Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction Questions in English

(B) The Fraunhofer lines are dark absorption lines observed in the solar spectrum.
When the continuous light emitted from the Sun's photosphere passes through the cooler outer layer known as the chromosphere,specific wavelengths of light are absorbed by the atoms present in the chromosphere.
This selective absorption of light results in the appearance of dark lines in the otherwise continuous spectrum of the Sun,which are known as Fraunhofer lines.

(C) Fraunhofer lines are dark absorption lines observed in the solar spectrum.
These lines are produced when the continuous spectrum emitted by the hot photosphere of the sun passes through the cooler gases and vapours of elements present in the sun's chromosphere.
As the light passes through these cooler vapours,specific wavelengths corresponding to the characteristic absorption spectra of these elements are absorbed,resulting in the dark lines known as Fraunhofer lines.
Therefore,the correct option is $C$.

(A) The Fraunhofer lines are observed in the solar spectrum.
These lines are formed when the atoms in the cooler outer layers of the Sun (the chromosphere) absorb specific wavelengths of light emitted by the hotter inner layers (the photosphere).
Since these atoms absorb discrete wavelengths,they produce dark lines in the continuous spectrum of the Sun.
Therefore,the Fraunhofer spectrum is a line absorption spectrum.

(A) For Fraunhofer diffraction at a single slit,the condition for the $n^{th}$ minimum is given by $d \sin \theta = n \lambda$.
Here,$d$ is the slit width,$\lambda$ is the wavelength of light,and $\theta$ is the angle of diffraction.
For the first minimum,we set $n = 1$.
Given values are $\lambda = 550 \,nm = 550 \times 10^{-9} \,m$ and $d = 0.55 \,mm = 0.55 \times 10^{-3} \,m$.
Substituting these values into the formula: $\sin \theta = \frac{n \lambda}{d} = \frac{1 \times 550 \times 10^{-9}}{0.55 \times 10^{-3}}$.
$\sin \theta = \frac{550 \times 10^{-9}}{550 \times 10^{-6}} = 10^{-3}$.
Since $\theta$ is very small,$\sin \theta \approx \theta$.
Therefore,$\theta = 0.001 \,rad$.

(B) In the case of diffraction by a circular disc (Poisson's spot),the intensity at the center depends on the number of Half-Period Zones $(HPZ)$ blocked by the disc.
The number of $HPZ$ blocked is given by $n = \frac{A}{\pi \lambda d}$,where $A$ is the area of the disc,$\lambda$ is the wavelength,and $d$ is the distance between the disc and the screen.
Since $A, \pi,$ and $\lambda$ are constants,we have $n \propto \frac{1}{d}$.
When the distance $d$ is decreased,the number of blocked $HPZ$ $(n)$ increases.
As more $HPZ$ are blocked,the destructive interference at the center increases,which leads to a decrease in the intensity of the central bright spot.

(D) For a single slit of width $a$,the condition for the $n^{th}$ secondary maxima in Fraunhofer diffraction is given by the path difference:
$a \sin \theta = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, ...$
For the first secondary maximum,we substitute $n = 1$:
$a \sin \theta = (2(1) + 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$
Thus,the correct option is $D$.

(B) Fraunhofer lines are dark absorption lines observed in the solar spectrum.
These lines are formed because the cooler gases in the Sun's chromosphere absorb specific wavelengths of light emitted by the hotter photosphere beneath it.
Therefore,they indicate the absorption of certain wavelengths by the elements present in the Sun's chromosphere.

(B) Given: Wavelength $\lambda = 6000 \, \mathring A = 6000 \times 10^{-10} \, m$.
Slit width $a = 0.2 \, mm = 0.2 \times 10^{-3} \, m$.
Distance of screen $D = 2 \, m$.
The angular width of the central maximum is given by $2\theta = \frac{2\lambda}{a}$.
The linear width of the central maximum is $W = D \times (2\theta) = \frac{2\lambda D}{a}$.
Substituting the values: $W = \frac{2 \times 6000 \times 10^{-10} \times 2}{0.2 \times 10^{-3}}$.
$W = \frac{24000 \times 10^{-10}}{0.2 \times 10^{-3}} = \frac{2.4 \times 10^{-6}}{0.2 \times 10^{-3}} = 12 \times 10^{-3} \, m = 12 \, mm$.

