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

(C) The intensity of light in an interference pattern is given by $I_{res} = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
At a path difference of $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
The intensity is $I = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Now,for a path difference of $\Delta x = \frac{\lambda}{4}$,the phase difference is $\phi' = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I' = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
Since $I = 4I_0$,then $I_0 = \frac{I}{4}$.
Substituting this into the expression for $I'$,we get $I' = 2(\frac{I}{4}) = 0.5I$.

(A) The maximum intensity $I_{max}$ and minimum intensity $I_{min}$ in an interference pattern are given by $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Given that $I_{max} = 4 I_{min}$,we have $(\sqrt{I_1} + \sqrt{I_2})^2 = 4(\sqrt{I_1} - \sqrt{I_2})^2$.
Taking the square root on both sides: $\sqrt{I_1} + \sqrt{I_2} = 2(\sqrt{I_1} - \sqrt{I_2})$.
Rearranging the terms: $\sqrt{I_1} + \sqrt{I_2} = 2\sqrt{I_1} - 2\sqrt{I_2}$.
This simplifies to $3\sqrt{I_2} = \sqrt{I_1}$.
Squaring both sides gives $9I_2 = I_1$,so the ratio $\frac{I_1}{I_2} = 9/1$.
The ratio of the intensities of the two waves is $I_2/I_1 = 1/9$.

(C) The position of the $n^{th}$ order maximum is $x_n = n \frac{D \lambda}{d}$.
The position of the $n^{th}$ order minimum is $x'_n = (n - 0.5) \frac{D \lambda}{d}$.
The distance between the $3^{rd}$ order maximum $(n=3)$ and the $3^{rd}$ order minimum $(n=3)$ is:
$\Delta x = x_3 - x'_3 = 3 \frac{D \lambda}{d} - (3 - 0.5) \frac{D \lambda}{d} = 0.5 \frac{D \lambda}{d} = \frac{\beta}{2}$.
Given $\Delta x = 0.18 \,cm = 1.8 \times 10^{-3} \,m$, $D = 1.2 \,m$, and $d = 0.02 \,cm = 2 \times 10^{-4} \,m$.
$\frac{\beta}{2} = 1.8 \times 10^{-3} \,m \implies \beta = 3.6 \times 10^{-3} \,m$.
Since $\beta = \frac{D \lambda}{d}$, we have $\lambda = \frac{\beta d}{D}$.
$\lambda = \frac{(3.6 \times 10^{-3} \,m) \times (2 \times 10^{-4} \,m)}{1.2 \,m} = 6 \times 10^{-7} \,m = 600 \,nm$.

(A) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.
Taking the logarithmic differentiation,we get the relative error formula: $\frac{\Delta \beta}{\beta} = \frac{\Delta D}{D} - \frac{\Delta d}{d}$.
Given that $D$ is increased by $0.5 \%$,so $\frac{\Delta D}{D} \times 100 = 0.5 \%$.
Given that $d$ is decreased by $0.3 \%$,so $\frac{\Delta d}{d} \times 100 = -0.3 \%$.
Substituting these values into the error formula:
$\frac{\Delta \beta}{\beta} \times 100 = 0.5 \% - (-0.3 \%) = 0.5 \% + 0.3 \% = 0.8 \%$.
Therefore,the fringe width increases by $0.8 \%$.

(C) In Young's double slit experiment,white light consists of wavelengths ranging from $4000 \ Å$ to $7000 \ Å$.
At the central position on the screen,the path difference for all wavelengths is zero.
Since the path difference is zero,all wavelengths interfere constructively at the center,resulting in a white fringe.
As we move away from the center,the path difference increases,causing different wavelengths to interfere constructively at different positions,resulting in coloured fringes.

