Thin film Interference, fresnel biprism, lloyd's mirror Questions in English

(A) In the standard Newton's rings experiment,the air film is between the lens $(n_l)$ and the glass plate $(n_g)$,where $n_l > n_{air}$ and $n_g > n_{air}$. The reflection at the bottom surface of the air film occurs at the glass-air interface (denser to rarer),while the top surface reflection occurs at the air-glass interface (rarer to denser). This introduces a phase change of $\pi$ for one ray,resulting in a dark central spot.
When the space is filled with a liquid of refractive index $n_{liq} > n_g$,the reflection at the bottom surface occurs at the glass-liquid interface (rarer to denser),and the reflection at the top surface occurs at the liquid-glass interface (denser to rarer).
Since both reflections now occur under similar conditions (either both undergo phase change or neither does),the path difference at the center is effectively zero,leading to constructive interference. Thus,the central spot becomes bright. Both the Assertion and the Reason are correct,and the Reason explains why the phase change conditions lead to a bright center.

(C) The Assertion is correct because thin films exhibit interference patterns when illuminated by white light,resulting in the observation of colours.
However,the Reason is incomplete and therefore incorrect. The phenomenon of interference in thin films occurs due to the superposition of light waves reflected from $BOTH$ the upper surface $AND$ the lower surface of the thin film,not just the upper surface. The path difference between these two reflected rays leads to constructive or destructive interference for different wavelengths,which creates the observed colours.

(C) The Assertion is correct. The colour of the Morpho butterfly's wing is not due to pigments,but due to structural colouration caused by the interference of light reflected from the microscopic scales on the wing surface.
When the wind moves the wing,the angle of incidence and the path difference between the light waves reflected from different layers of the scales change. This causes a shift in the interference pattern,resulting in a change in the observed colour.
The Reason is incorrect because the colour is not caused by pigments,but by the physical structure of the wing surface (thin-film interference).
Therefore,the correct option is $(C)$.

(N/A) For maximum transmission,the reflected rays must interfere destructively. This occurs when the path difference between the two reflected rays is an odd multiple of $\frac{\lambda}{2}$.
For normal incidence $(i = 0)$,the path difference is $\Delta x = 2nd$.
Since both reflections occur at a rarer to denser medium interface,there is no additional phase shift due to reflection.
Thus,the condition for destructive interference (minimum reflection,maximum transmission) is $2nd = (m + \frac{1}{2}) \lambda$,where $m = 0, 1, 2, \dots$.
For minimum thickness,we take $m = 0$,so $2nd = \frac{\lambda}{2}$.
$d = \frac{\lambda}{4n} = \frac{5500 \times 10^{-10} \, \text{m}}{4 \times 1.38} \approx 996.4 \times 10^{-10} \, \text{m} \approx 996.4 \, \mathring{A}$.

(A) The phenomenon of colours in thin films is due to the interference of light waves reflected from the top and bottom surfaces of the film.
For constructive interference to occur in the visible spectrum, the path difference must be related to the wavelength of light $(\lambda)$.
The condition for interference in a thin film of thickness $t$ is approximately $2\mu t \approx n\lambda$.
Since the wavelength of visible light ranges from approximately $400\,nm$ to $700\,nm$ (i.e., $4 \times 10^{-7}\,m$ to $7 \times 10^{-7}\,m$), the thickness $t$ of the oil film must be of the same order of magnitude as the wavelength of visible light.
Therefore, the order of magnitude of the thickness is $10^{-7}\,m$ to $10^{-6}\,m$.
Among the given options, $10^{-6}\,m$ is the most appropriate order of magnitude.

