Single Slit Diffraction of Light Questions in English

(N/A) $1$. Razor Blade Experiment: Place two razor blades such that their sharp edges are parallel and very close to each other,forming a narrow slit. When you look at a distant light bulb filament through this narrow slit,you will observe bright and dark fringes around the filament due to the diffraction of light.
$2$. Finger Experiment: Bring your fingers very close to each other and look at a distant light source through the narrow gap between them. You will observe a series of diffraction fringes.
$3$. Khadi Cloth Experiment: Look at a distant light bulb through a piece of fine-textured Khadi cloth. The fibers of the cloth act as a diffraction grating,and you will see bright and dark fringes around the light source.
$4$. Pinhole Experiment: Place an opaque screen with a tiny pinhole in front of a monochromatic light source (like a sodium lamp). When the light passes through this pinhole and falls on a distant screen,you will observe a diffraction pattern consisting of bright and dark fringes.

(C) The phenomenon responsible for the blurring of shadows of opaque objects is diffraction.
Diffraction is the bending of light waves around the corners of an obstacle or through an aperture.
When light encounters an opaque object,it does not travel in a perfectly straight line due to its wave nature.
Instead,the light waves bend around the edges of the object,causing the shadow to become fuzzy or blurred at the edges rather than having a sharp,distinct boundary.

(N/A) Diffraction is the phenomenon of bending of light around the corners of an obstacle or an aperture of the size comparable to the wavelength of light.
When light waves encounter an obstacle or an opening, they do not travel in a perfectly straight line but spread out into the region of the geometrical shadow.
This bending effect is more pronounced when the size of the obstacle or aperture is small, specifically when it is of the order of the wavelength of the incident light $\lambda$.

(A) In single slit diffraction,the condition for the $n^{th}$ order secondary maxima is given by the formula:
$a \sin \theta = (2n + 1) \frac{\lambda}{2}$
where:
$a$ is the width of the slit,
$\theta$ is the angle of diffraction,
$\lambda$ is the wavelength of the incident light,
$n$ is an integer $(n = 1, 2, 3, ...)$.

(A) The phenomenon of diffraction of light was first observed and described by the Italian scientist $Francesco \text{ } Maria \text{ } Grimaldi$ in the $17^{th}$ century. He coined the term 'diffraction' from the Latin word 'diffringere', meaning 'to break into pieces', to describe how light bends around obstacles.

(B) In a double-slit diffraction experiment,the interference pattern is modulated by the diffraction envelope. The number of interference fringes $n$ within the central diffraction maximum is given by the ratio of the width of the central diffraction maximum to the fringe width of the interference pattern.
The angular width of the central diffraction maximum is $\Delta \theta_{diff} = \frac{2\lambda}{a}$,where $a$ is the slit width.
The angular fringe width of the interference pattern is $\beta_{\theta} = \frac{\lambda}{d}$,where $d$ is the distance between the slits.
The number of interference fringes $n$ is the ratio of these two widths:
$n = \frac{2\lambda / a}{\lambda / d} = \frac{2d}{a}$.
Thus,the number of interference fringes depends on the ratio of the slit separation $d$ to the slit width $a$.

(A) The condition for significant diffraction is that the size of the obstacle or aperture $(d)$ must be comparable to the wavelength $(\lambda)$ of the wave, i.e., $\frac{\lambda}{d} \approx 1$.
For visible light, the wavelength $\lambda$ is approximately $6 \times 10^{-7} \,m$. Since most obstacles in our daily life are much larger than this (e.g., $d \approx 10^{-1} \,m$ to $1 \,m$), the ratio $\frac{\lambda}{d}$ is extremely small, making diffraction negligible.
For sound waves, the audible frequency range is $20 \,Hz$ to $20,000 \,Hz$. Taking a typical frequency of $332 \,Hz$ and the speed of sound $v = 332 \,m/s$, the wavelength is $\lambda = \frac{v}{f} = \frac{332}{332} = 1 \,m$.
Since the wavelength of sound $(1 \,m)$ is comparable to the size of common obstacles like doors or windows, the ratio $\frac{\lambda}{d}$ is significant, leading to prominent diffraction.
Therefore, we can hear sounds around corners or through doorways, but we cannot see light around them.

(A) The condition for diffraction minima in a single slit experiment is given by $d \sin \theta = n \lambda$,where $n = \pm 1, \pm 2, \pm 3, \dots$
Since $|\sin \theta| < 1$,we have $n < \frac{d}{\lambda}$.
Given $d = 0.6 \times 10^{-4} \ m = 6 \times 10^{-5} \ m$ and $\lambda = 6000 \times 10^{-10} \ m = 6 \times 10^{-7} \ m$.
Calculating the ratio: $\frac{d}{\lambda} = \frac{6 \times 10^{-5}}{6 \times 10^{-7}} = 100$.
Thus,$n < 100$. The possible integer values for $n$ on one side are $1, 2, \dots, 99$.
Therefore,the number of minima on one side is $99$.
The total number of minima on both sides of the central maximum is $99 + 99 = 198$.