(B) The condition for the first minimum in a single-slit Fraunhofer diffraction pattern is given by $a \sin \theta = n \lambda$. For the first minimum,$n = 1$,so $\sin \theta = \frac{\lambda}{a}$.
Here,the slit width $a = 12 \times 10^{-5} \text{ cm} = 12 \times 10^{-7} \text{ m}$.
The wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$.
Substituting the values: $\sin \theta = \frac{6 \times 10^{-7}}{12 \times 10^{-7}} = \frac{6}{12} = 0.5$.
Since $\sin \theta = 0.5$,the half-angular width $\theta = \arcsin(0.5) = 30^{\circ}$.

(C) The radius of the $n^{th}$ Fresnel zone is given by the formula $r_n = \sqrt{n b \lambda}$,where $b$ is the distance from the aperture to the screen and $\lambda$ is the wavelength of light.
Given that the radius of the $4^{th}$ zone for wavelength ${\lambda _1}$ is equal to the radius of the $5^{th}$ zone for wavelength ${\lambda _2}$,we have:
$r_4 = \sqrt{4 b \lambda_1}$
$r_5 = \sqrt{5 b \lambda_2}$
Since $r_4 = r_5$,we equate the expressions:
$\sqrt{4 b \lambda_1} = \sqrt{5 b \lambda_2}$
Squaring both sides,we get:
$4 b \lambda_1 = 5 b \lambda_2$
Dividing both sides by $b$ (where $b \neq 0$):
$4 \lambda_1 = 5 \lambda_2$
Therefore,the ratio is:
$\frac{\lambda_1}{\lambda_2} = \frac{5}{4}$

(B) According to the Rayleigh criterion,two point sources are said to be 'just resolved' when the central maximum of the diffraction pattern of one source falls exactly on the first minimum of the diffraction pattern of the other source.
In this condition,the resultant intensity curve shows a distinct dip between the two central maxima,which is approximately $80\%$ of the peak intensity.
Looking at the provided diagrams:
- Diagram $A$ shows two sources that are well-separated (resolved).
- Diagram $B$ shows the condition where the central maximum of one coincides with the first minimum of the other,which is the definition of 'just resolved'.
- Diagram $C$ shows the sources closer than the Rayleigh limit (unresolved).
- Diagram $D$ shows the sources very close,appearing as a single source (unresolved).
Therefore,the correct representation is Diagram $B$.

(C) The angular width of the central maxima in Fraunhofer diffraction is given by the formula: $\omega_{\theta} = \frac{2\lambda}{a}$.
Given:
Wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m}$.
Slit width $a = 12 \times 10^{-5} \, \text{cm} = 12 \times 10^{-7} \, \text{m}$.
Substituting the values into the formula:
$\omega_{\theta} = \frac{2 \times 6 \times 10^{-7} \, \text{m}}{12 \times 10^{-7} \, \text{m}}$.
$\omega_{\theta} = \frac{12 \times 10^{-7}}{12 \times 10^{-7}} = 1 \, \text{rad}$.
Thus,the angular width is $1 \, \text{rad}$.

(B) The Fresnel distance $(z_F)$ is the distance for which ray optics is a good approximation. It is given by the formula: $z_F = \frac{a^2}{\lambda}$.
Given:
Aperture width $(a)$ = $4 \, mm = 4 \times 10^{-3} \, m$.
Wavelength $(\lambda)$ = $500 \, nm = 500 \times 10^{-9} \, m$.
Substituting the values:
$z_F = \frac{(4 \times 10^{-3})^2}{500 \times 10^{-9}}$
$z_F = \frac{16 \times 10^{-6}}{5 \times 10^{-7}}$
$z_F = \frac{16}{5} \times 10^1 = 3.2 \times 10 = 32 \, m$.

(B) The angular width of the central maxima in Fraunhofer diffraction is given by $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.
Since $\theta \propto \lambda$,we have the ratio $\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1}$.
Given that the angular width decreases by $30 \%$,the new angular width $\theta_2 = \theta_0 - 0.30\theta_0 = 0.70\theta_0$.
Substituting the values: $\frac{0.70\theta_0}{\theta_0} = \frac{\lambda_2}{6000 \; \mathring{A}}$.
$0.70 = \frac{\lambda_2}{6000 \; \mathring{A}}$.
$\lambda_2 = 0.70 \times 6000 \; \mathring{A} = 4200 \; \mathring{A}$.