(D) For destructive interference in the reflected light from a thin film,the condition is given by $2 \mu t \cos r = n \lambda$,where $n = 1, 2, 3, \dots$.
Given: $\mu = 1.4$,$t = 10^{-4} \,cm = 10^{-6} \,m = 10^{-4} \times 10^{10} \, \mathring{A} = 10^6 \, \mathring{A}$.
For normal incidence,$r = 0^{\circ}$,so $\cos r = 1$.
The condition becomes $2 \times 1.4 \times 10^{-4} \,cm = n \lambda$.
$2.8 \times 10^{-4} \,cm = n \lambda$.
Converting to $\mathring{A}$s: $2.8 \times 10^{-4} \times 10^8 \, \mathring{A} = n \lambda$.
$28000 \, \mathring{A} = n \lambda$.
Thus,$\lambda = \frac{28000}{n} \, \mathring{A}$.
For $n = 4$,$\lambda = \frac{28000}{4} = 7000 \, \mathring{A}$.
For $n = 5$,$\lambda = \frac{28000}{5} = 5600 \, \mathring{A}$.
For $n = 7$,$\lambda = \frac{28000}{7} = 4000 \, \mathring{A}$.
Since all the given options satisfy the condition for destructive interference,they will not be seen in the reflected light.

(A) The correct answer is $A$.
Fresnel's Biprism is an optical device used to produce interference patterns.
It is specifically employed in optical testing to detect surface irregularities and imperfections in lenses by observing the distortion in the interference fringes produced by the device.

(A) The phenomenon of observing colours on thin soap films is caused by the interference of light waves.
When white light is incident on a thin soap film,light waves are reflected from both the outer and inner surfaces of the film.
These reflected waves undergo superposition,leading to constructive or destructive interference depending on the path difference and the wavelength of the light.
Since the thickness of the film varies and the path difference depends on the angle of incidence,different wavelengths (colours) interfere constructively at different points,resulting in the observed colourful patterns.

(D) Option $A$ is correct because the reflection of electromagnetic waves from a moving aeroplane causes interference,leading to ghost images.
Option $B$ is correct because thin-film interference is used to reduce reflection in solar cells,increasing light absorption.
Option $C$ is correct because non-reflective coatings (like $MgF_2$) use destructive interference to eliminate unwanted reflections in optical lenses.
Since all statements are scientifically accurate,the correct choice is $D$.

(A) The correct answer is $A$.
The wings of butterflies and moths consist of a colorless translucent membrane covered by a layer of scales. Each scale is a flattened outgrowth of a single cell and is about $100 \,\mu m$ long and $50 \,\mu m$ wide.
The scales overlap like roof tiles and completely cover the membrane,appearing as dust to the naked eye.
The iridescence in the wings is caused by thin-film interference. The structure of the scales contains microscopic layers that reflect light. When sunlight hits these layers,the light waves reflected from the top and bottom surfaces of these thin structures interfere with each other.
Depending on the thickness of these layers and the angle of viewing,certain wavelengths of light undergo constructive interference while others undergo destructive interference. This selective reinforcement of specific colors results in the beautiful iridescent colors observed on the wings.

(C) The phenomenon responsible for the change in colour of currency notes when tilted is $Interference$ of light.
Currency notes often use special inks containing thin-film layers. When light falls on these layers, it undergoes reflection from both the top and bottom surfaces of the thin film.
The path difference between the light waves reflected from the top and bottom surfaces depends on the angle of incidence (which changes as you tilt the note) and the thickness of the film.
Due to this path difference, constructive or destructive interference occurs for specific wavelengths of light, causing the observed colour to shift as the viewing angle changes.

(C) For constructive interference in a thin film (reflected light),the condition is $2 \mu t \cos r = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$
Given $\mu = 4/3$,$t = 3100 \,\mathring{A}$,and normal incidence ($r = 0$,so $\cos r = 1$).
The formula becomes $\lambda = \frac{4 \mu t}{2n - 1}$.
For $n = 1$: $\lambda = 4 \times (4/3) \times 3100 = 16533 \,\mathring{A}$ (Infrared).
For $n = 2$: $\lambda = \frac{4 \times (4/3) \times 3100}{3} = \frac{16533}{3} = 5511 \,\mathring{A}$.
This wavelength $(5511 \,\mathring{A})$ corresponds to the yellow region of the visible spectrum.
Therefore,the film appears yellow.

