,we consider the maxima that lie strictly within the central diffraction envelope. Thus,$12$ intensity maxima are observed.

(A) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance to the screen,$d$ is the slit separation,and $\lambda$ is the wavelength.
The number of fringes $N$ that can be observed in a field of view of width $W$ is given by $N = \frac{W}{\beta} = \frac{Wd}{D\lambda}$.
From this relation,$N \propto \frac{1}{\lambda}$. Thus,as the wavelength $\lambda$ increases,the fringe width $\beta$ increases,and the number of fringes $N$ observed in the field of view decreases. Conversely,for a shorter wavelength,the fringe width is smaller,allowing more fringes to be observed.
Therefore,Statement-$1$ is true,Statement-$2$ is true,and Statement-$2$ provides the correct explanation for Statement-$1$.

(D) For a wavelength $\lambda$,the condition for the $n$-th maximum is path difference $\Delta x = n\lambda$,and the condition for the $m$-th minimum is path difference $\Delta x = (m + 1/2)\lambda$.
Let the path difference be $\Delta x$. The condition for the interference pattern to be indistinct is that a maximum of $\lambda_1$ coincides with a minimum of $\lambda_2$,or vice versa.
Case $1$: Maximum of $\lambda_1$ coincides with minimum of $\lambda_2$:
$\Delta x = n\lambda_1 = (m + 1/2)\lambda_2$
$n(500) = (m + 1/2)(600)$
$5n = 6m + 3$
For the smallest integer values,if $m = 1$,$5n = 9$ (no integer $n$). If $m = 4$,$5n = 27$ (no). If $m = 1$,$5n = 9$. Let's test values: $5n - 6m = 3$. For $m=1, 5n=9$; $m=2, 5n=15 \Rightarrow n=3$.
Thus,$\Delta x = 3 \times 500 = 1500\,nm$.
Case $2$: Maximum of $\lambda_2$ coincides with minimum of $\lambda_1$:
$\Delta x = n\lambda_2 = (m + 1/2)\lambda_1$
$n(600) = (m + 1/2)(500)$
$6n = 5m + 2.5 \Rightarrow 12n = 10m + 5$.
Since $12n$ is even and $10m+5$ is odd,this case is impossible for integer $n, m$.
Therefore,the first point of indistinctness occurs at $\Delta x = 1500\,nm$.

(B) In a Young's double slit experiment,the condition for constructive interference (maxima) is given by $d \sin \theta = n \lambda$,where $d$ is the slit separation,$\lambda$ is the wavelength,and $n$ is the order of the maxima.
Given $d = 1.8 \lambda$,the condition becomes $1.8 \lambda \sin \theta = n \lambda$,which simplifies to $n = 1.8 \sin \theta$.
Since the maximum value of $\sin \theta$ is $1$,the maximum value of $n$ is $1.8 \times 1 = 1.8$.
Thus,the possible integer values for $n$ are $0, \pm 1$.
Therefore,the total number of maxima is $1$ (for $n=0$) $+ 2$ (for $n=1$ and $n=-1$),which equals $3$.

(C) We know that the ratio of maximum intensity to minimum intensity in an interference pattern is given by:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{\frac{\omega_1}{\omega_2}} + 1)^2}{(\sqrt{\frac{\omega_1}{\omega_2}} - 1)^2}$
where $\omega_1$ and $\omega_2$ are the widths of the two slits.
Given the ratio of slit widths $\frac{\omega_1}{\omega_2} = \frac{1}{25}$.
Substituting this value into the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{\frac{1}{25}} + 1)^2}{(\sqrt{\frac{1}{25}} - 1)^2} = \frac{(\frac{1}{5} + 1)^2}{(\frac{1}{5} - 1)^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(\frac{6}{5})^2}{(\frac{-4}{5})^2} = \frac{\frac{36}{25}}{\frac{16}{25}} = \frac{36}{16} = \frac{9}{4}$
Thus,the ratio of intensity at the maxima and minima is $9 : 4$.