(A) Given: Wavelength $(\lambda) = 5400 \ \mathring{A} = 5.4 \times 10^{-7} \ m$.
Slit width $(a) = 0.80 \ mm = 8 \times 10^{-4} \ m$.
Distance of screen $(D) = 1.4 \ m$.
The distance between the first two dark bands on each side of the central bright band is equivalent to the width of the central maximum.
The formula for the width of the central maximum is given by $w = \frac{2 \lambda D}{a}$.
Substituting the values: $w = \frac{2 \times 5.4 \times 10^{-7} \times 1.4}{8 \times 10^{-4}}$.
$w = \frac{15.12 \times 10^{-7}}{8 \times 10^{-4}} = 1.89 \times 10^{-3} \ m$.
Converting to $mm$,we get $w = 1.89 \ mm$.

(D) The angular width of the central diffraction maximum for a circular aperture is given by $\theta \approx \frac{1.22 \lambda}{D}$,where $D$ is the diameter of the pinhole and $\lambda$ is the wavelength of light.
As the diameter $D$ of the pinhole increases,the angular width $\theta$ decreases,which means the size of the diffraction pattern decreases.
Since the total amount of light passing through the pinhole increases as the area of the pinhole increases (area $\propto D^2$),and this light is now concentrated into a smaller area,the intensity of the diffraction pattern increases.
Therefore,the size decreases and the intensity increases.

(C) For a single slit diffraction,the condition for the $n^{th}$ maxima is given by $a \sin \theta = (n + \frac{1}{2}) \lambda$.
For the first maxima,$n = 1$,so $a \sin \theta = \frac{3 \lambda}{2}$.
Since $\theta$ is very small,$\sin \theta \approx \theta = \frac{y}{L}$,where $y$ is the distance from the central maxima and $L$ is the distance to the screen.
Thus,$y = \frac{3 \lambda L}{2 a}$.
The separation between the first maxima for two wavelengths $\lambda_1 = 650\, nm$ and $\lambda_2 = 655\, nm$ is $\Delta y = y_2 - y_1 = \frac{3 L}{2 a} (\lambda_2 - \lambda_1)$.
Given $L = 2.0\, m$,$a = 0.5\, mm = 0.5 \times 10^{-3}\, m$,$\lambda_1 = 650 \times 10^{-9}\, m$,and $\lambda_2 = 655 \times 10^{-9}\, m$.
$\Delta y = \frac{3 \times 2.0}{2 \times 0.5 \times 10^{-3}} \times (655 - 650) \times 10^{-9}$.
$\Delta y = \frac{6}{10^{-3}} \times 5 \times 10^{-9} = 6 \times 5 \times 10^{-6} = 30 \times 10^{-6} = 3 \times 10^{-5}\, m$.
Comparing with $x \times 10^{-5}\, m$,we get $x = 3$.

(C)
Among the given patterns,the colours observed on a compact disc are due to the diffraction of light.
The causes of the other patterns are as follows:
$(a)$ $A$ rainbow occurs because of the refraction,total internal reflection,and dispersion of light.
$(b)$ When white light passes through a prism,the colourful pattern formed is due to the dispersion of light.
$(d)$ The blue colour of the sky is due to the scattering of light.

(A) The angular width of the central diffraction maximum for a slit of width $a$ is given by $\theta = \frac{2\lambda}{a}$.
Given that the slit is rectangular with width $w$ and height $h$,where $w = 2h$.
The diffraction pattern is formed by the superposition of diffraction effects in the horizontal and vertical directions.
The angular width in the horizontal direction is $\theta_w = \frac{2\lambda}{w}$ and in the vertical direction is $\theta_h = \frac{2\lambda}{h}$.
Since $w = 2h$,we have $\theta_w = \frac{2\lambda}{2h} = \frac{\lambda}{h}$.
Comparing the two,$\theta_h = \frac{2\lambda}{h}$ and $\theta_w = \frac{\lambda}{h}$,it is clear that $\theta_h > \theta_w$.
Therefore,the central diffraction peak is wider in the vertical direction than in the horizontal direction.

(D) The phenomenon of diffraction is a general characteristic of all waves,including electromagnetic waves.
For diffraction to be observable,the size of the aperture or obstacle must be comparable to the wavelength of the incident wave.
Since infrared waves,microwaves,and $X$-rays are all types of electromagnetic waves,they all exhibit diffraction when interacting with appropriate obstacles or apertures.
Therefore,the correct option is $D$.

(B) .
Interference is the phenomenon resulting from the superposition of waves originating from two distinct,coherent sources.
Diffraction is the phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wavefront as they pass through an aperture or around an obstacle.

