Single Slit Diffraction of Light Questions in English

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by $\theta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.
From this relation,we see that $\theta \propto \lambda$.
Since the wavelength of blue light $(\lambda_{B})$ is smaller than the wavelength of yellow light $(\lambda_{y})$,i.e.,$\lambda_{B} < \lambda_{y}$.
Therefore,the angular width of the diffraction pattern for blue light will be smaller than that for yellow light,meaning the fringes will become narrower.

(B) For the first minima in single slit diffraction,the condition is given by $a \sin \theta = n \lambda$,where $n = 1$.
Given: $\theta = 30^o$,$\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
Substituting the values into the formula:
$a \sin(30^o) = 1 \times (5 \times 10^{-7} \, m)$
$a \times (0.5) = 5 \times 10^{-7} \, m$
$a = 10 \times 10^{-7} \, m = 10^{-6} \, m$.
Converting the width $a$ into centimeters:
$a = 10^{-6} \, m = 10^{-6} \times 10^2 \, cm = 10^{-4} \, cm$.
Thus,the width of the slit is $1.0 \times 10^{-4} \, cm$.

(C) For single slit diffraction,the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$.
Given: wavelength $\lambda = 6500 \, \mathring{A} = 6500 \times 10^{-10} \, \text{m}$,angle $\theta = 30^{\circ}$,and for the first minimum,$n = 1$.
Substituting the values into the formula:
$a \sin 30^{\circ} = 1 \times 6500 \, \mathring{A}$
$a \times 0.5 = 6500 \, \mathring{A}$
$a = 13000 \, \mathring{A}$
Converting to micrometers:
$a = 13000 \times 10^{-10} \, \text{m} = 1.3 \times 10^{-6} \, \text{m} = 1.3 \, \mu\text{m}$.

(C) The width of the central bright fringe in a single-slit diffraction pattern is given by the formula: $w = \frac{2 D \lambda}{a}$.
Given values:
Slit width $(a)$ $= 0.20 \, mm = 0.20 \times 10^{-3} \, m$.
Wavelength $(\lambda)$ $= 500 \, nm = 500 \times 10^{-9} \, m$.
Distance to screen $(D)$ $= 80 \, cm = 0.8 \, m$.
Substituting these values into the formula:
$w = \frac{2 \times 0.8 \times 500 \times 10^{-9}}{0.20 \times 10^{-3}}$
$w = \frac{1.6 \times 500 \times 10^{-9}}{0.20 \times 10^{-3}}$
$w = 8 \times 500 \times 10^{-6} \, m$
$w = 4000 \times 10^{-6} \, m = 4 \times 10^{-3} \, m$.
Converting to millimeters:
$w = 4 \, mm$.

(A) The condition for diffraction is that the size of the aperture or obstacle $(a)$ must be comparable to the wavelength of the incident light $(\lambda)$.
Mathematically,this is expressed as $a \approx \lambda$.
Therefore,the ratio $\frac{a}{\lambda} \approx 1$.

(B) The angular width of the central (principal) maxima in a single slit diffraction experiment is given by $\beta = \frac{2\lambda D}{d}$, where $d$ is the slit width, $\lambda$ is the wavelength, and $D$ is the distance of the screen from the slit.
As the slit width $d$ is reduced, the linear width of the principal maxima increases because the width is inversely proportional to $d$ $(\beta \propto \frac{1}{d})$.
However, as the slit width $d$ decreases, the amount of light passing through the slit decreases, which results in a decrease in the intensity of the diffraction pattern. Therefore, the principal maxima becomes less bright.

(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $a \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \tan \theta = \frac{x}{D}$.
Thus,the position of the $n^{th}$ minima is $x = \frac{n \lambda D}{a}$.
Given:
Wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$.
Slit width $a = 0.30 \, mm = 0.30 \times 10^{-3} \, m = 3 \times 10^{-4} \, m$.
Distance $D = 2 \, m$.
For the first minima,$n = 1$.
Substituting the values:
$x = \frac{1 \times 6 \times 10^{-7} \times 2}{3 \times 10^{-4}}$
$x = \frac{12 \times 10^{-7}}{3 \times 10^{-4}}$
$x = 4 \times 10^{-3} \, m$.