(D) The condition for bright fringes is $d \sin \theta = n \lambda$,where $n$ is an integer.
Given $d = 0.320 \, mm = 0.320 \times 10^{-3} \, m$ and $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$.
The maximum path difference at $\theta = 30^\circ$ is $\Delta X_{\max} = d \sin 30^\circ = 0.320 \times 10^{-3} \times 0.5 = 0.16 \times 10^{-3} \, m$.
The maximum order $n$ is given by $n = \frac{\Delta X_{\max}}{\lambda} = \frac{0.16 \times 10^{-3}}{500 \times 10^{-9}} = \frac{0.16 \times 10^6}{500} = 320$.
Since the range is $-30^\circ \le \theta \le 30^\circ$,we observe bright fringes for $n = 0, \pm 1, \pm 2, \dots, \pm 320$.
The total number of bright fringes is $2n + 1 = 2(320) + 1 = 641$.

(A) For bright fringes (maxima) in a Young's double slit experiment,the condition is $d \sin \theta = n \lambda$,where $n$ is the order of the fringe.
Since $\theta$ is small,$\sin \theta \approx \theta = \frac{1}{40}\, rad$.
Given $d = 0.1\, mm = 10^{-4}\, m$.
The condition becomes $d \theta = n \lambda$,so $\lambda = \frac{d \theta}{n}$.
Substituting the values: $\lambda = \frac{10^{-4} \times (1/40)}{n} = \frac{10^{-4}}{40n} = \frac{10^{-5}}{4n} = \frac{2500\, nm}{n}$.
For $\lambda$ to be in the visible range ($380\, nm$ to $740\, nm$):
If $n=4$,$\lambda = \frac{2500}{4} = 625\, nm$.
If $n=5$,$\lambda = \frac{2500}{5} = 500\, nm$.
Both $625\, nm$ and $500\, nm$ lie within the visible range. Thus,the wavelengths are $625\, nm$ and $500\, nm$.

(A) For the first minima to occur at point $P$ (directly in front of $S_1$),the path difference between the waves from $S_1$ and $S_2$ must be equal to $\frac{\lambda}{2}$.
The distance from $S_1$ to $P$ is $2d$.
The distance from $S_2$ to $P$ is $\sqrt{(2d)^2 + d^2} = \sqrt{4d^2 + d^2} = \sqrt{5}d$.
The path difference $\Delta x = S_2P - S_1P = \sqrt{5}d - 2d = d(\sqrt{5} - 2)$.
Setting the path difference equal to $\frac{\lambda}{2}$ for the first minima:
$d(\sqrt{5} - 2) = \frac{\lambda}{2}$
Therefore,the slit separation $d$ is:
$d = \frac{\lambda}{2(\sqrt{5} - 2)}$

(B) The path difference is given by $\Delta x = \frac{\lambda}{8}$.
The phase difference $\phi$ is related to the path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the value of $\Delta x$,we get $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4}$.
The resultant intensity $I$ at any point is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$,where $I_0$ is the intensity of each individual wave.
At the centre of a bright fringe,the phase difference is $0$,so the maximum intensity $I_{max} = 2I_0(1 + \cos 0) = 4I_0$.
At the given point,the intensity $I = 2I_0(1 + \cos(\pi/4)) = 2I_0(1 + \frac{1}{\sqrt{2}})$.
The ratio is $\frac{I}{I_{max}} = \frac{2I_0(1 + 1/\sqrt{2})}{4I_0} = \frac{1 + 0.707}{2} = \frac{1.707}{2} \approx 0.85$.

(A) The intensity of light $I$ is directly proportional to the slit width $w$,so $I_1/I_2 = w_1/w_2 = 4/1$.
Let $I_1 = 4I_0$ and $I_2 = I_0$.
The amplitudes are $a_1 = \sqrt{I_1} = 2\sqrt{I_0}$ and $a_2 = \sqrt{I_2} = \sqrt{I_0}$.
The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Substituting the values: $\frac{I_{\max}}{I_{\min}} = \left( \frac{2\sqrt{I_0} + \sqrt{I_0}}{2\sqrt{I_0} - \sqrt{I_0}} \right)^2 = \left( \frac{3\sqrt{I_0}}{\sqrt{I_0}} \right)^2 = (3)^2 = 9/1$.