(18 M) The Fresnel distance $(z_{F})$ is the distance for which ray optics is a good approximation. It is given by the formula $z_{F} = \frac{a^{2}}{\lambda}$,where $a$ is the aperture width and $\lambda$ is the wavelength.
Given: $a = 3\; mm = 3 \times 10^{-3}\; m$ and $\lambda = 500\; nm = 5 \times 10^{-7}\; m$.
Substituting the values:
$z_{F} = \frac{(3 \times 10^{-3})^{2}}{5 \times 10^{-7}}$
$z_{F} = \frac{9 \times 10^{-6}}{5 \times 10^{-7}}$
$z_{F} = 1.8 \times 10^{1} = 18\; m$.
Thus,ray optics is a good approximation for distances up to $18\; m$.

(40 M) Fresnel's distance $(Z_{F})$ is the distance for which ray optics is a good approximation.
It is given by the relation:
$Z_{F} = \frac{a^{2}}{\lambda}$
Where:
Aperture width,$a = 4 \; mm = 4 \times 10^{-3} \; m$
Wavelength of light,$\lambda = 400 \; nm = 400 \times 10^{-9} \; m$
Substituting the values:
$Z_{F} = \frac{(4 \times 10^{-3})^{2}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16 \times 10^{-6}}{400 \times 10^{-9}}$
$Z_{F} = \frac{16}{400} \times 10^{3} = 0.04 \times 1000 = 40 \; m$
Therefore,the distance for which ray optics is a good approximation is $40 \; m$.

(C) The distance between the two towers is $D = 40 \, km = 40,000 \, m$. The hill is located at the midpoint,so the distance from either tower to the hill is $Z_F = D/2 = 20,000 \, m$.
For radio waves to travel between the towers without significant diffraction,the Fresnel distance $Z_F$ must be at least equal to the distance to the obstacle. The condition for the Fresnel distance is given by $Z_F = \frac{a^2}{\lambda}$,where $a$ is the size of the aperture (or the clearance height) and $\lambda$ is the wavelength.
Here,the clearance height is $a = 50 \, m$.
Rearranging the formula to solve for the wavelength $\lambda$:
$\lambda = \frac{a^2}{Z_F}$
Substituting the values:
$\lambda = \frac{(50 \, m)^2}{20,000 \, m} = \frac{2500}{20,000} \, m = 0.125 \, m$.
Converting the wavelength to centimeters:
$\lambda = 0.125 \, m \times 100 \, cm/m = 12.5 \, cm$.
Thus,the longest wavelength is $12.5 \, cm$.

(N/A) In Fraunhofer diffraction,the graph of the intensity of light on the screen versus the diffraction angle $\theta$ is shown in the figure.
From the graph,it can be observed that the intensity of the central maxima is the highest.
$(i)$ All minima have zero light intensity.
$(ii)$ The intensity of the secondary maxima decreases as the angle of diffraction increases. Thus,the intensity decreases as the order increases,and the width of the maxima also decreases.
$(iii)$ As the ratio $\frac{\lambda}{a}$ is small,the diffraction effect is less,and if it is large,the diffraction effect is more.

(N/A) An aperture of size $a$ illuminated by a parallel beam sends diffracted light into an angle $\theta$,where the angular width is $\theta \approx \frac{\lambda}{a}$.
In travelling a distance $z$,the diffracted beam acquires a width $w = z \theta = \frac{z \lambda}{a}$ due to diffraction.
The Fresnel distance $z_F$ is defined as the distance at which the spreading due to diffraction becomes equal to the size of the aperture $a$.
Setting $a = \frac{z_F \lambda}{a}$,we get $z_F = \frac{a^2}{\lambda}$.
For distances $z \ll z_F$,the spreading due to diffraction is negligible compared to the size of the aperture,and the beam travels in a straight line,which is consistent with ray optics.
For distances $z > z_F$,the spreading due to diffraction dominates over the spreading due to ray optics. Thus,ray optics is an approximation valid only when the wavelength $\lambda \to 0$ or for distances $z \ll z_F$.

