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

(A) The intensity at any point on the screen is given by $I = 4I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
The phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x_1 = 0$,the phase difference $\phi_1 = 0$. Thus,$I_1 = 4I_0 \cos^2(0) = 4I_0$.
For path difference $\Delta x_2 = \frac{\lambda}{4}$,the phase difference $\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity $I_2 = 4I_0 \cos^2 \left( \frac{\pi/2}{2} \right) = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$ or $2:1$.

(B) The position of the $n^{th}$ dark fringe in Young's double slit experiment is given by:
$x_n = \frac{D}{d} (2n - 1) \frac{\lambda}{2}$
Rearranging for wavelength $\lambda$:
$\lambda = \frac{2 x_n d}{D(2n - 1)} \quad \dots (i)$
Given that the fifth dark fringe $(n=5)$ is formed opposite to one of the slits,the distance of this fringe from the central axis is half the distance between the slits:
$x_5 = \frac{d}{2} \implies 2x_5 = d$
Substituting $n=5$ and $2x_5 = d$ into equation $(i)$:
$\lambda = \frac{(2x_5) d}{D(2 \times 5 - 1)}$
$\lambda = \frac{d \cdot d}{D(10 - 1)}$
$\lambda = \frac{d^2}{9D}$

(C) The position of the slits is at $y = \pm d/2$. The second minimum occurs at $y = d/2$.
For a minimum in Young's double-slit experiment,the path difference is given by $\Delta x = (n - 1/2)\lambda$,where $n = 2$ for the second minimum.
Thus,$\Delta x = (2 - 1/2)\lambda = (3/2)\lambda$.
Also,the path difference is given by $\Delta x = d \sin \theta \approx d \tan \theta = d(y/D)$.
Substituting $y = d/2$,we get $\Delta x = d(d/2D) = d^2 / 2D$.
Equating the two expressions for path difference: $(3/2)\lambda = d^2 / 2D$.
Solving for $\lambda$,we get $\lambda = (d^2 / 2D) \times (2/3) = d^2 / 3D$.

(A) The position of the $n^{th}$ bright fringe (maxima) is given by the formula:
$y_n = \frac{n D \lambda}{d}$
Rearranging the formula to solve for wavelength $\lambda$:
$\lambda = \frac{y_n d}{n D}$
Given values:
$n = 3$
$y_n = 1.2 \ mm = 1.2 \times 10^{-3} \ m$
$D = 1 \ m$
$d = 1 \ mm = 1 \times 10^{-3} \ m$
Substituting these values into the formula:
$\lambda = \frac{(1.2 \times 10^{-3} \ m) \times (1 \times 10^{-3} \ m)}{3 \times 1 \ m}$
$\lambda = \frac{1.2 \times 10^{-6}}{3} \ m$
$\lambda = 0.4 \times 10^{-6} \ m = 4 \times 10^{-7} \ m$
Converting to $\mathring{A}$s $(\mathring{A})$, where $1 \ \mathring{A} = 10^{-10} \ m$:
$\lambda = 4000 \times 10^{-10} \ m = 4000 \ \mathring{A}$

(B) For constructive interference (bright bands),the path difference is $\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$
For destructive interference (dark bands),the path difference is $\Delta x = (n + \frac{1}{2})\lambda$,where $n = 0, 1, 2, \dots$
The $4^{\text{th}}$ bright band corresponds to $n = 4$ $(\Delta x = 4\lambda)$ and the $5^{\text{th}}$ bright band corresponds to $n = 5$ $(\Delta x = 5\lambda)$.
$A$ dark band formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band corresponds to $n = 4$ in the dark band formula: $\Delta x = (4 + \frac{1}{2})\lambda = 4.5\lambda$.
Given $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} = 6 \times 10^{-5} \text{ cm}$.
Path difference $\Delta x = 4.5 \times (6 \times 10^{-5} \text{ cm}) = 27 \times 10^{-5} \text{ cm} = 2.7 \times 10^{-4} \text{ cm}$.