(C) In a Young's double slit experiment $(YDSE)$,the angular width of a fringe $(\beta_{\theta})$ is given by the formula:
$\beta_{\theta} = \frac{\lambda}{d}$
Given:
Angular width $\beta_{\theta} = 0.1 \ radian$
Wavelength $\lambda = 6000 \ \mathring{A} = 6000 \times 10^{-10} \ m = 6 \times 10^{-7} \ m$
We need to find the distance between the slits $(d)$:
$d = \frac{\lambda}{\beta_{\theta}}$
$d = \frac{6 \times 10^{-7} \ m}{0.1}$
$d = 6 \times 10^{-6} \ m$
Since $1 \ \mu m = 10^{-6} \ m$,we have:
$d = 6 \ \mu m$
Therefore,the distance between the two slits is $6 \ \mu m$.

(A) The formula for fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of the light source,$D$ is the distance between the slits and the screen,and $d$ is the separation between the two slits.
From the formula,it is clear that the fringe width $\beta$ is inversely proportional to the slit separation $d$ (i.e.,$\beta \propto \frac{1}{d}$).
Therefore,if we decrease the separation of the slits $(d)$,the fringe width $(\beta)$ will increase.

(B) The fringe width $\beta$ in Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.
As the separation between the slits $d$ is increased,the fringe width $\beta$ decreases.
As $d$ continues to increase,the fringe width becomes so small that the fringes eventually become indistinguishable and disappear.

(D) The condition for the formation of bright fringes in a Young's Double Slit Experiment is given by $y = \frac{n D \lambda}{d}$.
Since the fringes are formed at the same spot,their positions are equal: $y_{\text{blue}} = y_{\text{red}}$.
For the $(n + 4)^{th}$ blue bright fringe: $y_{\text{blue}} = (n + 4) \frac{D \lambda_{\text{blue}}}{d}$.
For the $n^{th}$ red bright fringe: $y_{\text{red}} = n \frac{D \lambda_{\text{red}}}{d}$.
Equating the two: $(n + 4) \lambda_{\text{blue}} = n \lambda_{\text{red}}$.
Substituting the given values: $(n + 4) \times 5200\,\mathring{A} = n \times 7800\,\mathring{A}$.
Dividing both sides by $2600\,\mathring{A}$: $(n + 4) \times 2 = n \times 3$.
$2n + 8 = 3n$.
$n = 8$.

(C) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its velocity.
The fringe width $\beta$ in $Y.D.S.E.$ is given by $\beta = \frac{\lambda D}{d}$,where $D$ is the distance between the slits and the screen,and $d$ is the slit separation.
Substituting the expression for $\lambda$ into the fringe width formula,we get $\beta = \frac{hD}{mvd}$.
From this relation,it is clear that $\beta \propto \frac{1}{v}$.
Therefore,when the velocity $v$ of the electrons is increased,the fringe width $\beta$ decreases.

(A) The condition for the coincidence of the $n^{th}$ fringe of one wavelength with the $(n+1)^{th}$ fringe of another wavelength is given by $n_1 \lambda_1 = n_2 \lambda_2$.
Here,$n_1 = n$,$\lambda_1 = 7800 \, \mathring{A}$,$n_2 = n+1$,and $\lambda_2 = 5200 \, \mathring{A}$.
Substituting the values: $n \times 7800 = (n + 1) \times 5200$.
Dividing both sides by $2600$: $n \times 3 = (n + 1) \times 2$.
$3n = 2n + 2$.
$n = 2$.

(B) The formula for fringe width $\beta$ in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.
Given values are:
Wavelength $\lambda = 6000\,\mathring{A} = 6000 \times 10^{-10}\, m = 6 \times 10^{-7}\, m$.
Distance between slits $d = 4\, mm = 4 \times 10^{-3}\, m$.
Distance of screen $D = 2\, m$.
Substituting these values into the formula:
$\beta = \frac{6 \times 10^{-7} \times 2}{4 \times 10^{-3}}$
$\beta = \frac{12 \times 10^{-7}}{4 \times 10^{-3}}$
$\beta = 3 \times 10^{-4}\, m$.
Converting meters to millimeters:
$\beta = 3 \times 10^{-4} \times 10^3\, mm = 0.3\, mm$.