(C) In this setup,the point source $P$ and its virtual image $P'$ formed by the plane mirror act as two coherent point sources. The interference pattern produced by two coherent point sources on a screen perpendicular to the line joining them consists of concentric circles. Therefore,the interference fringes formed on the screen are circular.

(B) The phenomenon of interference in thin films occurs when the path difference between the reflected rays is comparable to the wavelength of visible light.
Visible light has a wavelength range of approximately $400\,nm$ to $700\,nm$, which is $4 \times 10^{-7}\,m$ to $7 \times 10^{-7}\,m$.
Converting this to millimeters, we get $4 \times 10^{-4}\,mm$ to $7 \times 10^{-4}\,mm$.
Among the given options, $10^{-3}\,mm$ (which is $10^{-6}\,m$ or $1\,\mu m$) is the only value that is of the same order of magnitude as the wavelength of visible light.
Therefore, the correct option is $B$.

(C) The shift in the central fringe due to the introduction of a mica sheet is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.
Given that the central fringe shifts to the position of the $n^{th}$ bright fringe, the shift is equal to $n \beta$, where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Equating the two: $(\mu - 1) t \frac{D}{d} = n \frac{\lambda D}{d}$.
This simplifies to: $\lambda = \frac{(\mu - 1) t}{n}$.
Given values: $\mu = 1.6$, $t = 7 \ \mu m = 7 \times 10^{-6} \ m$, and $n = 7$.
Substituting these values: $\lambda = \frac{(1.6 - 1) \times 7 \times 10^{-6}}{7}$.
$\lambda = 0.6 \times 10^{-6} \ m = 6 \times 10^{-7} \ m$.
Converting to $\mathring{A}$s: $\lambda = 6000 \ \mathring{A}$.

(C) For maximum transmission of light, the reflected light must undergo destructive interference (minima).
Since the refractive index of the film $(\mu_f = 2.0)$ is greater than that of air $(\mu_a = 1.0)$ and also greater than that of the glass slab $(\mu_g = 1.45)$, there is a phase change of $\pi$ at both the top and bottom surfaces of the film.
The condition for destructive interference (minima) in reflected light is given by $2 \mu_f t = n \lambda$, where $n = 1, 2, 3, \dots$
For minimum thickness, we take $n = 1$:
$2 \mu_f t = \lambda$
$t = \frac{\lambda}{2 \mu_f}$
Substituting the given values $\lambda = 550 \ \text{nm}$ and $\mu_f = 2.0$:
$t = \frac{550 \ \text{nm}}{2 \times 2.0} = \frac{550}{4} \ \text{nm} = 137.5 \ \text{nm}$.

(A) For a thin film, the condition for destructive interference (minima) in transmitted light is given by $2 \mu t = n \lambda$, where $n = 1, 2, 3, \dots$.
Here, $\mu = 1.4$ and $\lambda = 560 \times 10^{-9} \ m$.
The change in thickness $\Delta t$ for consecutive minima is $\Delta t = \frac{\lambda}{2 \mu}$.
Substituting the values: $\Delta t = \frac{560 \times 10^{-9}}{2 \times 1.4} = \frac{560 \times 10^{-9}}{2.8} = 200 \times 10^{-9} \ m = 2 \times 10^{-7} \ m$.
The area of the film is $A = \pi r^2 = \pi (1.8 \times 10^{-2} \ m)^2 = \pi \times 3.24 \times 10^{-4} \ m^2$.
The rate of evaporation is $R = \frac{A \times \Delta t}{\Delta T}$, where $\Delta T = 12 \ s$.
$R = \frac{\pi \times 3.24 \times 10^{-4} \times 2 \times 10^{-7}}{12} = \frac{\pi \times 6.48 \times 10^{-11}}{12} = 0.54 \times 10^{-11} \ m^3/s = 54 \times 10^{-13} \ m^3/s$.
Thus, the rate of evaporation is $54 \pi \times 10^{-13} \ m^3/s$.