(A) Given: Slit width $a = 2 \, mm = 2 \times 10^{-3} \, m$.
Wavelength $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$.
Distance of screen $D = 1 \, m$.
For single slit diffraction,the position of the $n^{th}$ minima is given by $y_n = \frac{n \lambda D}{a}$.
The distance between the first minima on both sides is the width of the central maximum,which is given by $x = 2y_1 = \frac{2 \lambda D}{a}$.
Substituting the values:
$x = \frac{2 \times (500 \times 10^{-9} \, m) \times (1 \, m)}{2 \times 10^{-3} \, m}$.
$x = 500 \times 10^{-6} \, m = 0.5 \times 10^{-3} \, m = 0.5 \, mm$.

(A) Given: Wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$.
Slit width $a = 0.3 \, mm = 0.3 \times 10^{-3} \, m = 3 \times 10^{-4} \, m$.
For the first minima in single slit diffraction,the condition is given by $a \sin \theta = n \lambda$.
For the first minima,$n = 1$.
So,$a \sin \theta = \lambda$.
Since $\theta$ is very small,$\sin \theta \approx \theta$.
Therefore,$\theta = \frac{\lambda}{a}$.
Substituting the values: $\theta = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3} \, rad$.

(D) In a single slit diffraction pattern,the condition for secondary maxima is given by the formula:
$a \sin \theta = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$
For the first secondary maximum,we substitute $n = 1$ into the equation:
$a \sin \theta = (2(1) + 1) \frac{\lambda}{2}$
$a \sin \theta = \frac{3\lambda}{2}$

(B) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin \theta = n \lambda$,where $a$ is the slit width,$\theta$ is the angular position,and $\lambda$ is the wavelength.
For the first minima,$n = 1$.
Given: $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$ and $\theta = 30^\circ$.
Substituting the values into the formula: $a \sin 30^\circ = 1 \times (5 \times 10^{-7} \, m)$.
Since $\sin 30^\circ = 0.5$,we have $a \times 0.5 = 5 \times 10^{-7} \, m$.
$a = \frac{5 \times 10^{-7}}{0.5} = 10 \times 10^{-7} \, m = 10^{-6} \, m$.
Converting to centimeters: $10^{-6} \, m = 10^{-6} \times 10^2 \, cm = 10^{-4} \, cm$.

(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $d \sin \theta = n \lambda$.
For small angles,$\sin \theta \approx \tan \theta = \frac{x}{D}$.
Thus,the position of the $n^{th}$ minima is $x = \frac{n \lambda D}{d}$.
Given: $\lambda = 5000 \, \mathring{A} = 5 \times 10^{-7} \, m$,$x = 5 \, mm = 5 \times 10^{-3} \, m$,$D = 2 \, m$,and $n = 1$ for the first minima.
Substituting these values into the formula: $d = \frac{n \lambda D}{x} = \frac{1 \times 5 \times 10^{-7} \times 2}{5 \times 10^{-3}}$.
$d = \frac{10 \times 10^{-7}}{5 \times 10^{-3}} = 2 \times 10^{-4} \, m$.
Converting to $mm$: $d = 2 \times 10^{-4} \times 10^3 \, mm = 0.2 \, mm$.

(C) When a narrow slit is illuminated by a parallel beam of monochromatic light,the light undergoes diffraction.
According to the theory of single-slit diffraction,the intensity distribution on the screen consists of a central maximum (bright fringe) which is the brightest.
On either side of this central maximum,there are secondary maxima and minima.
The intensity of these secondary maxima decreases rapidly as we move away from the center.
Therefore,a bright central fringe bordered on both sides by fringes of rapidly decreasing intensity is observed.

(D) For single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin \theta = n\lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $\theta$ is the angular position.
Given: $\lambda = 550\, nm = 550 \times 10^{-9}\, m$,$a = 22.0 \times 10^{-5}\, cm = 22.0 \times 10^{-7}\, m$,and $n = 2$ for the second minima.
Substituting the values into the formula:
$a \sin \theta = 2\lambda$
$\sin \theta = \frac{2\lambda}{a} = \frac{2 \times 550 \times 10^{-9}}{22.0 \times 10^{-7}}$
$\sin \theta = \frac{1100 \times 10^{-9}}{22.0 \times 10^{-7}} = \frac{1100}{2200} = 0.5$
Since $\sin \theta = 0.5$,we have $\theta = \arcsin(0.5) = \frac{\pi}{6}\, rad$.