(A) The intensity of light $I$ is directly proportional to the slit width $w$ $(I \propto w)$.
Given the ratio of slit widths is $w_1 / w_2 = 1 / 9$,the ratio of intensities of the two sources is $I_1 / I_2 = 1 / 9$.
Since intensity $I \propto a^2$ (where $a$ is the amplitude),the ratio of amplitudes is $a_1 / a_2 = \sqrt{I_1 / I_2} = \sqrt{1 / 9} = 1 / 3$.
Let $a_1 = a$ and $a_2 = 3a$.
The ratio of minimum intensity to maximum intensity is given by the formula:
$\frac{I_{\text{min}}}{I_{\text{max}}} = \left( \frac{a_1 - a_2}{a_1 + a_2} \right)^2$
Substituting the values:
$\frac{I_{\text{min}}}{I_{\text{max}}} = \left( \frac{a - 3a}{a + 3a} \right)^2 = \left( \frac{-2a}{4a} \right)^2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4}$.

(A) The fringe width in air is given by $\beta = \frac{\lambda D}{d} = 2 \,mm$.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Consequently,the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Given $\beta = 2 \,mm$ and $\mu = 1.33$,we have $\beta' = \frac{2}{1.33} \approx 1.5 \,mm$.
The change in fringe width is $\Delta \beta = \beta - \beta' = 2 \,mm - 1.5 \,mm = 0.5 \,mm$.

(A) The fringe width is given by $\beta = \frac{\lambda D}{d} = 0.4 \,mm$.
The position of the $n_{1}$-th bright fringe from the central maximum is given by $x_{n_1} = n_1 \beta$.
For the fifth bright fringe $(n_1 = 5)$:
$x_5 = 5 \beta = 5 \times 0.4 \,mm = 2.0 \,mm$.
The position of the $n_2$-th dark fringe from the central maximum is given by $x_{n_2} = (n_2 - 0.5) \beta$.
For the third dark fringe $(n_2 = 3)$:
$x_3 = (3 - 0.5) \beta = 2.5 \beta = 2.5 \times 0.4 \,mm = 1.0 \,mm$.
The distance between the fifth bright and third dark band on the same side is:
$\Delta x = x_5 - x_3 = 2.0 \,mm - 1.0 \,mm = 1.0 \,mm$.

(D) The condition for the $n^{th}$ bright fringe in a Young's double-slit experiment is given by $y = n \frac{\lambda D}{d}$.
Since the position $y$ is the same for both cases, we have $n_{1} \lambda_{1} = n_{2} \lambda_{2}$.
Given: $n_{1} = 3$, $\lambda_{1} = 700 \, nm$, and $n_{2} = 5$.
Substituting the values: $3 \times 700 = 5 \times \lambda_{2}$.
$\lambda_{2} = \frac{3 \times 700}{5} = 3 \times 140 = 420 \, nm$.

(B) In an interference experiment,the distance between two consecutive bright fringes (maxima) or two consecutive dark fringes (minima) is known as the fringe width,denoted by $\beta$.
According to the theory of Young's double-slit experiment,the fringe width is given by the formula:
$\beta = \frac{D \lambda}{d}$
where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength of the light used,and $d$ is the distance between the two slits.

(C) Given:
Slit separation $d = 0.28 \ mm = 0.28 \times 10^{-3} \ m$
Distance to screen $D = 1.4 \ m$
Distance of $n^{th}$ bright fringe $x_n = 1.2 \ cm = 1.2 \times 10^{-2} \ m$
Order of fringe $n = 4$
The formula for the position of the $n^{th}$ bright fringe is given by:
$x_n = \frac{n \lambda D}{d}$
Rearranging for wavelength $\lambda$:
$\lambda = \frac{x_n d}{n D}$
Substituting the values:
$\lambda = \frac{(1.2 \times 10^{-2} \ m) \times (0.28 \times 10^{-3} \ m)}{4 \times 1.4 \ m}$
$\lambda = \frac{0.336 \times 10^{-5}}{5.6} \ m$
$\lambda = 0.06 \times 10^{-5} \ m = 6 \times 10^{-7} \ m$
Converting to nanometers $(1 \ nm = 10^{-9} \ m)$:
$\lambda = 600 \times 10^{-9} \ m = 600 \ \text{nm}$
Therefore,the correct option is $C$.

(A) The fringe width (fringe separation) $\beta$ is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given:
Wavelength $\lambda = 7000 \ \mathring{A} = 7000 \times 10^{-10} \ m = 7 \times 10^{-7} \ m$.
Distance between slits $d = 10 \ mm = 10 \times 10^{-3} \ m = 10^{-2} \ m$.
Distance to the screen $D = 1.5 \ m$.
Substituting these values into the formula:
$\beta = \frac{7 \times 10^{-7} \times 1.5}{10^{-2}}$
$\beta = 7 \times 1.5 \times 10^{-5} \ m$
$\beta = 10.5 \times 10^{-5} \ m$
$\beta = 105 \times 10^{-6} \ m$
$\beta = 105 \ \mu m$.
Therefore,the correct option is $A$.