(D) The width of the central maximum in a single-slit diffraction pattern is given by the formula: $W = \frac{2D\lambda}{a}$.
Given values are:
$D = 2 \,m$
$\lambda = 580 \,nm = 580 \times 10^{-9} \,m$
$a = 0.30 \,mm = 0.30 \times 10^{-3} \,m$
Substituting these values into the formula:
$W = \frac{2 \times 2 \times 580 \times 10^{-9}}{0.30 \times 10^{-3}}$
$W = \frac{2320 \times 10^{-9}}{0.30 \times 10^{-3}}$
$W = \frac{2320}{0.30} \times 10^{-6} \,m$
$W \approx 7733.33 \times 10^{-6} \,m = 7.73 \times 10^{-3} \,m$.
Thus,the width of the central maximum is $7.73 \times 10^{-3} \,m$.

(A) The storage capacity of an optical disc is determined by the size of the pits and lands,which is limited by the diffraction limit of the laser beam used to read/write the data.
According to the diffraction limit,the minimum spot size $d$ is proportional to the wavelength $\lambda$ of the laser $(d \propto \lambda)$.
$A$ $DVD$ uses a red laser with a wavelength of approximately $6350 \,\mathring{A}$,whereas a $CD$ uses an infrared laser with a wavelength of $7800 \,\mathring{A}$.
Since the $DVD$ laser has a shorter wavelength,it can focus on a smaller spot size,allowing for a much higher density of pits and lands on the disc surface.
Consequently,a $DVD$ can store significantly more information (approximately $30$ times more) than a $CD$ of the same physical size.
Therefore,option $(A)$ is the correct explanation.

(A) Diffraction is the phenomenon of bending of waves around the corners of an obstacle or aperture. The condition for significant diffraction is that the wavelength of the wave $(\lambda)$ must be comparable to the size of the obstacle or aperture $(a)$.
Sound waves have wavelengths in the range of $0.1 \, m$ to $10 \, m$, which is comparable to the size of a door opening. Therefore, sound waves diffract easily around the edges of the door.
Light waves have very small wavelengths (in the range of $400 \, nm$ to $700 \, nm$). Since the door opening is much larger than the wavelength of light, light does not undergo significant diffraction and travels in a straight line, preventing us from seeing around the corner.

(B) The phenomenon of "floaters" is caused by small shadows cast on the retina by debris (such as collagen fibers or cells) floating in the vitreous humor of the eye. When light passes around these tiny particles, it undergoes diffraction. The diffraction patterns formed by these structures are perceived by the brain as floating specks or hair-like shapes. Therefore, this is a diffraction pattern. The correct option is $B$.

(A) The correct option is $A$.
Diffraction is the phenomenon of bending of light around the corners of an obstacle or an aperture.
For significant diffraction to occur,the size of the obstacle or the width of the aperture must be comparable to the wavelength $(\lambda)$ of the incident light.
If the obstacle is much larger than the wavelength,the light travels in a straight line,and diffraction is negligible.
Therefore,the size of the obstacle should be of the same order as the wavelength of the light used.

(C) The condition for significant diffraction is that the size of the obstacle or aperture must be comparable to the wavelength of the wave.
Sound waves have wavelengths in the range of $meters$ to $centimeters$,which is comparable to the size of common objects in our surroundings.
Light waves have very small wavelengths (in the range of $10^{-7} \ m$),which are much smaller than common objects.
Therefore,sound waves bend around obstacles more easily than light waves,making diffraction more noticeable for sound.

(C) The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2D\lambda}{b}$,where $D$ is the distance to the screen,$\lambda$ is the wavelength,and $b$ is the slit width.
The fringe width of a double-slit interference pattern is given by $\beta = \frac{D\lambda}{d}$,where $d$ is the slit separation.
It is given that $20$ maxima of the double-slit pattern lie within the central maximum of the diffraction pattern. The total width occupied by $20$ interference fringes is $20 \times \beta = 20 \frac{D\lambda}{d}$.
Equating the two widths: $20 \frac{D\lambda}{d} = \frac{2D\lambda}{b}$.
Simplifying the equation: $\frac{20}{d} = \frac{2}{b}$.
Rearranging for $b$: $b = \frac{2d}{20} = \frac{d}{10}$.
Given $d = 2 \, mm$,we get $b = \frac{2 \, mm}{10} = 0.2 \, mm$.

$(A)$ The width of the principal maxima in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{a}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $a$ is the slit width.
Given: $w = 2.5 \times 10^{-3}\,m$, $D = 2\,m$, and $a = 1 \times 10^{-3}\,m$.
Substituting the values into the formula: $2.5 \times 10^{-3} = \frac{2 \times \lambda \times 2}{1 \times 10^{-3}}$.
Solving for $\lambda$: $\lambda = \frac{2.5 \times 10^{-3} \times 10^{-3}}{4} = 0.625 \times 10^{-6}\,m$.
Converting to $\mathring{A}$s: $\lambda = 6.25 \times 10^{-7}\,m = 6250 \times 10^{-10}\,m = 6250\,\mathring{A}$.