(A) The limit of resolution of optical instruments like telescopes and microscopes is fundamentally imposed by the phenomenon of $Diffraction$.
When light passes through a circular aperture (like the objective lens of a telescope or microscope),it does not form a perfect point image due to the wave nature of light.
Instead,it forms a diffraction pattern consisting of a central bright spot (Airy disk) surrounded by concentric dark and bright rings.
According to the Rayleigh criterion,two point objects are just resolved when the central maximum of the diffraction pattern of one coincides with the first minimum of the diffraction pattern of the other.
Therefore,diffraction sets the ultimate limit on the resolving power of these instruments.

(N/A) The Fresnel distance is defined as the distance from a slit or aperture at which the spreading of light due to diffraction becomes comparable to the size of the aperture itself.
Beyond this distance,the ray optics approximation (rectilinear propagation of light) is no longer valid,and diffraction effects become significant.
It acts as a limit to the validity of ray optics.
The formula for Fresnel distance $(z_F)$ is given by:
$z_F = \frac{a^2}{\lambda}$
Where:
$a$ = size of the aperture or slit width
$\lambda$ = wavelength of the light used

(A) Given: Speed of student $v = 5.4 \,km/h = 5.4 \times \frac{5}{18} \,m/s = 1.5 \,m/s$.
Distance of pipe opening from the path $D = 8 \,m$.
Diameter of the pipe $d = 0.45 \,m$.
Frequency of sound $f = 1280 \,Hz$.
Speed of sound $v_s = 320 \,m/s$.
First,calculate the wavelength of the sound: $\lambda = \frac{v_s}{f} = \frac{320}{1280} = 0.25 \,m$.
The diffraction of sound waves at the circular opening of the pipe follows the condition for the first minimum: $\sin \theta = 1.22 \frac{\lambda}{d}$.
For small angles,$\sin \theta \approx \tan \theta = \frac{y}{D}$,where $y$ is the distance from the center of the central maximum to the first minimum.
$y = 1.22 \frac{\lambda}{d} D = 1.22 \times \frac{0.25}{0.45} \times 8 = 1.22 \times \frac{1}{1.8} \times 8 \approx 5.42 \,m$.
The width of the central maximum is $2y = 2 \times 5.42 = 10.84 \,m$.
The time $T$ for which the sound is heard is the time taken to cover this width: $T = \frac{2y}{v} = \frac{10.84}{1.5} \approx 7.23 \,s$.
This value lies in the range of $6-12$ seconds.

(C) In Fraunhofer diffraction,the source of light and the screen are effectively at an infinite distance from the aperture (slit).
Because the source is at an infinite distance,the wave front incident on the slit is a plane wave front.
Therefore,the correct condition is that the incident wave front must be plane.

(B) In Fraunhofer diffraction due to a single slit of width $a$,the angular position of the $n^{th}$ minima is given by $\sin \theta = \frac{n \lambda}{a}$. For small angles,$\theta \approx \frac{n \lambda}{a}$.
The angular width of the secondary maxima or minima is given by $\Delta \theta = \frac{\lambda}{a}$.
Given: $\lambda = 6328 \times 10^{-10} \ m$ and $a = 0.2 \times 10^{-3} \ m$.
$\Delta \theta = \frac{6328 \times 10^{-10}}{0.2 \times 10^{-3}} \ rad = 3.164 \times 10^{-3} \ rad$.
To convert radians to degrees,multiply by $\frac{180^{\circ}}{\pi}$:
$\Delta \theta = 3.164 \times 10^{-3} \times \frac{180}{3.14159} \approx 0.18^{\circ}$.

(A) The linear width of the central maximum in a Fraunhofer diffraction pattern is given by the formula $w = \frac{2 \lambda D}{d}$.
According to the problem,the linear width of the central maximum is equal to two times the width of the slit,so $w = 2d$.
Equating the two expressions: $\frac{2 \lambda D}{d} = 2d$.
Dividing both sides by $2$,we get $\frac{\lambda D}{d} = d$.
Solving for $D$,we find $D = \frac{d^2}{\lambda}$.