(C) The position of the $n^{th}$ bright fringe in a Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
Given:
$\lambda = 500 \, nm = 500 \times 10^{-9} \, m = 5 \times 10^{-7} \, m$
$D = 100 \, cm = 1 \, m$
$d = 1 \, mm = 10^{-3} \, m$
The distance between the fifth $(n=5)$ and third $(n=3)$ bright fringes is $\Delta y = y_5 - y_3$.
$\Delta y = \frac{5 \lambda D}{d} - \frac{3 \lambda D}{d} = \frac{2 \lambda D}{d}$.
Substituting the values:
$\Delta y = \frac{2 \times (5 \times 10^{-7} \, m) \times (1 \, m)}{10^{-3} \, m}$.
$\Delta y = 10 \times 10^{-4} \, m = 10^{-3} \, m$.
$\Delta y = 1 \, mm$.

(C) The fringe width (distance between two consecutive fringes) is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given values are:
$\lambda = 500 \ nm = 500 \times 10^{-9} \ m = 5 \times 10^{-7} \ m$
$D = 2 \ m$
$d = 3 \ mm = 3 \times 10^{-3} \ m$
Substituting these values into the formula:
$\beta = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-3}}$
$\beta = \frac{10 \times 10^{-7}}{3 \times 10^{-3}}$
$\beta = \frac{10}{3} \times 10^{-4} \ m$
$\beta = 3.33 \times 10^{-4} \ m$
$\beta = 0.333 \times 10^{-3} \ m = 0.33 \ mm$.
Therefore, the correct option is $C$.

(B) The condition for the $n$-th bright fringe in Young's Double Slit Experiment is given by $y_n = \frac{n \lambda D}{d}$.
Since the fringes superpose,their positions must be equal: $y_4 = y_5$.
Therefore,$n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 4$,$\lambda_1 = 5000 \ \mathring{A}$,and $n_2 = 5$.
Substituting the values: $4 \times 5000 = 5 \times \lambda_2$.
$\lambda_2 = \frac{4 \times 5000}{5} = 4000 \ \mathring{A}$.

(B) In an ideal Young's Double Slit Experiment $(YDSE)$,the intensity of all bright fringes is considered equal. However,in real-world interference patterns,the intensity of bright fringes decreases as we move away from the central maximum due to the diffraction effect caused by the finite width of the slits. Therefore,the statement that 'all bright fringes are equally bright' is technically incorrect in a practical physical context compared to the other fundamental properties of interference.

(B) For bright fringes in Young's double slit experiment,the condition is given by $d \sin \theta = n \lambda$,where $d$ is the slit separation,$\theta$ is the angle,$n$ is the order of the fringe,and $\lambda$ is the wavelength.
Given that $d = \lambda$,the equation becomes $\lambda \sin \theta = n \lambda$,which simplifies to $\sin \theta = n$.
Since the maximum value of $\sin \theta$ is $1$,we have $n \leq 1$.
The possible integer values for $n$ are $-1, 0, 1$.
Thus,there are $3$ bright fringes in total.

(A) The distance between any two consecutive bright fringes is known as fringe width,denoted by $\beta$.
The formula for fringe width is $\beta = \frac{\lambda D}{d}$.
Given values are:
Wavelength $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Distance between slits and screen $D = 1.5 \ m$.
Distance between slits $d = 0.2 \ mm = 0.2 \times 10^{-3} \ m$.
Substituting these values into the formula:
$\beta = \frac{600 \times 10^{-9} \times 1.5}{0.2 \times 10^{-3}}$
$\beta = \frac{900 \times 10^{-9}}{0.2 \times 10^{-3}}$
$\beta = 4500 \times 10^{-6} \ m$
$\beta = 4.5 \times 10^{-3} \ m = 4.5 \ mm$.
Thus,the distance between any two consecutive bright fringes is $4.5 \ mm$.