(B) For diffraction at a single slit,the condition for the $n^{\text{th}}$ minimum is given by $b \sin \theta = n \lambda$.
For small angles,$\sin \theta \approx \tan \theta = \frac{x}{D}$,where $x$ is the distance from the central maximum and $D$ is the distance of the screen.
Thus,$x_n = \frac{n \lambda D}{b}$.
Given for $n=2$,$x_2 = 3\,cm$ and for $n=4$,$x_4 = 6\,cm$.
Using the formula: $x_n = n \left( \frac{\lambda D}{b} \right)$.
For $n=2$: $3 = 2 \left( \frac{\lambda D}{b} \right) \Rightarrow \frac{\lambda D}{b} = 1.5\,cm$.
For $n=4$: $6 = 4 \left( \frac{\lambda D}{b} \right) \Rightarrow \frac{\lambda D}{b} = 1.5\,cm$.
The width of the central maximum is the distance between the first minima on either side,which is $w = 2x_1$.
Since $x_1 = 1 \left( \frac{\lambda D}{b} \right) = 1.5\,cm$,the width of the central maximum is $w = 2 \times 1.5 = 3\,cm$.

(B) Given: Slit width $a = 0.1\, mm = 10^{-4}\, m$.
Wavelength $\lambda = 6000\, \mathring{A} = 6000 \times 10^{-10}\, m = 6 \times 10^{-7}\, m$.
Distance of screen $D = 0.5\, m$.
For the $n^{th}$ dark band in single slit diffraction, the condition is $a \sin \theta = n \lambda$.
For small angles, $\sin \theta \approx \tan \theta = \frac{x}{D}$, where $x$ is the distance from the central bright band.
Thus, $a \left( \frac{x}{D} \right) = n \lambda \implies x = \frac{n \lambda D}{a}$.
For the $3^{rd}$ dark band $(n = 3)$:
$x = \frac{3 \times (6 \times 10^{-7}\, m) \times 0.5\, m}{10^{-4}\, m}$.
$x = \frac{9 \times 10^{-7}}{10^{-4}}\, m = 9 \times 10^{-3}\, m = 9\, mm$.

(B) For a single slit diffraction pattern,the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda_1$,where $a$ is the slit width and $\lambda_1$ is the wavelength.
For the first minimum $(n=1)$,$a \sin \theta = \lambda_R = 6600\,\mathring{A}$.
The condition for the $n^{th}$ secondary maximum is $a \sin \theta = (2n + 1) \frac{\lambda_2}{2}$.
For the first secondary maximum $(n=1)$,$a \sin \theta = (2(1) + 1) \frac{\lambda_2}{2} = \frac{3}{2} \lambda_2$.
Since the first minimum of red light coincides with the first secondary maximum of the other wavelength,we equate the two expressions:
$6600 = \frac{3}{2} \lambda_2$.
Solving for $\lambda_2$: $\lambda_2 = \frac{6600 \times 2}{3} = 2200 \times 2 = 4400\,\mathring{A}$.

(C) For the first diffraction minimum in a single slit experiment, the condition is given by $d \sin \theta = n\lambda$, where $n = 1$ for the first minimum.
Given:
$\lambda = 5000 \,\mathring{A} = 5000 \times 10^{-8} \,\text{cm} = 5 \times 10^{-5} \,\text{cm}$
$\theta = 30^o$
Substituting the values into the formula:
$d \sin 30^o = 5000 \times 10^{-8} \,\text{cm}$
$d \times (1/2) = 5000 \times 10^{-8} \,\text{cm}$
$d = 2 \times 5000 \times 10^{-8} \,\text{cm}$
$d = 10000 \times 10^{-8} \,\text{cm} = 10 \times 10^{-5} \,\text{cm}$

(B) Given: Distance of screen $D = 50 \, cm = 0.5 \, m$,wavelength $\lambda = 6000 \, \mathring{A} = 600 \times 10^{-9} \, m$,and distance between first and third minima $\Delta y = 3 \, mm = 3 \times 10^{-3} \, m$.
The position of the $n^{th}$ minima in a single slit diffraction pattern is given by $y_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.
The distance between the first $(n=1)$ and third $(n=3)$ minima is:
$\Delta y = y_3 - y_1 = \frac{3 D \lambda}{a} - \frac{1 D \lambda}{a} = \frac{2 D \lambda}{a}$.
Substituting the given values:
$3 \times 10^{-3} = \frac{2 \times 0.5 \times 600 \times 10^{-9}}{a}$.
Solving for $a$:
$a = \frac{2 \times 0.5 \times 600 \times 10^{-9}}{3 \times 10^{-3}} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = 200 \times 10^{-6} \, m = 0.2 \times 10^{-3} \, m = 0.2 \, mm$.