(C) For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \theta = \frac{y_n}{D}$,where $y_n$ is the distance from the central maximum.
Thus,$y_n = \frac{n \lambda D}{a}$.
The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.
Given: $a = 0.2 \ mm = 2 \times 10^{-4} \ m$,$D = 2 \ m$,and $2y_1 = 1 \ cm = 10^{-2} \ m$.
Substituting the values: $10^{-2} = \frac{2 \times \lambda \times 2}{2 \times 10^{-4}}$.
$10^{-2} = \frac{4 \lambda}{2 \times 10^{-4}} = 2 \times 10^4 \lambda$.
$\lambda = \frac{10^{-2}}{2 \times 10^4} = 0.5 \times 10^{-6} \ m = 5000 \times 10^{-10} \ m = 5000 \ Å$.

(D) Given: Slit width $a = 0.3 \ mm = 0.3 \times 10^{-3} \ m$,Screen distance $D = 1.5 \ m$,Wavelength $\lambda = 4500 \ Å = 4500 \times 10^{-10} \ m$.
For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$. For small $\theta$,$\sin \theta \approx \theta = \frac{y}{D}$.
Thus,$a \left( \frac{y}{D} \right) = n \lambda \implies y_n = \frac{n \lambda D}{a}$.
The distance of the first minimum $(n=1)$ from the central maximum is $y_1 = \frac{\lambda D}{a}$.
The distance between the first minimum on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.
Substituting the values: $2y_1 = \frac{2 \times 4500 \times 10^{-10} \times 1.5}{0.3 \times 10^{-3}}$.
$2y_1 = \frac{13500 \times 10^{-10}}{0.3 \times 10^{-3}} = 45000 \times 10^{-7} \ m = 4.5 \times 10^{-3} \ m = 4.5 \ mm$.

(C) The angular width of the central maximum in a Fraunhofer diffraction pattern is given by the formula: $\theta = \frac{2 \lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.
From this relation,we can see that $\theta \propto \lambda$.
Let the initial wavelength be $\lambda_1 = 6000 Å$ and the initial angular width be $\theta_1$.
When the wavelength is changed to $\lambda_2$,the angular width becomes $\theta_2 = \theta_1 - 0.20 \theta_1 = 0.80 \theta_1$.
Using the proportionality $\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1}$,we get:
$\frac{0.80 \theta_1}{\theta_1} = \frac{\lambda_2}{6000 Å}$
$0.80 = \frac{\lambda_2}{6000 Å}$
$\lambda_2 = 0.80 \times 6000 Å = 4800 Å$.

(C) Given: Slit width $a = 0.2 \, mm = 0.2 \times 10^{-3} \, m$.
Distance of screen $D = 2 \, m$.
Wavelength $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
The position of the $n^{th}$ minimum in a single slit diffraction pattern is given by $a \sin \theta = n \lambda$.
For small $\theta$, $\sin \theta \approx \theta = \frac{y}{D}$.
Thus, $y_n = \frac{n \lambda D}{a}$.
The distance between the first minima on either side of the central maximum is the width of the central maximum, given by $w = y_1 - (-y_1) = 2 y_1 = \frac{2 \lambda D}{a}$.
Substituting the values: $w = \frac{2 \times (5 \times 10^{-7} \, m) \times (2 \, m)}{0.2 \times 10^{-3} \, m} = \frac{20 \times 10^{-7}}{0.2 \times 10^{-3}} = 100 \times 10^{-4} \, m = 10^{-2} \, m$.

(C) The distance of the $1^{st}$ minima from the central maxima is given by $y_{1d} = \frac{\lambda D}{a}$.
Here,$\lambda = 5500 \ Å = 5500 \times 10^{-10} \ m$,$D = 2 \ m$,and $a = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
The distance between two minima on either side of the central maxima is $2 y_{1d} = \frac{2 \lambda D}{a}$.
Substituting the values: $2 y_{1d} = \frac{2 \times 5500 \times 10^{-10} \times 2}{0.5 \times 10^{-3}} = \frac{22000 \times 10^{-10}}{0.5 \times 10^{-3}} = 44000 \times 10^{-7} \ m = 4.4 \times 10^{-3} \ m = 4.4 \ mm$.