(D) The fringe width is given as $\beta = 0.002 \text{ cm}$.
For a Young's double-slit interference pattern,the distance of the $n^{\text{th}}$ dark fringe from the central bright fringe is given by the formula $x_n = (n - 0.5) \beta$,where $n$ is the order of the dark fringe $(n = 1, 2, 3, \dots)$.
For the $5^{\text{th}}$ dark fringe,$n = 5$.
Substituting the values: $x_5 = (5 - 0.5) \times 0.002 \text{ cm}$.
$x_5 = 4.5 \times 0.002 \text{ cm} = 0.009 \text{ cm}$.
Converting to scientific notation: $0.009 \text{ cm} = 0.9 \times 10^{-2} \text{ cm}$.
Comparing this result with the given options,none of the options match the calculated value.
Therefore,the correct choice is $D$.

(C) Given: Wavelength $\lambda = 500 \ nm = 500 \times 10^{-9} \ m$,Distance $D = 0.5 \ m$,Slit separation $d = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
Position of $n^{th}$ maxima is $x_n = \frac{n \lambda D}{d}$. For $n=3$,$x_3 = \frac{3 \lambda D}{d}$.
Position of $m^{th}$ minima on the other side is $x'_m = \frac{(2m-1) \lambda D}{2d}$. For $m=2$,$x'_2 = \frac{(2 \times 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$.
The total distance between them is $x = x_3 + x'_2 = \frac{3 \lambda D}{d} + \frac{3 \lambda D}{2d} = \frac{9 \lambda D}{2d}$.
Substituting the values: $x = \frac{9 \times 500 \times 10^{-9} \times 0.5}{2 \times 0.5 \times 10^{-3}} = \frac{4500 \times 10^{-9}}{2 \times 10^{-3}} = 2250 \times 10^{-6} \ m = 2.25 \ mm$.

(B) In Young's double-slit experiment,the fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.
From this relation,we can see that the fringe width is directly proportional to the wavelength of the light used,i.e.,$\beta \propto \lambda$.
The wavelength of yellow light $(\lambda_{yellow})$ is greater than the wavelength of blue light $(\lambda_{blue})$.
Since the wavelength decreases when replacing yellow light with blue light,the fringe width $\beta$ will also decrease.
Therefore,the interference fringes become narrower.

(D) In Young's double slit experiment, the fringe width is given by $\beta = \frac{D \lambda}{d}$.
Since the experimental setup ($D$ and $d$) remains the same, $\beta \propto \lambda$.
According to the de Broglie hypothesis, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
For the same speed $v$, the wavelength is inversely proportional to the mass $m$ of the particle, i.e., $\lambda \propto \frac{1}{m}$.
Since the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$, we have $m_p > m_e$.
Therefore, the wavelength of the proton beam $(\lambda_2)$ will be smaller than the wavelength of the electron beam $(\lambda_1)$, i.e., $\lambda_2 < \lambda_1$.
Since $\beta \propto \lambda$, it follows that $\beta_2 < \beta_1$, or $\beta_1 > \beta_2$.

(C) Given,$I_0 = 2 \times 10^{-2} \ W \ m^{-2}$ and $x = \frac{\beta}{3}$.
In $YDSE$,the path difference $\Delta x$ at a point $x$ is given by $\Delta x = \frac{xd}{D}$.
Since the fringe width $\beta = \frac{\lambda D}{d}$,we have $\frac{d}{D} = \frac{\lambda}{\beta}$.
Substituting this into the path difference formula: $\Delta x = x \times \frac{\lambda}{\beta} = \frac{\beta}{3} \times \frac{\lambda}{\beta} = \frac{\lambda}{3}$.
The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
The resultant intensity $I$ is given by $I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)$.
Substituting the values: $I = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I = 4I_0 \times (\frac{1}{2})^2 = 4I_0 \times \frac{1}{4} = I_0$.
Therefore,$I = 2 \times 10^{-2} \ W \ m^{-2}$.

(A) We know that the fringe width $\beta$ is given by the formula: $\beta = \frac{D \lambda}{d}$, where $D$ is the distance between the screen and the source, $d$ is the distance between the slits, and $\lambda$ is the wavelength of light used.
From this relation, we can see that $\beta \propto \lambda$.
We know that the wavelength of red light $(\lambda_{\text{red}} \approx 700 \, nm)$ is approximately twice the wavelength of violet light $(\lambda_{\text{violet}} \approx 400 \, nm$, though the ratio is often approximated as $2$ in physics problems).
Therefore, the ratio of fringe widths is $\frac{\beta_{\text{red}}}{\beta_{\text{violet}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} \approx \frac{700}{400} \approx 1.75$, which is approximately $2$.
Thus, $\beta_{\text{red}} \approx 2 \beta_{\text{violet}}$.