(B) For diffraction to be significant and observable,the size of the aperture (slit) or obstacle must be comparable to the wavelength of the incident light wave. If the size is much larger than the wavelength,the light behaves according to ray optics and diffraction effects are negligible.

(B) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.
Given: $D = 50\,cm = 0.5\,m$,$\lambda = 690\,nm = 690 \times 10^{-9}\,m$,and the distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta x = x_3 - x_1 = 3.00\,mm = 3 \times 10^{-3}\,m$.
Using the formula: $\Delta x = (3 - 1) \frac{D \lambda}{a} = \frac{2 D \lambda}{a}$.
Rearranging for $a$: $a = \frac{2 D \lambda}{\Delta x}$.
Substituting the values: $a = \frac{2 \times 0.5 \times 690 \times 10^{-9}}{3 \times 10^{-3}} = \frac{690 \times 10^{-9}}{3 \times 10^{-3}} = 230 \times 10^{-6}\,m = 0.230\,mm$.

(B) The condition for the $n^{th}$ minima in single slit diffraction is given by $a \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \theta = \frac{y}{D}$.
Thus,the position of the first minima $(n=1)$ is $y = \frac{\lambda D}{a}$.
The distance between the first minima on either side of the central maximum is $2y = \frac{2 \lambda D}{a}$.
Given: $a = 2 \, mm = 2 \times 10^{-3} \, m$,$\lambda = 500 \, nm = 5 \times 10^{-7} \, m$,and $D = 1 \, m$.
Substituting the values:
$2y = \frac{2 \times (5 \times 10^{-7} \, m) \times (1 \, m)}{2 \times 10^{-3} \, m} = 5 \times 10^{-4} \, m$.
Converting to millimeters: $5 \times 10^{-4} \, m = 0.5 \, mm$.

(C) The width of the slit is $a = 10^4 \ \mathring{A} = 1000 \ nm$. The wavelength of visible light ranges from approximately $400 \ nm$ to $700 \ nm$. Since the slit width is comparable to the wavelength of light, diffraction occurs.
In single-slit diffraction, the central maximum is formed at the center for all wavelengths. Since sunlight consists of all colors (white light), all colors overlap at the central maximum, making the center appear white.
As we move away from the center, the position of the minima and secondary maxima depends on the wavelength $(\sin \theta = n\lambda / a)$. Different colors have different wavelengths, so they will diffract by different amounts, leading to the dispersion of colors at the edges of the central bright patch.

(C) The condition for the $n^{th}$ minimum in a single-slit diffraction pattern is given by $d \sin \theta = n \lambda$.
For the first minimum,$n = 1$.
Given: $\lambda = 6500\,\mathring{A} = 6500 \times 10^{-10}\, \text{m} = 6.5 \times 10^{-7}\, \text{m}$ and $\theta = 30^\circ$.
Substituting these values into the formula:
$d \sin 30^\circ = 1 \times (6.5 \times 10^{-7}\, \text{m})$
$d \times 0.5 = 6.5 \times 10^{-7}\, \text{m}$
$d = \frac{6.5 \times 10^{-7}}{0.5}\, \text{m} = 13 \times 10^{-7}\, \text{m} = 1.3 \times 10^{-6}\, \text{m}$.
Since $1\, \mu\text{m} = 10^{-6}\, \text{m}$,we have $d = 1.3\, \mu\text{m}$.

(B) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $d \sin \theta = n \lambda$.
For the first minimum,$n = 1$,so the equation becomes $d \sin \theta = \lambda$.
Given: $\lambda = 5000 \,\mathring{A} = 5000 \times 10^{-8} \, \text{cm} = 5 \times 10^{-5} \, \text{cm}$ and $\theta = 30^o$.
Substituting the values: $d \sin 30^o = 5 \times 10^{-5} \, \text{cm}$.
Since $\sin 30^o = 0.5$,we have $d \times 0.5 = 5 \times 10^{-5} \, \text{cm}$.
Therefore,$d = \frac{5 \times 10^{-5}}{0.5} = 10 \times 10^{-5} \, \text{cm} = 1.0 \times 10^{-4} \, \text{cm}$.