(C) In Young's double slit experiment,the intensity at any point on the screen is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta\phi)$.
Given $I_1 = I_2 = I_0$,the formula becomes $I = 2I_0 + 2I_0 \cos(\Delta\phi) = 4I_0 \cos^2(\frac{\Delta\phi}{2})$.
The position in front of one of the slits corresponds to $y = \frac{d}{2}$.
The path difference $\Delta x$ is given by $\Delta x = d \sin \theta \approx d \tan \theta = d \cdot \frac{y}{D}$.
Substituting $y = \frac{d}{2}$ and $D = 10d$,we get $\Delta x = d \cdot \frac{d/2}{10d} = \frac{d}{20}$.
Given $d = 5\lambda$,the path difference is $\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$.
The phase difference $\Delta\phi$ is $\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
Now,the intensity is $I = 2I_0(1 + \cos(\frac{\pi}{2})) = 2I_0(1 + 0) = 2I_0$.

(D) The intensity $I$ of a wave is proportional to the square of its amplitude,$I \propto A^2$.
Given the amplitudes of the two sources are $A_1 = 3a$ and $A_2 = a$.
The maximum intensity $(I_{\max})$ at bright fringes is given by $I_{\max} = (A_1 + A_2)^2 = (3a + a)^2 = (4a)^2 = 16a^2$.
The minimum intensity $(I_{\min})$ at dark fringes is given by $I_{\min} = (A_1 - A_2)^2 = (3a - a)^2 = (2a)^2 = 4a^2$.
The ratio of intensities of bright and dark fringes is $\frac{I_{\max}}{I_{\min}} = \frac{16a^2}{4a^2} = \frac{4}{1}$.

(C) The position of the central maxima is independent of the medium,so it remains at $Y = 2 \, cm$.
The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$. When immersed in a liquid of refractive index $n = 1.5$,the wavelength becomes $\lambda' = \frac{\lambda}{n}$.
Thus,the new fringe width $\beta'$ becomes $\beta' = \frac{\beta}{n} = \frac{\beta}{1.5}$.
The initial distance between the central maxima and the $10^{th}$ maxima is $\Delta Y = 5 \, cm - 2 \, cm = 3 \, cm$.
In the liquid,the new distance $\Delta Y'$ is $\Delta Y' = \frac{\Delta Y}{n} = \frac{3 \, cm}{1.5} = 2 \, cm$.
Therefore,the new position of the $10^{th}$ maxima is $Y' = Y_{central} + \Delta Y' = 2 \, cm + 2 \, cm = 4 \, cm$.

(C) The intensity at any point in a double-slit experiment is given by $I = I_{max} \cos^2(\phi / 2)$,where $\phi$ is the phase difference.
Given $I = I_{max} / 4$,we have $I_{max} / 4 = I_{max} \cos^2(\phi / 2)$.
This simplifies to $\cos^2(\phi / 2) = 1/4$,so $\cos(\phi / 2) = 1/2$.
Thus,$\phi / 2 = 60^{\circ}$ or $\pi / 3$ radians,which means the phase difference $\phi = 120^{\circ}$ or $2\pi / 3$ radians.
The path difference $\Delta$ is related to the phase difference by $\Delta = (\lambda / 2\pi) \times \phi$.
Substituting $\phi = 2\pi / 3$,we get $\Delta = (\lambda / 2\pi) \times (2\pi / 3) = \lambda / 3$.
For small angles,the path difference $\Delta = d \sin \theta$,where $d$ is the slit separation.
Therefore,$d \sin \theta = \lambda / 3$,which gives $\sin \theta = \lambda / (3d)$.
Hence,the angular position is $\theta = \sin ^{-1}(\lambda / 3d)$.

(A) The formula for fringe width in air is $\beta = \frac{\lambda D}{d}$.
Given: Slit separation $d = 0.2 \, cm = 2 \times 10^{-3} \, m$,wavelength $\lambda = 600 \, nm = 600 \times 10^{-9} \, m$,and screen distance $D = 1 \, m$.
First,calculate the fringe width in air: $\beta = \frac{600 \times 10^{-9} \times 1}{2 \times 10^{-3}} = 300 \times 10^{-6} \, m = 0.3 \, mm$.
When the system is immersed in a medium of refractive index $\mu$,the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$,so the new fringe width is $\beta' = \frac{\beta}{\mu}$.
Given $\mu = \frac{4}{3}$,the new fringe width is $\beta' = \frac{0.3}{4/3} = 0.3 \times \frac{3}{4} = 0.225 \, mm$.