(A) When a thin film of oil spreads over the surface of water and is viewed in broad daylight,brilliant colours are observed.
These colours arise due to the interference of sunlight reflected from the upper and lower surfaces of the thin oil film.
Diffraction is the bending of light rays around the corners of obstacles.
Dispersion is the splitting of white light into its constituent colours.
Polarisation is the restriction of the vibrations of a transverse wave to a single plane.

(D) Let $t$ be the thickness of the plate.
When light passes through a plate of thickness $t$ and refractive index $\mu$,the optical path length is $\mu t$.
The optical path length in air for the same distance $t$ is $t$.
The change in optical path (optical path difference) is given by $\Delta = \mu t - t = (\mu - 1)t$.
According to the problem,this change in optical path is equal to half the wavelength,i.e.,$\Delta = \frac{\lambda}{2}$.
Equating the two expressions: $(\mu - 1)t = \frac{\lambda}{2}$.
Solving for $t$,we get $t = \frac{\lambda}{2(\mu - 1)}$.

(B) The formula for fringe width in a biprism experiment is given by $W = \frac{\lambda D}{d}$.
Since the wavelength $\lambda$ and the distance between coherent sources $d$ are kept constant,the ratio $\frac{\lambda}{d}$ is a constant.
Therefore,$\frac{W}{D} = \frac{\lambda}{d} = \text{constant}$.
This implies that $\frac{W_1}{D_1} = \frac{W_2}{D_2} = \frac{W_3}{D_3} = \frac{W_4}{D_4}$.

(C) The position of the $n^{\text{th}}$ bright fringe is given by $y_n = \frac{n \lambda_1 D}{d}$.
The position of the $m^{\text{th}}$ dark fringe is given by $y'_m = \frac{(2m - 1) \lambda_2 D}{2d}$.
Given that the $5^{\text{th}}$ bright band coincides with the $6^{\text{th}}$ dark band,we set $n = 5$ and $m = 6$:
$\frac{5 \lambda_1 D}{d} = \frac{(2 \times 6 - 1) \lambda_2 D}{2d}$
$\frac{5 \lambda_1 D}{d} = \frac{11 \lambda_2 D}{2d}$
Canceling $D$ and $d$ from both sides:
$5 \lambda_1 = \frac{11 \lambda_2}{2}$
$\frac{\lambda_1}{\lambda_2} = \frac{11}{5 \times 2} = \frac{11}{10}$

(C) The position of the $n^{th}$ bright fringe is given by $x_n = n \beta$,where $\beta$ is the fringe width.
For the $8^{th}$ bright fringe $(n=8)$: $x_8 = 8 \beta = 8 \times 0.6 \ mm = 4.8 \ mm$.
The position of the $n^{th}$ dark fringe is given by $x'_n = (n - 0.5) \beta$.
For the $6^{th}$ dark fringe $(n=6)$: $x'_6 = (6 - 0.5) \beta = 5.5 \beta = 5.5 \times 0.6 \ mm = 3.3 \ mm$.
The distance between the $8^{th}$ bright fringe and the $6^{th}$ dark fringe is $\Delta x = x_8 - x'_6 = 4.8 \ mm - 3.3 \ mm = 1.5 \ mm$.

(B) The position of the $n^{\text{th}}$ bright band is given by $y_n = \frac{n \lambda D}{d}$.
For the $6^{\text{th}}$ bright band with wavelength $\lambda_1$,the position is $y_6 = \frac{6 \lambda_1 D}{d}$.
The position of the $m^{\text{th}}$ dark band is given by $y_m = \frac{(m - 0.5) \lambda D}{d}$.
For the $7^{\text{th}}$ dark band with wavelength $\lambda_2$,the position is $y_7 = \frac{(7 - 0.5) \lambda_2 D}{d} = \frac{6.5 \lambda_2 D}{d}$.
Since the bands coincide,we equate the positions:
$\frac{6 \lambda_1 D}{d} = \frac{6.5 \lambda_2 D}{d}$.
Canceling the common terms $\frac{D}{d}$,we get $6 \lambda_1 = 6.5 \lambda_2$.
Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{6.5}{6} = \frac{13}{12}$.