(A) When electrons pass through a narrow slit of width $'d'$,they undergo diffraction due to their wave nature. The intensity distribution of the diffraction pattern on the screen is given by the single-slit diffraction formula,where the intensity $I$ is proportional to $N$. The central maximum occurs at $y = 0$. The first minima occur at positions given by $y = \pm \frac{\lambda D}{d}$. Since the slit width $'d'$ is comparable to the de-Broglie wavelength $\lambda$,the diffraction pattern is spread out. The graph representing this intensity distribution shows a central peak at $y = 0$ with secondary maxima on either side,which corresponds to the pattern shown in Graph $A$.

(A) For a single slit diffraction pattern,the condition for the $n^{th}$ order secondary maxima is given by $a \sin \theta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$.
For the first-order secondary maximum,we set $n = 1$.
Thus,$a \sin \theta = (1 + \frac{1}{2}) \lambda = \frac{3}{2} \lambda$.
Since $\theta$ is very small,$\sin \theta \approx \theta$.
Therefore,$\theta = \frac{3 \lambda}{2 a}$.
Given: $\lambda = 5890 \times 10^{-10}\, m$ and $a = 0.25 \times 10^{-3}\, m$.
Substituting the values: $\theta = \frac{3 \times 5890 \times 10^{-10}}{2 \times 0.25 \times 10^{-3}}$.
$\theta = \frac{17670 \times 10^{-10}}{0.5 \times 10^{-3}} = 35340 \times 10^{-7} = 3.534 \times 10^{-3}\, rad$.

(B) For a single slit diffraction pattern,the position of the $n^{th}$ minimum is given by $y_n = \frac{n \lambda D}{a}$,where $n$ is the order of the minimum,$\lambda$ is the wavelength,$D$ is the distance to the screen,and $a$ is the slit width.
The distance between the $1^{st}$ and $6^{th}$ minima is $\Delta y = y_6 - y_1 = \frac{6 \lambda D}{a} - \frac{1 \lambda D}{a} = \frac{5 \lambda D}{a}$.
Given: $\Delta y = 0.5 \, mm = 0.5 \times 10^{-3} \, m$,$D = 0.5 \, m$,and $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
Substituting the values into the formula:
$0.5 \times 10^{-3} = \frac{5 \times (5 \times 10^{-7}) \times 0.5}{a}$
Solving for $a$:
$a = \frac{5 \times 5 \times 10^{-7} \times 0.5}{0.5 \times 10^{-3}}$
$a = 25 \times 10^{-4} \, m = 2.5 \times 10^{-3} \, m = 2.5 \, mm$.

(B) The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2D \lambda}{a}$,where $D$ is the distance to the screen,$\lambda$ is the wavelength,and $a$ is the slit width.
Since the wavelength of blue light $(\lambda_{blue})$ is smaller than the wavelength of red light $(\lambda_{red})$,the fringe width $w$ is directly proportional to the wavelength $\lambda$.
Therefore,when red light is replaced by blue light,the wavelength decreases,causing the diffraction bands to become narrower and more crowded together.

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by $\theta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.
From this relation,we see that $\theta \propto \lambda$.
Since the wavelength of blue light $(\lambda_B)$ is less than the wavelength of yellow light $(\lambda_Y)$,i.e.,$\lambda_B < \lambda_Y$,the angular width of the diffraction pattern will decrease.
Therefore,the fringes will become narrower.

(B) The angular width of the central maximum in a single slit diffraction is given by $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.
Since $\theta \propto \frac{1}{d}$,if the slit width $d$ is doubled,the angular width $\theta$ becomes half,meaning the central maximum becomes narrower.
As the width of the central maximum decreases,the energy is concentrated in a smaller area,which increases the intensity of the central maximum,making it brighter.

(B) In the diffraction pattern produced by a single slit of width $a$,the condition for minima is given by $a \sin \theta = n\lambda$ (where $n = \pm 1, \pm 2, \dots$).
Secondary maxima occur approximately midway between the minima.
The condition for secondary maxima is given by $a \sin \theta = (2n + 1)\frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ represents the order of the secondary maxima.