(B) The condition for maxima in $Y.D.S.E.$ is given by $d \sin \theta = n \lambda$,where $n$ is the order of the maxima.
For the maximum possible order $n$,we consider the limit $\theta \to 90^{\circ}$,so $\sin \theta \to 1$.
Thus,$n < \frac{d}{\lambda}$.
Given $d = 700 \ nm$ and $\lambda = 200 \ nm$,we have $n < \frac{700}{200} = 3.5$.
The possible integer values for $n$ are $0, \pm 1, \pm 2, \pm 3$.
Total number of maxima $= 2n_{max} + 1 = 2(3) + 1 = 7$.

(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe (maximum) from the central maximum is given by the formula: $x_n = \frac{n \lambda D}{d}$.
Here,$n = 4$ (for the fourth maximum),$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
Since $n$,$D$,and $d$ are constant for both experiments,the position $x$ is directly proportional to the wavelength $\lambda$ $(x \propto \lambda)$.
Given $\lambda_{blue} = 4360 \, \mathring{A}$ and $\lambda_{green} = 5460 \, \mathring{A}$.
Since $\lambda_{blue} < \lambda_{green}$,it follows that $X_{blue} < X_{green}$.

(C) The fringe width $\beta$ in Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the separation between the slits.
To maintain the same fringe spacing $\beta$,the ratio $\frac{D}{d}$ must remain constant.
If the slit separation $d$ is doubled (i.e.,$d' = 2d$),then to keep $\beta$ constant,we must have $\frac{D'}{d'} = \frac{D}{d}$.
Substituting $d' = 2d$,we get $\frac{D'}{2d} = \frac{D}{d}$,which implies $D' = 2D$.
Therefore,the distance $D$ must be doubled.

(A) When a glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the beams,the path difference introduced is $\Delta x = (\mu - 1)t$.
The condition for the intensity to remain unchanged at the position of the original central maximum is that the new path difference must be an integer multiple of the wavelength $\lambda$,such that the phase difference is a multiple of $2\pi$.
For the intensity to remain the same (i.e.,constructive interference),the path difference must be $\Delta x = n\lambda$,where $n = 1, 2, 3, ...$.
Setting the path difference equal to the first order of constructive interference $(n = 1)$:
$(\mu - 1)t = \lambda$
Given $\mu = 1.5$:
$(1.5 - 1)t = \lambda$
$0.5t = \lambda$
$t = 2\lambda$
Thus,the minimum thickness is $2\lambda$.

(C) The angular width of a fringe in $YDSE$ is given by $\beta_{\theta} = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit separation.
Given: $\beta_{\theta} = 1^o = \frac{\pi}{180} \text{ radians}$ and $d = 0.01\,mm = 10^{-5}\,m$.
Substituting the values: $\lambda = \beta_{\theta} \times d = \frac{\pi}{180} \times 10^{-5}\,m$.
$\lambda = 0.01745 \times 10^{-5}\,m = 1.745 \times 10^{-7}\,m$.
Converting to micrometers: $\lambda = 0.1745 \times 10^{-6}\,m = 0.1745\,\mu m \approx 0.174\,\mu m$.

(C) Given: Slit separation $d = 1 \, mm = 10^{-3} \, m$,screen distance $D = 1 \, m$,wavelength $\lambda = 5 \times 10^{-7} \, m$.
The position of the $n^{th}$ bright fringe is given by $y_{bright} = n \frac{D \lambda}{d}$.
For the $5^{th}$ bright fringe $(n=5)$: $y_1 = 5 \frac{D \lambda}{d}$.
The position of the $n^{th}$ dark fringe is given by $y_{dark} = (n - 0.5) \frac{D \lambda}{d}$.
For the $3^{rd}$ dark fringe $(n=3)$: $y_2 = (3 - 0.5) \frac{D \lambda}{d} = 2.5 \frac{D \lambda}{d}$.
The distance between them is $\Delta y = |y_1 - y_2| = |5 - 2.5| \frac{D \lambda}{d} = 2.5 \frac{D \lambda}{d}$.
Substituting the values:
$\Delta y = 2.5 \times \frac{1 \times 5 \times 10^{-7}}{10^{-3}} \, m$
$\Delta y = 2.5 \times 5 \times 10^{-4} \, m = 12.5 \times 10^{-4} \, m = 1.25 \times 10^{-3} \, m = 1.25 \, mm$.