(C) Bragg's law for diffraction by a crystal is given by the equation:
$n\lambda = 2d \sin\theta$
where $n$ is the order of diffraction,$\lambda$ is the wavelength,$d$ is the interatomic distance,and $\theta$ is the glancing angle.
Since the maximum value of $\sin\theta$ is $1$,the equation becomes $n\lambda \leq 2d$.
For the first order of diffraction $(n=1)$,the condition is $\lambda \leq 2d$.
Therefore,for diffraction to occur,the wavelength $\lambda$ must be smaller than or equal to $2d$.

(B) For the first minimum in a single-slit diffraction pattern, the condition is given by the formula:
$a \sin \theta = n \lambda$
For the first minimum, $n = 1$, so the equation becomes:
$a \sin \theta = \lambda$
Given:
$\lambda = 600 \, nm = 600 \times 10^{-9} \, m$
$\theta = 30^{\circ}$
Substituting the values into the equation:
$a \sin 30^{\circ} = 600 \times 10^{-9} \, m$
Since $\sin 30^{\circ} = 0.5$:
$a \times 0.5 = 600 \times 10^{-9} \, m$
$a = 1200 \times 10^{-9} \, m$
$a = 1.2 \times 10^{-6} \, m$
$a = 1.2 \, \mu m$

(C) 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 $a \sin \theta = \lambda$.
Given:
Slit width $a = 0.001 \text{ mm} = 1 \times 10^{-6} \text{ m}$.
Wavelength $\lambda = 5000 \mathring{A} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$.
Substituting the values:
$\sin \theta = \frac{\lambda}{a} = \frac{5 \times 10^{-7}}{1 \times 10^{-6}} = 0.5$.
Since $\sin \theta = 0.5$,the angle of diffraction is $\theta = 30^{\circ}$.

(C) For the $n^{\text{th}}$ minima in a single slit diffraction pattern,the condition is given by:
$b \sin \theta = n \lambda$
Since $\lambda$ is very small,$\sin \theta \approx \tan \theta = \frac{y_n}{D}$.
Thus,the position of the $n^{\text{th}}$ minima is $y_n = \frac{n \lambda D}{b}$.
The position of the first minima $(n=1)$ is $y_1 = \frac{\lambda D}{b}$.
The position of the third minima $(n=3)$ is $y_3 = \frac{3 \lambda D}{b}$.
The distance between the first and third minima is $\Delta y = y_3 - y_1 = \frac{2 \lambda D}{b}$.
Given $\Delta y = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$,$\lambda = 6000 \mathring A = 6000 \times 10^{-10} \text{ m}$,and $D = 50 \text{ cm} = 0.5 \text{ m}$.
Substituting the values:
$3 \times 10^{-3} = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{b}$
$b = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}}$
$b = \frac{6000 \times 10^{-10}}{3 \times 10^{-3}} = 2000 \times 10^{-7} = 2 \times 10^{-4} \text{ m}$.
Therefore,the width of the slit is $2 \times 10^{-4} \text{ m}$. The correct option is $C$.

(A) The width of the $n^{\text{th}}$ secondary maxima in a single-slit diffraction pattern is given by $\beta = \frac{\lambda D}{a}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen (focal length of the lens),and $a$ is the slit width.
Given:
$\lambda = 400 \ nm = 400 \times 10^{-9} \ m$
$a = 0.2 \ mm = 0.2 \times 10^{-3} \ m$
$D = 100 \ cm = 1 \ m$
Substituting the values:
$\text{Width} = \frac{400 \times 10^{-9} \ m \times 1 \ m}{0.2 \times 10^{-3} \ m}$
$= \frac{400}{0.2} \times 10^{-6} \ m$
$= 2000 \times 10^{-6} \ m$
$= 2 \times 10^{-3} \ m = 2 \ mm$.

(B) The linear width of the central maximum in a single-slit diffraction pattern is given by the formula:
$W = \frac{2 \lambda f}{a}$
Where:
$\lambda = 6000 \text{ Å} = 6 \times 10^{-7} \text{ m}$
$a = 0.01 \text{ mm} = 1 \times 10^{-5} \text{ m}$
$f = 20 \text{ cm} = 0.2 \text{ m}$
Substituting the values:
$W = \frac{2 \times (6 \times 10^{-7} \text{ m}) \times (0.2 \text{ m})}{1 \times 10^{-5} \text{ m}}$
$W = \frac{2.4 \times 10^{-7}}{10^{-5}} \text{ m}$
$W = 2.4 \times 10^{-2} \text{ m} = 24 \text{ mm}$

(C) For the first diffraction minima,the condition is given by $a \sin \theta = n \lambda$. For the first minimum,$n = 1$.
Given: wavelength $\lambda = 2.0 \ cm$ and slit width $a = 4.0 \ cm$.
Substituting the values: $\sin \theta = \frac{\lambda}{a} = \frac{2.0}{4.0} = 0.5$.
Thus,$\theta = \arcsin(0.5) = 30^{\circ}$.
The angular spread of the central maxima is the angle between the first minima on either side of the central maximum,which is $2\theta$.
Therefore,angular spread $= 2 \times 30^{\circ} = 60^{\circ}$.