(A) The fringe width $(z)$ in a biprism experiment (or Young's Double Slit Experiment) is given by the formula:
$z = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the slit and the eyepiece (screen),and $d$ is the distance between the two coherent sources.
Given that the wavelength $\lambda$ and the distance between the coherent sources $d$ are kept constant,we can write:
$z \propto D$
This implies that the ratio $\frac{z}{D}$ is a constant.
Therefore,$\frac{z_{1}}{D_{1}} = \frac{z_{2}}{D_{2}} = \frac{z_{3}}{D_{3}} = \frac{z_{4}}{D_{4}}$.
Thus,option $A$ is correct.

(B) Let the incident intensity be $I$.
At point $A$,$25\%$ of $I$ is reflected as ray $AB$. So,$I_1 = 0.25I = I/4$.
The intensity of the refracted ray is $0.75I$.
At point $C$,this ray is reflected. $25\%$ of the incident energy $(0.75I)$ is reflected. So,the intensity of the ray reaching $A^1$ is $0.25 \times 0.75I = 0.1875I = (3/16)I$.
At point $A^1$,this ray is refracted out as $A^1B^1$. Since $25\%$ is reflected at $A^1$,$75\%$ of the incident energy is transmitted.
So,$I_2 = 0.75 \times (3/16)I = (3/4) \times (3/16)I = (9/64)I$.
The amplitudes are $a_1 = \sqrt{I_1} = \sqrt{I/4} = (1/2)\sqrt{I}$ and $a_2 = \sqrt{I_2} = \sqrt{9I/64} = (3/8)\sqrt{I}$.
The ratio of amplitudes is $a_1/a_2 = (1/2) / (3/8) = 4/3$.
$I_{\text{max}} / I_{\text{min}} = (a_1 + a_2)^2 / (a_1 - a_2)^2 = ((4+3)/ (4-3))^2 = (7/1)^2 = 49/1$.

(B) The formula for fringe width in a biprism experiment is given by $Z = \frac{\lambda D}{d}$.
Since the wavelength $\lambda$ and the distance between the coherent sources $d$ are kept constant,the ratio $\frac{\lambda}{d}$ is a constant value.
Therefore,we can write $\frac{Z}{D} = \frac{\lambda}{d} = \text{constant}$.
This implies that for different values of $D$ and $Z$,the ratio remains the same:
$\frac{Z_1}{D_1} = \frac{Z_2}{D_2} = \frac{Z_3}{D_3} = \frac{Z_4}{D_4}$.

(A) The width of the region $L$ is constant and is given by the product of the number of fringes $n$ and the fringe width $\beta$.
$L = n_1 \beta_1 = n_2 \beta_2$
Since the fringe width $\beta = \frac{\lambda D}{d}$,we have:
$n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$
$n_1 \lambda_1 = n_2 \lambda_2$
Given $n_1 = 21$,$\lambda_1 = 4800 \text{ Å}$,and $\lambda_2 = 5600 \text{ Å}$:
$21 \times 4800 = n_2 \times 5600$
$n_2 = \frac{21 \times 4800}{5600} = \frac{21 \times 48}{56} = \frac{21 \times 6}{7} = 3 \times 6 = 18$
Thus,the number of fringes observed is $18$.