(B) The path difference between the waves reaching point $P$ from the two slits is given by $\Delta x = \frac{xd}{D}$.
The corresponding phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{xd}{D}\right)$.
The resultant intensity $I_P$ at point $P$ for two coherent sources of equal intensity $I_0$ is given by the formula $I_P = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$.
Substituting the value of $\phi$:
$I_P = 4I_0 \cos^2\left(\frac{1}{2} \cdot \frac{2\pi xd}{\lambda D}\right)$
$I_P = 4I_0 \cos^2\left(\frac{\pi dx}{\lambda D}\right)$.

(D) Given: Distance between slits and screen $D = 1.2 \, m$, distance between slits $d = 2.4 \, mm = 2.4 \times 10^{-3} \, m$, thickness of mica sheet $t = 1 \, \mu m = 1 \times 10^{-6} \, m$, and refractive index $\mu = 1.5$.
When a transparent sheet is introduced in the path of one of the interfering beams, the shift in the position of the central bright fringe is given by the formula:
$y = (\mu - 1) t \frac{D}{d}$
Substituting the given values:
$y = (1.5 - 1) \times (1 \times 10^{-6} \, m) \times \frac{1.2 \, m}{2.4 \times 10^{-3} \, m}$
$y = 0.5 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}}$
$y = 0.5 \times 10^{-6} \times 0.5 \times 10^{3}$
$y = 0.25 \times 10^{-3} \, m = 0.25 \, mm$
Thus, the shift in the position of the central bright fringe is $0.25 \, mm$.

(D) The intensity $ I $ at any point is given by $ I = I_{max} \cos^2 \left( \frac{\phi}{2} \right) $, where $ \phi $ is the phase difference.
Given that at path difference $ \Delta x = \lambda $, the intensity is $ K $. Since $ \Delta x = \lambda $ corresponds to a phase difference $ \phi = 2\pi $, we have $ K = I_{max} \cos^2 \left( \frac{2\pi}{2} \right) = I_{max} \cos^2(\pi) = I_{max} $.
Now, for a path difference $ \Delta x = \frac{\lambda}{3} $, the phase difference is $ \phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} $.
The intensity at this point is $ I = I_{max} \cos^2 \left( \frac{\phi}{2} \right) = K \cos^2 \left( \frac{2\pi/3}{2} \right) $.
$ I = K \cos^2 \left( \frac{\pi}{3} \right) = K \left( \frac{1}{2} \right)^2 = \frac{K}{4} $.

(C) The condition for the $n^{\text{th}}$ bright fringe in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
Given that the $n^{\text{th}}$ bright band of $\lambda_{1}$ coincides with the $(n+1)^{\text{th}}$ bright band of $\lambda_{2}$,we have:
$\frac{n \lambda_{1} D}{d} = \frac{(n+1) \lambda_{2} D}{d}$
Canceling the common terms $\frac{D}{d}$,we get:
$n \lambda_{1} = (n+1) \lambda_{2}$
Substituting the given values $\lambda_{1} = 780 \ nm$ and $\lambda_{2} = 520 \ nm$:
$n(780) = (n+1)(520)$
$780n = 520n + 520$
$780n - 520n = 520$
$260n = 520$
$n = \frac{520}{260} = 2$
Thus,the value of $n$ is $2$.

(A) The position of the $n^{th}$ bright fringe is given by the formula:
$y = \frac{n \lambda D}{d}$
where $\lambda$ is the wavelength,$D$ is the distance of the screen from the slits,and $d$ is the separation between the slits.
For the fourth bright fringe $(n=4)$,the positions for the two wavelengths are:
$y_1 = \frac{4 \lambda_1 D}{d}$ and $y_2 = \frac{4 \lambda_2 D}{d}$
The separation between these two fringes is:
$\Delta y = y_1 - y_2 = \frac{4 D}{d} (\lambda_1 - \lambda_2)$
Given values:
$n = 4$
$D = 1.2 \,m$
$d = 2 \,mm = 2 \times 10^{-3} \,m$
$\lambda_1 = 6500 \text{ Å} = 6500 \times 10^{-10} \,m$
$\lambda_2 = 5200 \text{ Å} = 5200 \times 10^{-10} \,m$
Substituting these values into the formula:
$\Delta y = \frac{4 \times 1.2}{2 \times 10^{-3}} \times (6500 - 5200) \times 10^{-10} \,m$
$\Delta y = \frac{4.8}{2 \times 10^{-3}} \times 1300 \times 10^{-10} \,m$
$\Delta y = 2.4 \times 10^3 \times 1300 \times 10^{-10} \,m$
$\Delta y = 3120 \times 10^{-7} \,m = 0.312 \times 10^{-3} \,m$
$\Delta y = 0.312 \,mm$
Thus,the separation between the fourth bright fringes is $0.312 \,mm$.