(A) For a single slit diffraction, the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$.
Given that the angle $\theta$ is very small, we can approximate $\sin \theta \approx \tan \theta = \frac{x}{f}$, where $x$ is the distance from the central maximum and $f$ is the focal length.
Substituting this into the condition for the $n^{th}$ minimum: $a \left( \frac{x}{f} \right) = n \lambda$.
For the third minimum, $n = 3$. Given $a = 0.3 \times 10^{-3} \, m$, $x = 5 \times 10^{-3} \, m$, and $f = 1 \, m$.
Rearranging for $\lambda$: $\lambda = \frac{a x}{3 f}$.
$\lambda = \frac{0.3 \times 10^{-3} \times 5 \times 10^{-3}}{3 \times 1} = \frac{1.5 \times 10^{-6}}{3} = 0.5 \times 10^{-6} \, m$.
Converting to $\mathring{A}$: $\lambda = 0.5 \times 10^{-6} \times 10^{10} \, \mathring{A} = 5000 \, \mathring{A}$.

(D) The width of the diffraction bands in a single-slit diffraction pattern is given by the formula $\beta = \frac{2\lambda D}{a}$,where $\lambda$ is the wavelength of the light,$D$ is the distance of the screen from the slit,and $a$ is the width of the slit.
Since the wavelength of blue light $(\lambda_{blue})$ is smaller than the wavelength of red light $(\lambda_{red})$,the width of the diffraction bands $\beta$ will decrease.
Therefore,when red light is replaced by blue light,the diffraction bands become narrower and more crowded together.

(A) muslin cloth consists of a fine mesh of threads,creating a large number of tiny,closely spaced slits.
When white light passes through these fine slits,it undergoes diffraction.
Since the diffraction pattern depends on the wavelength of light,different colours are diffracted at different angles.
This separation of white light into its constituent colours results in the observation of a coloured spectrum.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) When light from a distant source encounters a small circular obstacle,the light waves diffract around the edges of the obstacle.
According to the Huygens-Fresnel principle,every point on the edge of the obstacle acts as a source of secondary wavelets.
These secondary wavelets travel to the centre of the geometric shadow and arrive in phase with each other.
Because they arrive in phase,they undergo constructive interference,resulting in a bright spot at the centre of the shadow,which is known as the Poisson spot or Arago spot.
Therefore,the Assertion is correct,but the Reason is incorrect because constructive interference,not destructive interference,occurs at the centre.

(C) The Assertion is correct because clouds appear white due to the scattering of light by water droplets.
The size of water droplets in clouds is much larger than the wavelength of visible light.
According to the theory of scattering, when the size of the scattering particle $(a)$ is much larger than the wavelength of light $(\lambda)$, i.e., $a \gg \lambda$, the scattering is independent of the wavelength.
Since all wavelengths of visible light are scattered equally, the combined effect is perceived as white light.
Diffraction is a phenomenon that occurs when light encounters an obstacle or aperture of a size comparable to its wavelength. Since cloud particles are much larger than the wavelength of light, they do not cause significant diffraction of visible light.
Therefore, the Reason is incorrect.

(A) For diffraction to occur,the size of the obstacles or the spacing between the slits in a diffraction grating must be of the order of the wavelength of the incident radiation.
The wavelength of $X-$rays is typically in the range of $0.01 \ nm$ to $10 \ nm$.
Standard optical diffraction gratings have a grating spacing (distance between adjacent lines) of the order of $10^3 \ nm$ to $10^4 \ nm$.
Since the grating spacing is much larger than the wavelength of $X-$rays,the diffraction effect is negligible,and these gratings cannot be used to discriminate between $X-$ray wavelengths.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) Diffraction is a general characteristic of all types of waves,including mechanical (like sound) and non-mechanical (like light),as well as transverse and longitudinal waves. Thus,the Assertion is correct.
For diffraction to be significant or perceptible,the size of the obstacle or aperture must be of the order of the wavelength of the wave. If the wavelength is much smaller than the obstacle,the wave behaves like a ray and diffraction is negligible. Thus,the Reason is also correct and explains why diffraction is observed in specific conditions. Therefore,the correct option is $A$.

(D) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $n = 1, 2, 3, ...$.
For the second minimum $(n = 2)$,the angle is $\theta_2 = 60^{\circ}$.
Thus,$a \sin 60^{\circ} = 2 \lambda$.
$a \left( \frac{\sqrt{3}}{2} \right) = 2 \lambda \implies \frac{\lambda}{a} = \frac{\sqrt{3}}{4}$.
For the first minimum $(n = 1)$,the condition is $a \sin \theta_1 = 1 \lambda$.
$\sin \theta_1 = \frac{\lambda}{a} = \frac{\sqrt{3}}{4}$.
$\sin \theta_1 \approx \frac{1.732}{4} = 0.433$.
$\theta_1 = \arcsin(0.433) \approx 25.6^{\circ}$.
Rounding to the nearest integer,$\theta_1 \approx 25^{\circ}$.