(A) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$,Wavelength $\lambda = 5000 \text{ Å} = 5 \times 10^{-7} \,m$,Change in fringe width $\Delta \beta = \beta_{2} - \beta_{1} = 12.5 \times 10^{-5} \,m$.
Fringe width formula is $\beta = \frac{\lambda D}{d}$.
The change in fringe width is given by $\Delta \beta = \frac{\lambda}{d} \Delta D$,where $\Delta D = D_{2} - D_{1}$ is the shift in the screen position.
Rearranging for $\Delta D$: $\Delta D = \frac{d \cdot \Delta \beta}{\lambda}$.
Substituting the values: $\Delta D = \frac{10^{-3} \times 12.5 \times 10^{-5}}{5 \times 10^{-7}}$.
$\Delta D = \frac{12.5 \times 10^{-8}}{5 \times 10^{-7}} = 2.5 \times 10^{-1} \,m = 0.25 \,m = 25 \,cm$.
Therefore,the screen should be moved away or towards the slit by $25 \,cm$.

(D) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the source and the screen,and $d$ is the distance between the two virtual sources.
Since $D$ and $d$ remain constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.
Given $\lambda_1 = 5000 \ Å$ and $\lambda_2 = 6400 \ Å$.
The ratio of the new fringe width $\beta_2$ to the initial fringe width $\beta_1$ is $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} = \frac{6400}{5000} = 1.28$.
This means $\beta_2 = 1.28 \beta_1$.
The percentage increase in fringe width is given by $\frac{\beta_2 - \beta_1}{\beta_1} \times 100\% = (1.28 - 1) \times 100\% = 0.28 \times 100\% = 28\%$.
Therefore,the fringe width increases by $28\%$.

(A) The distance of the $n^{th}$ dark fringe from the central fringe is given by $y_n = (n - 1/2) \beta$,where $\beta$ is the fringe width.
For the second dark fringe $(n = 2)$:
$y_2 = (2 - 1/2) \beta = 1.5 \beta = 3 \ mm$.
Therefore,$\beta = 3 / 1.5 = 2 \ mm$.
The distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = n \beta$.
For the sixth bright fringe $(n = 6)$:
$y_6 = 6 \times \beta = 6 \times 2 \ mm = 12 \ mm$.

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident light is $I_0$,the intensity of the first reflected ray is $I_1 = 0.25 I_0 = \frac{I_0}{4}$.
The intensity of light reaching the lower surface of the plate is $I_0 - 0.25 I_0 = 0.75 I_0 = \frac{3}{4} I_0$.
Since $50 \%$ of this intensity is reflected from the lower surface,the intensity of the second reflected ray is $I_2 = 0.50 \times \frac{3}{4} I_0 = \frac{3}{8} I_0$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}$
Substituting the values of $I_1$ and $I_2$:
$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{I_0}{4}}+\sqrt{\frac{3 I_0}{8}}\right)^2}{\left(\sqrt{\frac{I_0}{4}}-\sqrt{\frac{3 I_0}{8}}\right)^2} = \left(\frac{\frac{1}{2} \sqrt{I_0} + \sqrt{\frac{3}{8}} \sqrt{I_0}}{\frac{1}{2} \sqrt{I_0} - \sqrt{\frac{3}{8}} \sqrt{I_0}}\right)^2 = \left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$

(A) The formula for fringe width $W$ in a Biprism experiment is given by $W = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slit and the screen,and $d$ is the slit separation.
Since the fringe width $W$ and the wavelength $\lambda$ are constant for both setups,we have $W = \frac{\lambda D_1}{d_1} = \frac{\lambda D_2}{d_2}$.
This implies that $\frac{D_1}{d_1} = \frac{D_2}{d_2}$,or $\frac{D_1}{D_2} = \frac{d_1}{d_2}$.
Given the ratio of slit separation $\frac{d_1}{d_2} = \frac{2}{3}$,it follows that the ratio of the distances is $\frac{D_1}{D_2} = \frac{2}{3}$.