(C) For a single slit diffraction experiment, the condition for the $n^{th}$ order minima is given by $a \sin \theta = n \lambda$.
For small angles, $\sin \theta \approx \tan \theta = \frac{y}{D}$.
Thus, the distance of the $n^{th}$ minima from the central maximum is $y_n = \frac{n \lambda D}{a}$.
Given: $\lambda = 550 \,nm = 550 \times 10^{-9} \,m$, $a = 0.20 \,mm = 0.20 \times 10^{-3} \,m$, $D = 100 \,cm = 1.0 \,m$, and $n = 1$.
Substituting these values:
$y_1 = \frac{1 \times 550 \times 10^{-9} \times 1.0}{0.20 \times 10^{-3}} \,m$
$y_1 = \frac{550}{0.20} \times 10^{-6} \,m$
$y_1 = 2750 \times 10^{-6} \,m = 275 \times 10^{-5} \,m$.
Comparing this with $x \times 10^{-5} \,m$, we get $x = 275$.

(A) For a single slit diffraction, the condition for $n^{th}$ order minima is given by $b \sin \theta = n \lambda$.
For small angles, $\sin \theta \approx \theta$, so $\theta = \frac{n \lambda}{b}$.
Given: Wavelength $\lambda = 600 \,nm = 600 \times 10^{-9} \,m$, Slit width $b = 0.4 \,mm = 4 \times 10^{-4} \,m$, and order $n = 2$.
The angular position of the second-order minimum is $\theta = \frac{2 \times 600 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3} \,rad$.
The total angular divergence between the second-order minima on both sides of the central maximum is $2\theta$.
Total divergence $= 2 \times (3 \times 10^{-3} \,rad) = 6 \times 10^{-3} \,rad$.

(B) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$, screen distance $D = 1 \,m$, wavelength $\lambda = 500 \,nm = 5 \times 10^{-7} \,m$.
The width of the central maximum of a single slit diffraction pattern is given by $w = \frac{2 \lambda D}{a}$, where $a$ is the width of each slit.
The fringe width of the double slit interference pattern is $\beta = \frac{\lambda D}{d}$.
We are given that $10$ maxima of the double slit pattern fit within the central maximum of the single slit pattern. The central maximum spans from the first minimum on one side to the first minimum on the other, covering a width of $2 \beta_{total}$ or simply the range of $10$ fringes. Specifically, the condition is $10 \times \beta = \frac{2 \lambda D}{a}$.
Substituting $\beta = \frac{\lambda D}{d}$:
$10 \times \frac{\lambda D}{d} = \frac{2 \lambda D}{a}$
Simplifying the equation:
$\frac{10}{d} = \frac{2}{a}$
$a = \frac{2d}{10} = \frac{d}{5}$
Given $d = 1 \,mm = 10 \times 10^{-4} \,m$:
$a = \frac{10 \times 10^{-4} \,m}{5} = 2 \times 10^{-4} \,m$.
Thus, the value is $2$.

(C) For a single slit diffraction pattern,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 $\theta$ is the angular position.
For the second minimum $(n=2)$,$\sin \theta_1 = \frac{2 \lambda}{a}$.
For the third minimum $(n=3)$,$\sin \theta_2 = \frac{3 \lambda}{a}$.
The total angular separation is $\theta_1 + \theta_2 = 30^{\circ} = \frac{\pi}{6} \text{ radians}$.
Using the small angle approximation,$\sin \theta \approx \theta$ (in radians),we have:
$\theta_1 \approx \frac{2 \lambda}{a}$ and $\theta_2 \approx \frac{3 \lambda}{a}$.
Adding these,we get $\theta_1 + \theta_2 \approx \frac{2 \lambda}{a} + \frac{3 \lambda}{a} = \frac{5 \lambda}{a}$.
Given $\theta_1 + \theta_2 = 30^{\circ} = \frac{\pi}{6} \text{ radians}$,we have $\frac{5 \lambda}{a} = \frac{\pi}{6}$.
Substituting $\lambda = 628 \ nm = 0.628 \ \mu m$:
$a = \frac{5 \times 0.628 \times 6}{\pi} \approx \frac{18.84}{3.14} \approx 6 \ \mu m$.

(A) For a single slit diffraction experiment, the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda_1$, where $n = 1, 2, 3, ...$.
For the first minimum $(n=1)$, the position is $a \sin \theta = 1 \times \lambda_1 = 660 \ nm$.
The condition for the $m^{th}$ secondary maximum is given by $a \sin \theta = (2m + 1) \frac{\lambda_2}{2}$, where $m = 1, 2, 3, ...$.
For the first secondary maximum $(m=1)$, the position is $a \sin \theta = (2(1) + 1) \frac{\lambda_2}{2} = \frac{3}{2} \lambda_2$.
Since the positions coincide, we equate the two expressions:
$660 \ nm = \frac{3}{2} \lambda_2$.
Solving for $\lambda_2$:
$\lambda_2 = 660 \times \frac{2}{3} = 440 \ nm$.
Converting to $\mathring{A}$s $(\mathring{A})$:
$440 \ nm = 440 \times 10 \mathring{A} = 4400 \mathring{A}$.