(D) The conditions for sustained interference are as follows: The sources must be coherent,meaning they should have the same frequency (wavelength) and a constant phase difference.
In this experiment,one slit is covered with a red filter (transmitting only red light) and the other with a blue filter (transmitting only blue light).
Since the wavelengths of red light and blue light are significantly different,the two sources are incoherent.
Because the sources are incoherent,they cannot produce a stable interference pattern on the screen.
Therefore,no interference pattern will be observed.

(C) The fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
When the screen distance $D$ is changed by $\Delta D$, the change in fringe width $\Delta \beta$ is given by $\Delta \beta = \frac{\lambda \Delta D}{d}$.
Given:
$\Delta D = 5 \times 10^{-2} \,m$
$\Delta \beta = 3 \times 10^{-5} \,m$
$d = 1 \times 10^{-3} \,m$
Rearranging the formula for wavelength $\lambda$:
$\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$
Substituting the values:
$\lambda = \frac{(3 \times 10^{-5} \,m) \times (1 \times 10^{-3} \,m)}{5 \times 10^{-2} \,m}$
$\lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} \,m$
$\lambda = 0.6 \times 10^{-6} \,m = 6 \times 10^{-7} \,m$
Converting to nanometers $(1 \,nm = 10^{-9} \,m)$:
$\lambda = 600 \times 10^{-9} \,m = 600 \,nm$.

(B) For sustained interference,the sources must be coherent,which is a necessary condition for a stable interference pattern.
Additionally,for high-contrast fringes (where the intensity at the minima is zero),the intensities of the two sources must be equal.
Therefore,both conditions are essential for ideal,sustained interference fringes.

(A) In Young's double slit experiment,the intensity of light $I$ is proportional to the width of the slit $w$ $(I \propto w)$.
If the slits have unequal widths,the intensities of the light waves from the two slits,$I_{1}$ and $I_{2}$,will be unequal $(I_{1} \neq I_{2})$.
The intensity of the interference pattern is given by $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.
The minimum intensity is given by $I_{\min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^2$.
Since $I_{1} \neq I_{2}$,the value of $I_{\min}$ is greater than zero $(I_{\min} > 0)$.
Therefore,the dark fringes are not perfectly dark.

(D) In Young's double slit experiment,the primary requirement for interference is the presence of coherent sources.
Statement $(1)$ is correct: An ordinary extended source of light is incoherent,so a single slit $S$ is required to act as a point source to ensure spatial coherence.
Statement $(2)$ is incorrect: Even if a beam is well-collimated,it does not guarantee the spatial coherence required for the two slits to act as secondary coherent sources.
Statement $(3)$ is correct: If the source is already spatially coherent,the light waves arriving at the two slits maintain a constant phase difference,making the initial slit $S$ redundant.
Therefore,statements $(1)$ and $(3)$ are correct.

(B) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,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 of thickness $t$ and refractive index $\mu$ is introduced in the path of light from one of the slits,the optical path length increases by $(\mu - 1)t$.
This causes the entire interference pattern to shift by a distance $y_0 = \frac{D}{d}(\mu - 1)t$.
However,the fringe width $\beta$ depends only on the wavelength $\lambda$,the distance $D$,and the slit separation $d$. Since these parameters remain unchanged,the fringe width remains constant.
Therefore,the fringe width remains $0.3 \,mm$.