(B) The angular width of the central maximum of a single slit diffraction pattern is given by $2\theta = \frac{2\lambda}{a}$,where $a$ is the slit width.
The angular separation between adjacent maxima in a double-slit interference pattern is $\Delta\theta = \frac{\lambda}{d}$,where $d$ is the distance between the slits.
We are given that $10$ maxima of the double-slit pattern lie within the central maximum of the single-slit pattern. This means the angular width of the central maximum $(2\lambda/a)$ must contain $10$ interference fringes. Since the central maximum spans from $-\lambda/a$ to $+\lambda/a$,the total number of fringes $N$ is given by the condition that the width of $10$ fringes equals the width of the central maximum: $10 \times (\frac{\lambda}{d}) = \frac{2\lambda}{a}$.
Simplifying the equation: $\frac{10}{d} = \frac{2}{a}$.
Given $d = 1 \; mm = 10^{-3} \; m$,we solve for $a$:
$a = \frac{2d}{10} = \frac{d}{5} = \frac{1 \; mm}{5} = 0.2 \; mm$.

(N/A) In a single slit diffraction experiment, if the width of the slit $(a)$ is doubled, the angular width of the central maximum $(2\lambda/a)$ reduces to half. The intensity of the central maximum is proportional to the square of the slit width $(I \propto a^2)$, so it increases by a factor of $4$.
$(b)$ The interference pattern in a double-slit experiment is the result of the superposition of two waves, but it is modulated by the diffraction pattern produced by each individual slit. The intensity distribution is the product of the interference pattern and the diffraction envelope.
$(c)$ This phenomenon is known as Poisson's spot. Light waves diffract around the edges of the circular obstacle and reach the center of the shadow in phase, resulting in constructive interference that creates a bright spot.
$(d)$ Diffraction is significant only when the wavelength $(\lambda)$ is comparable to the size of the obstacle $(d)$. For light, $\lambda$ is very small $(\sim 500 \; nm)$, so it does not diffract significantly around a wall. For sound, $\lambda$ is in the range of $0.1 \; m$ to $1 \; m$, which is comparable to the size of the wall, allowing it to bend around the obstacle.
$(e)$ The ray optics approximation is valid when the size of the apertures or obstacles is much larger than the wavelength of light. In most optical instruments, the apertures are large enough that diffraction effects are negligible, making the straight-line propagation assumption valid.

(B) Given:
Wavelength of light beam,$\lambda = 500\; nm = 500 \times 10^{-9}\; m$
Distance of the screen from the slit,$D = 1\; m$
Distance of the first minimum from the centre of the screen,$x = 2.5\; mm = 2.5 \times 10^{-3}\; m$
For the first minimum in single slit diffraction,the condition is given by $a \sin \theta = n \lambda$,where $a$ is the slit width and $n = 1$.
For small angles,$\sin \theta \approx \tan \theta = \frac{x}{D}$.
Thus,$a \left( \frac{x}{D} \right) = n \lambda$.
Rearranging for slit width $a$:
$a = \frac{n \lambda D}{x} = \frac{1 \times 500 \times 10^{-9} \times 1}{2.5 \times 10^{-3}}$
$a = \frac{500 \times 10^{-9}}{2.5 \times 10^{-3}} = 200 \times 10^{-6}\; m = 0.2 \times 10^{-3}\; m = 0.2\; mm$.
Therefore,the width of the slit is $0.2\; mm$.

(N/A) Consider a single slit of width $a$. To find the condition for minima,we divide the slit into $2n$ equal parts.
The path difference between the secondary wavelets from the corresponding points of the upper and lower halves of the slit is $\Delta x = (a/2) \sin \theta$.
For the first minimum,we set the path difference between the top edge and the center of the slit to be $\lambda / 2$. Thus,$(a/2) \sin \theta = \lambda / 2$,which gives $a \sin \theta = \lambda$.
In general,for the $n^{th}$ minimum,we divide the slit into $2n$ equal parts. The path difference between the wavelets from the corresponding points of the two adjacent parts is $\lambda / 2$.
Since every point in the upper half of the slit has a corresponding point in the lower half such that their path difference is $\lambda / 2$,the waves from these pairs interfere destructively.
Consequently,the resultant intensity at these angles $\theta = n\lambda / a$ becomes zero.