(B) For single slit diffraction,the condition for the $n^{th}$ minima is given by $d \sin \theta = n \lambda$.
For the first minima,$n = 1$,so the condition becomes $d \sin \theta = \lambda$.
Given wavelength $\lambda = 600 \ \mu m = 600 \times 10^{-6} \ m = 6 \times 10^{-4} \ m$.
Given slit width $d = 1.2 \ mm = 1.2 \times 10^{-3} \ m = 12 \times 10^{-4} \ m$.
Substituting the values into the formula: $\sin \theta = \frac{\lambda}{d} = \frac{6 \times 10^{-4}}{12 \times 10^{-4}} = \frac{1}{2}$.
Since $\sin \theta = \frac{1}{2}$,the angle $\theta = \arcsin(0.5) = \frac{\pi}{6} \text{ radians}$.

(C) The width of the central maxima in a single-slit diffraction pattern is given by the formula: $W = \frac{2 \lambda D}{a}$.
Given values:
Slit width $a = 0.15 \ cm = 1.5 \times 10^{-3} \ m$.
Distance from screen $D = 2.1 \ m$.
Wavelength $\lambda = 5 \times 10^{-5} \ cm = 5 \times 10^{-7} \ m$.
Substituting these values into the formula:
$W = \frac{2 \times (5 \times 10^{-7} \ m) \times (2.1 \ m)}{1.5 \times 10^{-3} \ m}$.
$W = \frac{21 \times 10^{-7}}{1.5 \times 10^{-3}} \ m$.
$W = 14 \times 10^{-4} \ m = 1.4 \times 10^{-3} \ m = 1.4 \ mm$.

(D) The slit width is given by $b = \frac{m \lambda D}{d}$.
Given: $m = 3$,$\lambda = 600 \ nm = 600 \times 10^{-6} \ mm$,$D = 1 \ m = 1000 \ mm$,$d = 5 \ mm$.
Least counts: $\Delta D = 1 \ cm = 10 \ mm$,$\Delta d = 1 \ mm$.
Calculating $b$: $b = \frac{3 \times 600 \times 10^{-6} \times 1000}{5} = 0.36 \ mm = 360 \ \mu m$.
Using the method of maximum error: $b_{max} = \frac{m \lambda (D + \Delta D)}{(d - \Delta d)} = \frac{3 \times 600 \times 10^{-6} \times 1010}{4} = 0.4545 \ mm = 454.5 \ \mu m$.
$b_{min} = \frac{m \lambda (D - \Delta D)}{(d + \Delta d)} = \frac{3 \times 600 \times 10^{-6} \times 990}{6} = 0.297 \ mm = 297 \ \mu m$.
Errors: $\Delta b_1 = |454.5 - 360| = 94.5 \ \mu m$ and $\Delta b_2 = |297 - 360| = 63 \ \mu m$.
Using the differential method: $\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d} = \frac{10}{1000} + \frac{1}{5} = 0.01 + 0.2 = 0.21$.
$\Delta b = 0.21 \times 360 = 75.6 \ \mu m$.
Thus,the possible errors are $75.6 \ \mu m$ or $94.5 \ \mu m$.

(D) In a single slit diffraction pattern,the central maximum is the brightest and widest. As we move away from the center,the intensity of the secondary maxima decreases rapidly. The width of the central maximum is $2\lambda D/a$,while the width of the secondary maxima is $\lambda D/a$. Therefore,the fringes have unequal width and unequal intensity. Option $D$ states that the fringes have equal width and equal intensity,which is incorrect.

(C) For a single slit diffraction pattern,the condition for the $n^{th}$ secondary maximum is given by: $a \sin \theta = (n + \frac{1}{2}) \lambda$,where $a$ is the slit width,$\theta$ is the angle of diffraction,and $\lambda$ is the wavelength.
Since $\theta$ is very small,$\sin \theta \approx \tan \theta = \frac{y}{D}$,where $y$ is the distance from the center and $D$ is the distance to the screen.
For the second maximum,$n = 2$. Thus,$a \frac{y}{D} = (2 + \frac{1}{2}) \lambda = \frac{5}{2} \lambda$.
Given: $a = 0.65 \ mm = 0.65 \times 10^{-3} \ m$,$D = 1.3 \ m$,and $y = 2.6 \ mm = 2.6 \times 10^{-3} \ m$.
Substituting the values: $(0.65 \times 10^{-3}) \times \frac{2.6 \times 10^{-3}}{1.3} = \frac{5}{2} \lambda$.
$(0.65 \times 10^{-3}) \times (2 \times 10^{-3}) = 2.5 \lambda$.
$1.3 \times 10^{-6} = 2.5 \lambda$.
$\lambda = \frac{1.3 \times 10^{-6}}{2.5} = 0.52 \times 10^{-6} \ m = 5200 \ \times 10^{-10} \ m = 5200 \ Å$.