(D) In Young's double slit experiment,the interference pattern is formed by the superposition of light waves from two coherent sources. If a third slit is introduced between the two slits,it acts as an additional source of light. This additional light contributes to the background illumination on the screen,which does not participate in the interference pattern. As a result,the intensity of the dark fringes increases,while the intensity of the bright fringes remains relatively unchanged. This leads to a decrease in the visibility or contrast between the bright and dark fringes.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the slit separation.
When the screen is moved by $\Delta D = 5 \times 10^{-2} \, m$, the change in fringe width is $\Delta \beta = 3 \times 10^{-5} \, m$.
From the formula, the change in fringe width is $\Delta \beta = \frac{\lambda}{d} \Delta D$.
Rearranging for $\lambda$, we get $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.
Substituting the given values:
$\lambda = \frac{(3 \times 10^{-5} \, m) \times (10^{-3} \, m)}{5 \times 10^{-2} \, m}$.
$\lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} \, m = 0.6 \times 10^{-6} \, m = 6 \times 10^{-7} \, m$.
Converting to $\text{Å}$ $(1 \, \text{Å} = 10^{-10} \, m)$:
$\lambda = 6000 \times 10^{-10} \, m = 6000 \, \text{Å}$.

(A) In Young's double slit experiment,the half angular width $\theta$ of the central maximum is given by the relation $\sin \theta = \frac{\lambda}{d}$.
Given wavelength $\lambda = 589 \ nm = 589 \times 10^{-9} \ m$.
Given slit separation $d = 0.589 \ mm = 0.589 \times 10^{-3} \ m$.
Substituting the values:
$\sin \theta = \frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}$
$\sin \theta = \frac{589 \times 10^{-9}}{589 \times 10^{-6}} = 10^{-3} = 0.001$.
Therefore,$\theta = \sin^{-1}(0.001)$.

(B) The condition for the position of the $n^{\text{th}}$ bright fringe in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the $n^{\text{th}}$ bright fringe of red light $(\lambda_r = 7.5 \times 10^{-5} \text{ cm})$,the position is $y_r = \frac{n \lambda_r D}{d}$.
For the $(n+1)^{\text{th}}$ bright fringe of blue light $(\lambda_b = 5 \times 10^{-5} \text{ cm})$,the position is $y_b = \frac{(n+1) \lambda_b D}{d}$.
Since the fringes coincide,$y_r = y_b$,which implies $n \lambda_r = (n+1) \lambda_b$.
Substituting the given values: $n(7.5 \times 10^{-5}) = (n+1)(5 \times 10^{-5})$.
Dividing both sides by $10^{-5}$,we get $7.5n = 5(n+1)$.
$7.5n = 5n + 5$.
$2.5n = 5$.
$n = \frac{5}{2.5} = 2$.
Thus,the value of $n$ is $2$.

(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,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
Given that $\lambda' = \lambda + 0.20\lambda = 1.20\lambda$ and $d' = d - 0.25d = 0.75d$.
The new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d'} = \frac{1.20\lambda D}{0.75d}$.
Substituting $\beta = \frac{\lambda D}{d}$,we get $\beta' = \frac{1.20}{0.75} \beta = 1.6 \beta$.
Given the initial fringe width $\beta = 0.3 \ mm$,the final fringe width is $\beta' = 1.6 \times 0.3 \ mm = 0.48 \ mm$.

(A) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Here,the wavelength $\lambda = 5000 \ Å = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.
The distance between the screen and the slits $D = 1 \ m$.
The distance between the slits $d = 0.2 \ cm = 0.2 \times 10^{-2} \ m = 2 \times 10^{-3} \ m$.
Substituting these values into the formula:
$\beta = \frac{5 \times 10^{-7} \times 1}{2 \times 10^{-3}}$
$\beta = 2.5 \times 10^{-4} \ m$
Converting this to millimeters:
$\beta = 2.5 \times 10^{-4} \times 10^3 \ mm = 0.25 \ mm$.
Therefore,the distance between two consecutive dark fringes is $0.25 \ mm$.

(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 slits and the screen,and $d$ is the distance between the two slits.
From the formula,we can see that the fringe width is inversely proportional to the slit separation,i.e.,$\beta \propto \frac{1}{d}$.
Let the initial slit separation be $d_1 = d$ and the initial fringe width be $\beta_1 = \beta$.
Given that the final slit separation is $d_2 = 3d$.
Therefore,the final fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D}{3d} = \frac{\beta}{3}$.
The ratio of the initial fringe width to the final fringe width is $\frac{\beta_1}{\beta_2} = \frac{\beta}{\beta/3} = 3:1$.