(B) The Fresnel distance $(Z_F)$ is the distance for which ray optics is a good approximation. It is given by the formula: $Z_F = \frac{a^2}{\lambda}$.
Given:
Aperture,$a = 4 \,mm = 4 \times 10^{-3} \,m$.
Wavelength,$\lambda = 400 \,nm = 400 \times 10^{-9} \,m$.
Substituting the values into the formula:
$Z_F = \frac{(4 \times 10^{-3})^2}{400 \times 10^{-9}}$
$Z_F = \frac{16 \times 10^{-6}}{4 \times 10^{-7}}$
$Z_F = 4 \times 10^1 = 40 \,m$.
Thus,ray optics is a good approximation for a distance of $40 \,m$.

(B) For the first diffraction minimum,the condition is given by $a \sin \theta = n\lambda$. For $n=1$,$a \sin \theta = \lambda$.
Given $\theta = 30^{\circ}$,we have $a = \frac{\lambda}{\sin 30^{\circ}} = 2\lambda$.
For the first secondary maximum,the condition is $a \sin \theta' = \frac{3\lambda}{2}$.
Substituting the value of $a = 2\lambda$ into the equation:
$2\lambda \sin \theta' = \frac{3\lambda}{2}$
$\sin \theta' = \frac{3\lambda}{2 \times 2\lambda} = \frac{3}{4}$.
Therefore,$\theta' = \sin^{-1}\left(\frac{3}{4}\right)$.

(C) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $x_n = \frac{(2n+1) \lambda D}{2a}$,where $n$ is the order of the secondary maximum.
For the second secondary maximum $(n=2)$ of red light $(\lambda_1 = 6500 Å)$: $x_2 = \frac{(2(2)+1) \lambda_1 D}{2a} = \frac{5 \lambda_1 D}{2a}$.
For the third secondary maximum $(n=3)$ of an unknown wavelength $(\lambda_2)$: $x_3 = \frac{(2(3)+1) \lambda_2 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
Since the maxima coincide,$x_2 = x_3$,which implies $\frac{5 \lambda_1 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
Simplifying,we get $5 \lambda_1 = 7 \lambda_2$.
Substituting $\lambda_1 = 6500 Å$,we have $5 \times 6500 = 7 \times \lambda_2$.
$\lambda_2 = \frac{32500}{7} Å \approx 4642.8 Å$.

(B) In a Fraunhofer diffraction pattern,the condition for secondary maxima is given by the formula:
$\theta = (2n + 1) \frac{\lambda}{2a}$
For the first secondary maximum,we take $n = 1$:
$\theta = (2(1) + 1) \frac{\lambda}{2a} = \frac{3 \lambda}{2a}$
The distance $x$ of the secondary maximum from the central maximum on a screen at distance $D$ is given by $x = D \tan \theta$. For small angles,$\tan \theta \approx \theta$:
$x = D \theta = D \left( \frac{3 \lambda}{2a} \right) = \frac{3 D \lambda}{2a}$

(A) The Fresnel distance $(z_F)$ is the distance up to which ray optics is a good approximation. It is given by the formula:
$z_F = \frac{a^2}{\lambda}$
where $a$ is the aperture size and $\lambda$ is the wavelength of light.
Given:
Aperture $a = 5 \times 10^{-3} \ m$
Fresnel distance $z_F = 50 \ m$
Substituting the values into the formula:
$50 = \frac{(5 \times 10^{-3})^2}{\lambda}$
$50 = \frac{25 \times 10^{-6}}{\lambda}$
$\lambda = \frac{25 \times 10^{-6}}{50}$
$\lambda = 0.5 \times 10^{-6} \ m$
$\lambda = 5 \times 10^{-7} \ m$
Converting to $\text{Å}$s $(\text{Å})$:
$1 \ \text{Å} = 10^{-10} \ m$
$\lambda = 5 \times 10^{-7} \times 10^{10} \ \text{Å} = 5000 \ \text{Å}$
Therefore, the correct option is $A$.