(B) For a single slit diffraction pattern,the condition for secondary maxima is given by $a \sin \theta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$ is the order of the secondary maximum.
For the second secondary maximum $(n = 2)$ with wavelength $\lambda_1 = 6384 Å$,the condition is $a \sin \theta = (2 + \frac{1}{2}) \lambda_1 = \frac{5}{2} \lambda_1$.
For the third secondary maximum $(n = 3)$ with wavelength $\lambda_0$,the condition is $a \sin \theta = (3 + \frac{1}{2}) \lambda_0 = \frac{7}{2} \lambda_0$.
Since the positions coincide,we equate the two expressions: $\frac{5}{2} \lambda_1 = \frac{7}{2} \lambda_0$.
This simplifies to $5 \lambda_1 = 7 \lambda_0$.
Substituting $\lambda_1 = 6384 Å$,we get $\lambda_0 = \frac{5 \times 6384}{7} = \frac{31920}{7} = 4560 Å$.

(C) The width of the central maximum in a single slit diffraction experiment is given by the formula: $y = \frac{2D\lambda}{a}$,where '$D$' is the distance of the screen from the slit,'$\lambda$' is the wavelength of light,and '$a$' is the width of the slit.
Initially,$y = \frac{2D\lambda}{a}$.
In the second case,the slit width is halved,so the new width is $a' = \frac{a}{2}$.
The new wavelength is $\lambda' = 1.5\lambda = \frac{3}{2}\lambda$.
The new width of the central maximum '$y'$' is given by: $y' = \frac{2D\lambda'}{a'} = \frac{2D(1.5\lambda)}{a/2}$.
Simplifying this,we get: $y' = \frac{2D(3/2\lambda)}{a/2} = \frac{3D\lambda}{a/2} = \frac{6D\lambda}{a}$.
Since $y = \frac{2D\lambda}{a}$,we can substitute this into the expression for $y'$:
$y' = 3 \times (\frac{2D\lambda}{a}) = 3y$.
Therefore,the new width of the central maximum is $3y$.

(B) The condition for the $n^{\text{th}}$ secondary maximum in a single slit diffraction pattern is given by: $a \sin \theta = (n + \frac{1}{2}) \lambda'$,where $n$ is the order of the secondary maximum.
For the $4^{\text{th}}$ secondary maximum of wavelength $\lambda_1$,the condition is: $a \sin \theta_1 = (4 + \frac{1}{2}) \lambda_1 = \frac{9}{2} \lambda_1$.
For the $3^{\text{rd}}$ secondary maximum of wavelength $\lambda$,the condition is: $a \sin \theta_2 = (3 + \frac{1}{2}) \lambda = \frac{7}{2} \lambda$.
Since the maxima coincide,$\theta_1 = \theta_2$,so $a \sin \theta_1 = a \sin \theta_2$.
Equating the two expressions: $\frac{9}{2} \lambda_1 = \frac{7}{2} \lambda$.
Solving for $\lambda_1$: $\lambda_1 = \frac{7}{9} \lambda$.

(B) The width of the diffraction bands (or the distance between successive minima/maxima) is given by the formula $\beta = \frac{\lambda D}{a}$.
From this relation,we can see that the band width $\beta$ is directly proportional to the wavelength $\lambda$ of the light used,i.e.,$\beta \propto \lambda$.
We know that the wavelength of blue light is smaller than the wavelength of red light,i.e.,$\lambda_{\text{blue}} < \lambda_{\text{red}}$.
Since $\lambda$ decreases,the band width $\beta$ also decreases.
Therefore,the diffraction bands become narrow and crowded together.

(B) For the $n^{\text{th}}$ secondary maximum in a single slit diffraction pattern,the position is given by $x_n = \frac{(2n+1) \lambda D}{2a}$.
For the second secondary maximum $(n=2)$ with wavelength $\lambda = 6195 Å$,the position is $x_2 = \frac{(2 \times 2 + 1) \lambda D}{2a} = \frac{5 \lambda D}{2a}$.
For the third secondary maximum $(n=3)$ with wavelength $\lambda_0$,the position is $x_3 = \frac{(2 \times 3 + 1) \lambda_0 D}{2a} = \frac{7 \lambda_0 D}{2a}$.
Since the positions coincide,$x_2 = x_3$,which implies $\frac{5 \lambda D}{2a} = \frac{7 \lambda_0 D}{2a}$.
Simplifying this,we get $5 \lambda = 7 \lambda_0$.
Therefore,$\lambda_0 = \frac{5 \lambda}{7} = \frac{5 \times 6195 Å}{7} = 5 \times 885 Å = 4425 Å$.