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen from the slits,and $d$ is the distance between the slits.
Initially,$d_1 = 2 \ mm$,$D_1 = 100 \ cm = 1000 \ mm$,and $\beta_1 = 0.36 \ mm$.
Thus,$0.36 = \frac{\lambda \times 1000}{2} \implies \lambda = \frac{0.36 \times 2}{1000} = 0.00072 \ mm$.
Now,the new distance between the slits is $d_2 = 2 \ mm - 0.5 \ mm = 1.5 \ mm$.
The new distance of the screen is $D_2 = 100 \ cm + 50 \ cm = 150 \ cm = 1500 \ mm$.
The new fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D_2}{d_2}$.
Substituting the values: $\beta_2 = \frac{0.00072 \times 1500}{1.5} = \frac{0.00072 \times 1000}{1} = 0.72 \ mm$.

(B) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$.
Given initial fringe width $\beta_1 = \frac{\lambda D}{d} = 3 \text{ mm}$.
According to the problem, the new distance $D' = D + 0.5D = 1.5D$ and the new slit separation $d' = d - 0.1d = 0.9d$.
The new fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D'}{d'} = \frac{\lambda (1.5D)}{0.9d}$.
Simplifying this, $\beta_2 = \frac{1.5}{0.9} \times \frac{\lambda D}{d} = \frac{15}{9} \times \beta_1 = \frac{5}{3} \times 3 \text{ mm}$.
Therefore, $\beta_2 = 5 \text{ mm}$.

(D) In Young's double slit experiment $(YDSE)$,the path difference is given as $\Delta x = \frac{\lambda}{6}$.
Assuming the intensities of the two slits are equal,let $I_1 = I_2 = I_s$.
The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
The resultant intensity $I$ at any point is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\Delta \phi)$.
Substituting the values: $I = I_s + I_s + 2 \sqrt{I_s I_s} \cos(\frac{\pi}{3}) = 2 I_s + 2 I_s (\frac{1}{2}) = 2 I_s + I_s = 3 I_s$.
The maximum intensity $I_0$ occurs when $\cos(\Delta \phi) = 1$,so $I_0 = I_1 + I_2 + 2 \sqrt{I_1 I_2} = (\sqrt{I_s} + \sqrt{I_s})^2 = (2 \sqrt{I_s})^2 = 4 I_s$.
Therefore,the ratio $\frac{I}{I_0} = \frac{3 I_s}{4 I_s} = \frac{3}{4}$.

(C) In Young's double slit experiment $(YDSE)$,the fringe width $\beta$ is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given values are:
Slit separation,$d = 2 \ mm = 2 \times 10^{-3} \ m$.
Distance to screen,$D = 2 \ m$.
Wavelength of light,$\lambda = 400 \ nm = 400 \times 10^{-9} \ m$.
Substituting these values into the formula:
$\beta = \frac{(400 \times 10^{-9} \ m) \times (2 \ m)}{2 \times 10^{-3} \ m}$.
$\beta = \frac{800 \times 10^{-9}}{2 \times 10^{-3}} \ m$.
$\beta = 400 \times 10^{-6} \ m = 0.4 \times 10^{-3} \ m$.

(C) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{D \lambda}{d}$.
Given that the wavelength is increased by $50 \%$,the new wavelength is $\lambda' = \lambda + 0.50 \lambda = 1.50 \lambda$.
The distance between the slits is doubled,so $d' = 2d$.
The new fringe width $\beta'$ is $\beta' = \frac{D \lambda'}{d'} = \frac{D (1.50 \lambda)}{2d} = 0.75 \beta$.
The percentage change in fringe width is given by $\frac{\beta' - \beta}{\beta} \times 100 \%$.
Substituting the values: $\frac{0.75 \beta - \beta}{\beta} \times 100 \% = -0.25 \times 100 \% = -25 \%$.
The magnitude of the percentage change is $25 \%$.

(B) The fringe width $\beta$ in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
Given $\lambda_1 = 560 \,nm$ and $\beta_1 = 7.2 \,mm$.
For the second light,$\beta_2 = 8.1 \,mm$.
Since $D$ and $d$ are constant,we have $\beta \propto \lambda$,which implies $\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$.
Rearranging for $\lambda_2$,we get $\lambda_2 = \lambda_1 \times \frac{\beta_2}{\beta_1}$.
Substituting the values: $\lambda_2 = 560 \,nm \times \frac{8.1 \,mm}{7.2 \,mm} = 560 \times \frac{81}{72} = 560 \times \frac{9}{8} = 70 \times 9 = 630 \,nm$.
Thus,the wavelength of the second light is $630 \,nm$.