(D) The formula for fringe width is given by $\beta = \frac{\lambda D}{d}$,where $d$ is the distance between the two virtual sources.
For the first position: $W_1 = \frac{\lambda D_1}{d}$
For the second position: $W_2 = \frac{\lambda D_2}{d}$
Subtracting the two equations: $W_1 - W_2 = \frac{\lambda}{d}(D_1 - D_2)$
Rearranging for $d$: $d = \frac{\lambda(D_1 - D_2)}{(W_1 - W_2)}$

(D) Concept: In a biprism experiment,the position of the $n^{\text{th}}$ dark fringe from the central axis is given by the formula: $y_n = (2n - 1) \frac{\lambda D}{2d}$.
Given that the $4^{\text{th}}$ dark band is formed opposite to one of the slits,the distance of the slit from the central axis is $y = \frac{d}{2}$.
For $n = 4$,the position is $y_4 = (2(4) - 1) \frac{\lambda D}{2d} = \frac{7 \lambda D}{2d}$.
Equating the two expressions for $y_4$: $\frac{7 \lambda D}{2d} = \frac{d}{2}$.
Solving for $\lambda$: $7 \lambda D = d^2 \implies \lambda = \frac{d^2}{7D}$.
Thus,option $(D)$ is correct.

(B) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slit and the eyepiece,and $d$ is the distance between the two coherent sources.
Initially,$\beta = \frac{\lambda D}{d}$.
In the new situation,the distance between the sources becomes $d' = 2d$ and the distance between the slit and the eyepiece becomes $D' = 2D$.
The new fringe width $\beta'$ is given by $\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(2d)} = \frac{\lambda D}{d}$.
Therefore,$\beta' = \beta$,which means the fringe width remains unchanged.

(D) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $D = 1.2 \ m$ is the distance between the source and the eyepiece,and $d = 0.84 \ mm = 0.84 \times 10^{-3} \ m$ is the distance between the virtual sources.
When the eyepiece is moved transversely by a distance $y = 2.799 \ cm = 2.799 \times 10^{-2} \ m$,the number of fringes shifted is $n = 30$.
The relationship is $y = n \beta = n \frac{\lambda D}{d}$.
Substituting the values: $2.799 \times 10^{-2} = 30 \times \frac{\lambda \times 1.2}{0.84 \times 10^{-3}}$.
$\lambda = \frac{2.799 \times 10^{-2} \times 0.84 \times 10^{-3}}{30 \times 1.2}$.
$\lambda = \frac{2.35116 \times 10^{-5}}{36} = 0.06531 \times 10^{-5} \ m = 6531 \times 10^{-10} \ m$.
Therefore,$\lambda = 6531 \ \mathring{A}$.

(A) The condition for constructive interference in reflected light from a thin film is given by $2 \mu t = (2n + 1) \frac{\lambda}{2}$, where $n = 0, 1, 2, \dots$ and $\mu$ is the refractive index, $t$ is the thickness, and $\lambda$ is the wavelength.
For the minimum thickness, we take $n = 0$.
Substituting the values: $2 \mu t = \frac{\lambda}{2}$.
Rearranging for $t$: $t = \frac{\lambda}{4 \mu}$.
Given $\lambda = 600 \, nm$ and $\mu = 1.5$, we get $t = \frac{600}{4 \times 1.5} = \frac{600}{6} = 100 \, nm$.

(C) For a thin film,the condition for constructive interference (maxima) in reflected light is $2 \mu t \cos r = (n + 1/2) \lambda_1$,where $n$ is an integer.
For destructive interference (minima),the condition is $2 \mu t \cos r = m \lambda_2$,where $m$ is an integer.
Given $\lambda_1 = 600 \ nm$ and $\lambda_2 = 450 \ nm$. Assuming normal incidence $(\cos r = 1)$:
$2 \mu t = (n + 1/2) \lambda_1 = (2n + 1) \frac{\lambda_1}{2} = (2n + 1) \times 300 \ nm$
$2 \mu t = m \lambda_2 = m \times 450 \ nm$
Equating the two: $(2n + 1) \times 300 = m \times 450 \implies (2n + 1) \times 2 = 3m \implies 4n + 2 = 3m$.
For the smallest thickness with no minimum in between,we test $n=1$: $4(1) + 2 = 6 = 3m \implies m = 2$.
Substituting $n=1$ into the maxima condition:
$2 \times 1.5 \times t = (1 + 0.5) \times 600 \ nm$
$3t = 1.5 \times 600 \ nm = 900 \ nm$
$t = 300 \ nm = 3 \times 10^{-7} \ m$.
Thus,the thickness is $3$ units.