(D) The formula for fringe width in $YDSE$ is given by $\omega = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.
Given that the new separation $d' = \frac{d}{2}$ and the new distance $D' = 2D$.
The new fringe width $\omega'$ will be $\omega' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(d/2)} = 4 \times \frac{\lambda D}{d} = 4\omega$.
Therefore,the fringe width becomes $4$ times the original value.

(A) The position of the $n$-th order bright fringe in a Young's double-slit experiment is given by $y_n = n \lambda \frac{D}{d}$.
For the third-order bright fringe of wavelength $\lambda_1$,the position is $y_1 = 3 \lambda_1 \frac{D}{d}$.
For the fourth-order bright fringe of wavelength $\lambda_2$,the position is $y_2 = 4 \lambda_2 \frac{D}{d}$.
Given that the fringes coincide,we have $y_1 = y_2$.
Therefore,$3 \lambda_1 \frac{D}{d} = 4 \lambda_2 \frac{D}{d}$.
Simplifying the equation,we get $3 \lambda_1 = 4 \lambda_2$.
Thus,the ratio is $\frac{\lambda_1}{\lambda_2} = \frac{4}{3}$.

(A) The formula for fringe width (fringe separation) in Young's double-slit experiment is given by $\beta = \frac{D \lambda}{d}$.
Given:
Distance between slits,$d = 1\, mm = 1 \times 10^{-3}\, m$.
Distance of the screen,$D = 1\, m$.
Wavelength of light,$\lambda = 500\, nm = 500 \times 10^{-9}\, m = 5 \times 10^{-7}\, m$.
Substituting these values into the formula:
$\beta = \frac{1 \times 5 \times 10^{-7}}{1 \times 10^{-3}} = 5 \times 10^{-4}\, m$.
Thus,the fringe separation is $5 \times 10^{-4}\, m$.

(A) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 = \frac{25}{9}$.
Taking the square root on both sides,we get $\frac{A_1 + A_2}{A_1 - A_2} = \frac{5}{3}$.
By applying componendo and dividendo,$\frac{A_1}{A_2} = \frac{5+3}{5-3} = \frac{8}{2} = \frac{4}{1}$.
The ratio of the widths of the slits $(w_1/w_2)$ is proportional to the ratio of the squares of their amplitudes $(A_1^2/A_2^2)$.
Therefore,$\frac{w_1}{w_2} = \left( \frac{A_1}{A_2} \right)^2 = \left( \frac{4}{1} \right)^2 = \frac{16}{1}$.

(A) To obtain a stable interference pattern in Young's double-slit experiment,the light sources must be coherent.
$(1)$ An ordinary extended source of light emits waves with random phase differences. $A$ single narrow slit acts as a secondary source,ensuring spatial coherence,which is necessary for interference.
$(2)$ $A$ well-collimated beam of light (like a laser) is inherently coherent. Therefore,the initial slit is not required to establish coherence.
$(3)$ $A$ spatially coherent point source already provides waves with a constant phase relationship. Thus,no additional slit is needed to create coherence.
Since all three statements describe valid conditions for achieving coherence in interference experiments,all statements are correct.

(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe for a wavelength $\lambda$ is given by $y_n = \frac{n \lambda D}{d}$.
For two wavelengths $\lambda_1 = 2500 \,\mathring{A}$ and $\lambda_2 = 3500 \,\mathring{A}$ to coincide at the same position $y$,we have:
$n_1 \lambda_1 = n_2 \lambda_2$
Substituting the given values:
$n_1 \times 2500 = n_2 \times 3500$
$\frac{n_1}{n_2} = \frac{3500}{2500} = \frac{7}{5}$
This implies that the $7^{th}$ order fringe of the $1^{st}$ source $(\lambda_1 = 2500 \,\mathring{A})$ coincides with the $5^{th}$ order fringe of the $2^{nd}$ source $(\lambda_2 = 3500 \,\mathring{A})$.