(N/A) The bending of waves around the corners of an obstacle or through an aperture is called diffraction.
Accurate definition: Diffraction is the physical effect produced by the limited part of a wavefront.
Diffraction was first discovered by the scientist Francesco Maria Grimaldi.
Diffraction occurs in all types of waves,including sound waves,light waves,water waves,and matter waves.
Diffraction depends on the ratio of the wavelength of the wave $\lambda$ to the width of the slit $d$,denoted as $\frac{\lambda}{d}$.
Due to diffraction,light spreads into the geometric shadow region of an opaque object,creating dark and bright fringes.
Optical instruments like telescopes and microscopes have a limit on their resolving power due to diffraction.
The iridescent colours appearing on a $CD$ are due to diffraction effects.
Diffraction is explained by wave theory and is categorized into two types based on the wavefront incident on the slit:
$(i)$ If spherical wavefronts are incident on the slit,it is called Fresnel diffraction.
$(ii)$ If plane wavefronts are incident on the slit,it is called Fraunhofer diffraction.

(N/A) The condition for significant diffraction is that the wavelength of the wave $(\lambda)$ must be comparable to the size of the obstacle or aperture $(d)$.
For light, the average wavelength is $\lambda \approx 6 \times 10^{-7} \, m$. The width of a door is $d \approx 1 \, m$. The ratio $\frac{\lambda}{d} \approx 6 \times 10^{-7}$, which is extremely small. Therefore, light waves do not bend significantly around the corners of the door, making the diffraction of light negligible, and the people cannot see each other.
For sound, the frequency of human speech is typically between $100 \, Hz$ and $400 \, Hz$. Taking a frequency of $330 \, Hz$ and the speed of sound $v = 330 \, m/s$, the wavelength is $\lambda = \frac{v}{f} = \frac{330}{330} = 1 \, m$. Since the wavelength $\lambda = 1 \, m$ is comparable to the door width $d = 1 \, m$, the ratio $\frac{\lambda}{d} = 1$. This leads to significant diffraction of sound waves, allowing the sound to bend around the corners of the door and reach the person inside.

(N/A) As shown in the figure,by placing a monochromatic light source at the focus of a convex lens,the light rays emerging from it become parallel,forming a plane wavefront that is incident on the slit $LN$.
According to Huygens' principle,every point on the plane wavefront within the slit acts as an independent source of secondary spherical wavelets. As these wavelets emerge from the slit,they undergo interference with each other. This superposition results in constructive and destructive interference patterns.
Consequently,a diffraction pattern consisting of bright and dark fringes is obtained on a screen placed at the focal plane of a second convex lens,as illustrated in the figure.

(N/A) Richard Feynman noted that almost no one has ever been able to define the difference between interference and diffraction in a completely satisfactory way.
He suggested that it is primarily a matter of usage and terminology,rather than a fundamental physical distinction.
According to his perspective,when we deal with a small number of sources (e.g.,two interfering sources),the resulting phenomenon is typically referred to as interference. Conversely,when we deal with a large number of sources (such as the continuous distribution of wavelets across a slit),the term diffraction is more commonly used.

(B) In Young's double-slit experiment,if one of the two slits is closed,the experimental setup effectively becomes a single-slit diffraction experiment.
Since interference requires two coherent sources,the interference pattern will disappear.
Instead,the light passing through the remaining single slit will undergo diffraction,resulting in a diffraction pattern on the screen.
The central maximum of this diffraction pattern will be located on the screen at a point directly in front of the open slit.

(N/A) The width of the central maximum is defined as the distance between the first minima on either side of the central maximum.
Let the width of the slit be $a$ and the distance between the slit and the screen be $D$.
For the first minimum on one side of the central maximum,the condition for diffraction is given by $a \sin \theta = \lambda$. Since $\theta$ is very small,$\sin \theta \approx \theta$,so $\theta = \frac{\lambda}{a}$.
Here,$\theta$ represents half the angular width of the central maximum.
Therefore,the total angular width of the central maximum is $2\theta = \frac{2\lambda}{a}$.
Now,for the linear width $\beta_0$,we use the relation between arc length,radius,and angle: $\text{arc} = \text{radius} \times \text{angle}$.
Here,the arc length is $\beta_0$,the radius is $D$,and the angle is $2\theta$.
Thus,$\beta_0 = D \times (2\theta) = D \times \frac{2\lambda}{a}$.
Therefore,the linear width of the central maximum is $\beta_0 = \frac{2D\lambda}{a}$.