(A) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{d}$,where $D$ is the distance to the screen,$\lambda$ is the wavelength,and $d$ is the slit width.
Given:
$D = 50 \ cm = 0.5 \ m$
$\lambda = 600 \ nm = 600 \times 10^{-9} \ m$
Separation between $1^{st}$ and $3^{rd}$ minima: $\Delta x = x_3 - x_1 = 3 \ mm = 3 \times 10^{-3} \ m$
Using the formula:
$x_3 - x_1 = (3 - 1) \frac{D \lambda}{d} = \frac{2 D \lambda}{d}$
Rearranging for $d$:
$d = \frac{2 D \lambda}{\Delta x}$
$d = \frac{2 \times 0.5 \times 600 \times 10^{-9}}{3 \times 10^{-3}}$
$d = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = 200 \times 10^{-6} \ m = 0.2 \ mm$.

(C) The width of the central diffraction maximum is given by $W = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit width.
Initially,$W = Y = \frac{2 \times 400 \times D}{d}$.
When half the slit is covered,the new slit width $d' = \frac{d}{2}$.
The new wavelength $\lambda' = 600 \ nm$.
The new width $W'$ is given by $W' = \frac{2 \lambda' D}{d'} = \frac{2 \times 600 \times D}{d/2} = \frac{4 \times 600 \times D}{d}$.
Taking the ratio: $\frac{W'}{Y} = \frac{4 \times 600 \times D / d}{2 \times 400 \times D / d} = \frac{2400}{800} = 3$.
Therefore,$W' = 3 Y$.

(A) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $\sin \theta = (2n + 1) \frac{\lambda}{2a}$.
For the first secondary maximum $(n = 1)$,$\sin \theta = \frac{3\lambda}{2a}$.
Since the angle $\theta$ is very small,$\sin \theta \approx \tan \theta = \frac{x}{D}$,where $x$ is the distance from the central maximum and $D$ is the distance to the screen.
Thus,$x = \frac{3\lambda D}{2a}$.
The separation between the positions of the first secondary maxima for two wavelengths $\lambda_1$ and $\lambda_2$ is $\Delta x = \frac{3D}{2a} (\lambda_2 - \lambda_1)$.
Given: $\lambda_1 = 590 \times 10^{-9} \ m$,$\lambda_2 = 596 \times 10^{-9} \ m$,$D = 2 \ m$,$a = 2.4 \times 10^{-3} \ m$.
Substituting the values:
$\Delta x = \frac{3 \times 2 \times (596 - 590) \times 10^{-9}}{2 \times 2.4 \times 10^{-3}}$
$\Delta x = \frac{6 \times 6 \times 10^{-9}}{4.8 \times 10^{-3}} = \frac{36 \times 10^{-9}}{4.8 \times 10^{-3}} = 7.5 \times 10^{-6} \ m$.

(B) For the $n^{\text{th}}$ minimum in a single-slit diffraction pattern,the condition is $a \sin \theta_n = n \lambda$.
For the $n^{\text{th}}$ secondary maximum,the condition is $a \sin \theta_n = (2n + 1) \frac{\lambda}{2}$.
Given for the first minimum $(n=1)$: $a \sin 30^{\circ} = 1 \cdot \lambda \Rightarrow a \cdot \frac{1}{2} = \lambda \Rightarrow a = 2\lambda$.
For the first secondary maximum $(n=1)$: $a \sin \theta = (2(1) + 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$.
Substituting $a = 2\lambda$ into the equation: $(2\lambda) \sin \theta = \frac{3\lambda}{2}$.
$\sin \theta = \frac{3}{4} \Rightarrow \theta = \sin^{-1}\left(\frac{3}{4}\right)$.

(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $\theta = \frac{\lambda}{d}$.
For the first case: $\theta = \frac{\lambda}{d} \implies d = \frac{\lambda}{\theta}$.
For the second case: $q\theta = \frac{p\lambda}{d'} \implies d' = \frac{p\lambda}{q\theta}$.
The half-angular width of the $n^{th}$ secondary maxima is given by $\theta_n = \frac{(2n+1)\lambda}{2d}$.
For the first secondary maxima $(n=1)$,$\theta_{s1} = \frac{3\lambda}{2d}$ and $\theta_{s2} = \frac{3(p\lambda)}{2d'}$.
The ratio is $\frac{\theta_{s1}}{\theta_{s2}} = \frac{3\lambda / 2d}{3p\lambda / 2d'} = \frac{1}{d} \cdot \frac{d'}{p} = \frac{d'}{pd}$.
Substituting $d' = \frac{p\lambda}{q\theta}$ and $d = \frac{\lambda}{\theta}$,we get $\frac{\theta_{s1}}{\theta_{s2}} = \frac{p\lambda / q\theta}{p(\lambda / \theta)} = \frac{1}{q}$.
Thus,the ratio is $1:q$,which corresponds to the inverse of the given options if interpreted as $q:1$ or $1:q$. Given the standard form,the ratio of widths is $1:q$.