(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 the entire arrangement is placed in a liquid of refractive index $n$, the wavelength of light changes to $\lambda' = \frac{\lambda}{n}$.
Since $D$ and $d$ remain unchanged, the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{(\lambda / n) D}{d} = \frac{1}{n} \left( \frac{\lambda D}{d} \right) = \frac{\beta}{n}$.
Therefore, the new fringe width is $\frac{\beta}{n}$.

(A) The condition for the $n$-th bright fringe is given by $y = n \frac{\lambda D}{d}$.
For the bright fringes of both wavelengths $\lambda_1 = 650 \ nm$ and $\lambda_2 = 550 \ nm$ to coincide,their positions must be equal: $y_1 = y_2$.
$n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$.
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$.
Since we want the least distance,we take the smallest integers $n_1 = 11$ and $n_2 = 13$.
Substituting $n_1 = 11$ into the position formula:
$y = 11 \times \frac{650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$.
$y = 11 \times 325 \times 1.2 \times 10^{-6} = 4290 \times 10^{-6} \ m = 429 \times 10^{-5} \ m$.

(B) Given: Slit separation $d = 2 \ mm = 2 \times 10^{-3} \ m$,Screen distance $D = 10 \ m$,Wavelength $\lambda = 6000 \ \mathring{A} = 6 \times 10^{-7} \ m$.
In $Young's$ double slit experiment,the intensity at any point on the screen is given by $I = 4 I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
The path difference $\Delta x$ at a position $y$ on the screen is $\Delta x = \frac{yd}{D}$.
For a point in front of one of the slits,$y = \frac{d}{2}$.
Thus,$\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.
Substituting the values: $\Delta x = \frac{(2 \times 10^{-3})^2}{2 \times 10} = \frac{4 \times 10^{-6}}{20} = 2 \times 10^{-7} \ m$.
The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$.
The intensity $I = I_{max} \cos^2(\frac{\phi}{2}) = (4 I_0) \cos^2(\frac{2\pi/3}{2}) = 4 I_0 \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $I = 4 I_0 (\frac{1}{2})^2 = 4 I_0 \times \frac{1}{4} = I_0$.

(C) The angular fringe width $\theta$ in a Young's double slit experiment is given by the formula $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of the light used and $d$ is the distance between the two slits.
Statement $I$: The angular fringe width depends only on the wavelength $\lambda$ and the slit separation $d$. It is independent of the distance $D$ between the screen and the slits. Therefore,moving the screen away does not change the angular separation. Thus,Statement $I$ is false.
Statement $II$: Since $\theta = \frac{\lambda}{d}$,the angular fringe width is directly proportional to the wavelength $\lambda$. If the source is replaced by one with a higher wavelength,$\theta$ will increase. Thus,Statement $II$ is true.
Conclusion: Statement $I$ is false and Statement $II$ is true.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{D\lambda}{d}$.
Given that the fringe widths are equal,we have $\beta_1 = \beta_2$.
Therefore,$\frac{D_1 \lambda_1}{d_1} = \frac{D_2 \lambda_2}{d_2}$.
Rearranging the terms to find the ratio $\frac{D_1}{D_2}$,we get $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}$.
Given the ratios of slit separations $\frac{d_1}{d_2} = 2$ and the ratio of wavelengths $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$,which implies $\frac{\lambda_2}{\lambda_1} = 2$.
Substituting these values,we get $\frac{D_1}{D_2} = 2 \times 2 = 4$.

(B) Given: Slit separation $d = 0.54$ mm $= 0.54 \times 10^{-3}$ m.
Distance of screen $D = 1.8$ m.
Distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = \frac{n\lambda D}{d}$.
For the $6^{th}$ bright fringe,$n = 6$ and $y_6 = 1.2$ cm $= 1.2 \times 10^{-2}$ m.
Rearranging the formula for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.2 \times 10^{-2} \times 0.54 \times 10^{-3}}{6 \times 1.8}$.
$\lambda = \frac{0.648 \times 10^{-5}}{10.8} = 0.06 \times 10^{-5}$ m.
$\lambda = 600 \times 10^{-9}$ m $= 600$ nm.