(B) The Fresnel distance $(z_F)$ is the distance at which ray optics is a good approximation for an aperture of size $a$ and wavelength $\lambda$. It is given by the formula: $z_F = \frac{a^2}{\lambda}$.
Given:
Aperture $a = 0.3 \ cm = 0.3 \times 10^{-2} \ m = 3 \times 10^{-3} \ m$.
Wavelength $\lambda = 6000 \ Å = 6000 \times 10^{-10} \ m = 6 \times 10^{-7} \ m$.
Substituting the values into the formula:
$z_F = \frac{(3 \times 10^{-3})^2}{6 \times 10^{-7}}$
$z_F = \frac{9 \times 10^{-6}}{6 \times 10^{-7}}$
$z_F = 1.5 \times 10^1 = 15 \ m$.
Therefore,the correct option is $B$.

(B) The formula for Fresnel distance $(Z_F)$ is given by $Z_F = \frac{a^2}{\lambda}$, where $a$ is the slit width and $\lambda$ is the wavelength of light.
Given:
Slit width $a = 2 \, mm = 2 \times 10^{-3} \, m$
Wavelength $\lambda = 4000 \, Å = 4 \times 10^{-7} \, m$
Substituting these values into the formula:
$Z_F = \frac{(2 \times 10^{-3})^2}{4 \times 10^{-7}}$
$Z_F = \frac{4 \times 10^{-6}}{4 \times 10^{-7}}$
$Z_F = 10 \, m$
Therefore, the Fresnel distance is $10 \, m$.

(B) In Fresnel diffraction,the source of light and the screen are at finite distances from the obstacle or aperture. No lenses are required to make the rays parallel. The resulting diffraction pattern at any point on the screen is determined by the superposition of wavelets originating from different half-period zones of the wavefront. The intensity at a specific point depends on the number of half-period zones that contribute to the superposition at that point,as these zones can interfere constructively or destructively.

(C) Fraunhofer diffraction occurs when the source of light and the screen are effectively at an infinite distance from the obstacle or aperture.
In practical terms,this is achieved when the Fresnel distance $z_F = \frac{b^2}{\lambda}$ is much smaller than the distance $L$ between the obstacle and the screen.
Therefore,the condition is $L \gg \frac{b^2}{\lambda}$,which can be rearranged as $\frac{b^2}{L \lambda} \ll 1$.

(A) Statement $A$ is true: In Fresnel diffraction,the source of light and the screen are at a finite distance from the diffracting aperture or obstacle. This is in contrast to Fraunhofer diffraction,where both are at an infinite distance.
Statement $B$ is true: $X$-ray diffraction (a form of diffraction) is a standard technique used to determine the molecular structure of complex biological molecules,including the helical structure of nucleic acids like $DNA$.
Therefore,both statements $A$ and $B$ are correct.

(B) The ray optics approximation is valid up to the Fresnel distance $(z_f)$, which is defined as the distance at which the spreading of the beam due to diffraction becomes comparable to the size of the aperture.
It is given by the formula: $z_f = \frac{a^2}{\lambda}$.
Given values: Fresnel distance $z_f = 5 \,m$, wavelength $\lambda = 800 \,nm = 800 \times 10^{-9} \,m$.
Rearranging the formula for slit width $a$: $a = \sqrt{z_f \cdot \lambda}$.
Substituting the values: $a = \sqrt{5 \times 800 \times 10^{-9} \,m^2}$.
$a = \sqrt{4000 \times 10^{-9} \,m^2} = \sqrt{4 \times 10^{-6} \,m^2}$.
$a = 2 \times 10^{-3} \,m = 2 \,mm$.

(B) In Fraunhofer diffraction,the source of light and the screen are effectively at an infinite distance from the diffracting aperture or obstacle.
This ensures that the wavefronts incident on the aperture are plane wavefronts,and the diffracted rays reaching the screen are parallel to each other.
Therefore,both the source and the screen must be at infinity relative to the diffracting element.

(C) In Fraunhofer diffraction,the source of light and the screen are effectively at an infinite distance from the aperture (slit).
This implies that the wavefronts incident on the slit are plane wavefronts,which means the incident beam must be parallel.
This condition is practically achieved by placing the light source at the focus of a converging lens,or by placing the source at an infinite distance from the slit.
Therefore,the correct condition is that the source is at infinity and the incident beam is parallel.