(D) The condition for constructive interference in a thin film for reflected light (considering a phase change of $\pi$ at the first surface) is given by $2 \mu d \cos r = (n + 1/2) \lambda$. However,for an air film between two glass plates or similar,the path difference is $2 \mu d \cos r = n \lambda$ for constructive interference if we consider the specific setup of thin film interference.
Given thickness $d = 2.9 \times 10^{-4} \ mm = 2.9 \times 10^{-7} \ m$.
For normal incidence,$\cos r = 1$ and $\mu = 1$ (air).
The path difference is $\Delta = 2d = 2 \times 2.9 \times 10^{-7} \ m = 5.8 \times 10^{-7} \ m = 580 \ nm$.
For constructive interference,the condition is $2d = n \lambda$.
For $n = 1$,$\lambda = 580 \ nm$.
Since the wavelength of yellow light is $580 \ nm$,it undergoes constructive interference and is reflected strongly.
Thus,the reflected light appears yellow.

(B) The colours observed in a soap bubble are produced due to the phenomenon of interference.
When light falls on the thin film of the soap bubble,it reflects from both the outer and inner surfaces.
These reflected waves undergo superposition,leading to constructive or destructive interference depending on the thickness of the film and the wavelength of the light.
This interference pattern results in the appearance of various colours.

(A) Let $I_0$ be the intensity of the incident light.
Intensity reflected from the upper surface,$I_1 = 25 \% \text{ of } I_0 = \frac{I_0}{4}$.
Intensity transmitted into the glass plate = $I_0 - \frac{I_0}{4} = \frac{3I_0}{4}$.
Intensity reflected from the lower surface,$I_2 = 50 \% \text{ of } \frac{3I_0}{4} = \frac{1}{2} \times \frac{3I_0}{4} = \frac{3I_0}{8}$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \left( \frac{\sqrt{\frac{I_0}{4}} + \sqrt{\frac{3I_0}{8}}}{\sqrt{\frac{I_0}{4}} - \sqrt{\frac{3I_0}{8}}} \right)^2 = \left( \frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}} \right)^2$.

(B) For constructive interference in reflected light from a thin film, the condition is given by $2nt = (m + 1/2)\lambda$, where $m = 0, 1, 2, \dots$ and $n$ is the refractive index of the film.
To find the minimum thickness, we set $m = 0$.
Thus, $2nt = \lambda/2$.
Rearranging for $t$, we get $t = \lambda / (4n)$.
Substituting the given values: $\lambda = 532 \,nm$ and $n = 1.33$.
$t = 532 / (4 \times 1.33) = 532 / 5.32 = 100 \,nm$.
Therefore, the minimum thickness is $100 \,nm$.

(D) For reflected light,the condition for destructive interference (dark fringe) is given by:
$2 \mu t \cos r = n \lambda$
where $\mu$ is the refractive index,$t$ is the thickness,$r$ is the angle of refraction,and $n$ is an integer $(n = 1, 2, 3, ...)$.
For the smallest thickness,we take $n = 1$:
$2 \mu t \cos r = \lambda$
Given values:
$\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$
$\mu = 1.5$
$r = 60^{\circ} \Rightarrow \cos 60^{\circ} = 0.5$
Substituting these values into the equation:
$2 \times 1.5 \times t \times 0.5 = 6 \times 10^{-7}$
$1.5 \times t = 6 \times 10^{-7}$
$t = \frac{6 \times 10^{-7}}{1.5} = 4 \times 10^{-7} \text{ m}$