(B) Given: Slit separation $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$.
Distance to screen $D = 2.0 \text{ m}$.
Order of fringe $n = 3$.
Distance of $n^{th}$ bright fringe $y_n = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$.
The formula for the position of the $n^{th}$ bright fringe is $y_n = \frac{n \lambda D}{d}$.
Rearranging for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.5 \times 10^{-2} \times 2 \times 10^{-4}}{3 \times 2.0}$.
$\lambda = \frac{3 \times 10^{-6}}{6} = 0.5 \times 10^{-6} \text{ m}$.
Converting to $ \mathring{A}$s: $\lambda = 0.5 \times 10^{-6} \times 10^{10} \mathring{A} = 5000 \mathring{A}$.
Thus,the wavelength of light used is $5000 \mathring{A}$.

(B) The fringe width $\beta$ in a medium of refractive index $\mu$ is given by the formula $\beta' = \frac{\beta}{\mu}$.
Given the initial fringe width $\beta = 2.4 \text{ } \mu\text{m}$ and the refractive index of the new medium $\mu = 1.2$.
Substituting these values into the formula,we get $\beta' = \frac{2.4}{1.2} = 2 \text{ } \mu\text{m}$.
Therefore,the new fringe width is $2 \text{ } \mu\text{m}$.

(C) Let $I_s$ be the intensity due to an individual slit.
The resultant intensity $I_R$ is given by the formula: $I_R = I_s + I_s + 2 \sqrt{I_s I_s} \cos \phi = 2I_s + 2I_s \cos \phi = 2I_s(1 + \cos \phi) = 4I_s \cos^2(\phi/2)$.
Given that the resultant intensity $I_R$ at point $P$ is equal to the intensity due to an individual slit $(I_s)$,we have:
$4I_s \cos^2(\phi/2) = I_s$
$\cos^2(\phi/2) = 1/4$
$\cos(\phi/2) = 1/2$
This implies $\phi/2 = \pi/3$,so the phase difference $\phi = 2\pi/3$.
The path difference $\Delta x$ is related to the phase difference $\phi$ by the formula $\Delta x = (\lambda / 2\pi) \phi$.
Substituting the given values: $\Delta x = (6000 \mathring{A} / 2\pi) \times (2\pi/3) = 2000 \mathring{A}$.

(C) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given $I = \frac{3}{4} I_{max}$,we have $\frac{3}{4} I_{max} = I_{max} \cos^2(\frac{\phi}{2})$.
This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{3}{4}$,which implies $\cos(\frac{\phi}{2}) = \frac{\sqrt{3}}{2}$.
Therefore,$\frac{\phi}{2} = \frac{\pi}{6}$,which gives the phase difference $\phi = \frac{\pi}{3}$.
The relationship between path difference $\Delta x$ and phase difference $\phi$ is $\Delta x = \frac{\lambda}{2\pi} \phi$.
Substituting $\phi = \frac{\pi}{3}$,we get $\Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{3} = \frac{\lambda}{6}$.
Comparing this with $\frac{\lambda}{x}$,we find $x = 6$.

(C) The intensity of light in Young's double slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = \frac{\lambda}{3}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
The intensity at this point is $I_1 = I_{max} \cos^2(\frac{2\pi/3}{2}) = I_{max} \cos^2(\frac{\pi}{3}) = I_{max} (\frac{1}{2})^2 = \frac{I_{max}}{4}$.
Given $I_1 = K$,we have $K = \frac{I_{max}}{4}$,which implies $I_{max} = 4K$.
For path difference $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity at this point is $I_2 = I_{max} \cos^2(\frac{2\pi}{2}) = I_{max} \cos^2(\pi) = I_{max} (-1)^2 = I_{max}$.
Since $I_{max} = 4K$,the intensity at path difference $\lambda$ is $4K$.