Wave Optics

The wavelength of first light is $\lambda_1=650 \mathrm{~nm}=650 \times 10^{-9} \mathrm{~m}$

The wavelength of second light is $\lambda_2=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m}$

Slit separation is $\mathrm{d}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$

Screen distance is $\mathrm{D}=1.2 \mathrm{~m}$

For bright fringes of different wavelengths to coincide at the same distance $(y)$ from the central maximum, their positions must be equal :

$ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{\mathrm{n}_2 \lambda_2 \mathrm{D}}{\mathrm{~d}} $

$ \mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2 $

Where $n_1$ and $n_2$ are the orders of the bright fringes for $\lambda_1$ and $\lambda_2$ respectively.

$ \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_2}{\lambda_1}=\frac{550}{650}=\frac{11}{13} $

Since we are looking for the least distance, we take the smallest integers that satisfy this ratio $\mathrm{n}_1=11$ and $\mathrm{n}_2=13$.

This means the $11^{\text {th }}$ bright fringe of 650 nm coincides with the $13^{\text {th }}$ bright fringe of 550 nm .

Using $\mathrm{n}_1=11$ and $\lambda_1=650 \mathrm{~nm}$ :

$\Rightarrow $ $ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}} $

$\Rightarrow $ $ \mathrm{y}=\frac{11 \times\left(650 \times 10^{-9}\right) \times 1.2}{2 \times 10^{-3}} \mathrm{~m}=429 \times 10^{-5} \mathrm{~m} $

The least distance from the central maximum where the bright fringes coincide is $429 \times 10^{-5} \mathrm{~m}$.

Hence, the correct answer is 429 .

In two Young's Double Slit Experiments (YDSE) the fringe widths are equal ( $\beta_1=\beta_2$ ).

Ratio of slit separations (d) is $\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{2}{1}=2$

Ratio of wavelengths ( $\lambda$ ) is $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$

The formula for fringe width ( $\beta$ ) in a YDSE setup is:

$ \beta=\frac{\lambda D}{d} $

where:

• $\lambda$ is the wavelength of light.

• D is the distance between the slits and the screen.

• d is the separation between the slits.

$ \beta_1=\beta_2 \Rightarrow \frac{\lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\lambda_2 \mathrm{D}_2}{\mathrm{~d}_2} $

$ \Rightarrow \frac{\mathrm{D}_1}{\mathrm{D}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \times\left(\frac{\mathrm{d}_1}{\mathrm{~d}_2}\right) $

Since $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$, so, $\frac{\lambda_2}{\lambda_1}=2$

Putting the values into the equation :

$ \frac{\mathrm{D}_1}{\mathrm{D}_2}=(2) \times(2)=4 $

The corresponding ratio of the distances between the slits and the respective screens is $4: 1$.

Hence, the correct answer is $\mathbf{4}$.

When a transparent sheet of thickness t and refractive index $\mu$ is placed in front of one slit, the optical path length of the light from that slit increases.

Extra path length introduced is $\Delta \mathrm{x}=(\mu-1) \mathrm{t} \ldots$ (i)

In YDSE, the path length at a point is given as $\Delta \mathrm{x}=\mathrm{d} \sin \theta \Rightarrow \sin \theta=\frac{\Delta \mathrm{x}}{\mathrm{d}} \ldots$ (ii)

For small angle, $\sin \theta \approx \tan \theta \approx \theta$ (in radians).

Also, $\tan \theta=\frac{\Delta y}{D} \ldots$ (iii)

From (i), (ii) and (iii);

$ \frac{\Delta y}{D}=\frac{\Delta x}{d} $

$\Rightarrow \Delta y=\frac{D \Delta x}{d}=\frac{D(\mu-1) t}{d} $

$\Rightarrow $ $\Delta y=\frac{D(\mu-1) t}{d} \Rightarrow(\mu-1)=\frac{\Delta y \cdot d}{D \cdot t}$

$ \Rightarrow \mu=\frac{\Delta y \cdot d}{D \cdot t}+1 $

Where :

• $\mathrm{D}=$ distance between the slits and the screen $=1 \mathrm{~m}$.

• $\mathrm{d}=$ distance between the two slits $=0.4 \mathrm{~mm}=0.4 \times 10^{-3} \mathrm{~m}$.

• Sheet thickness is $\mathrm{t}=20 \mu \mathrm{~m}=20 \times 10^{-6} \mathrm{~m}$

• Fringe shift is $\Delta y=20 \mathrm{~mm}=20 \times 10^{-3} \mathrm{~m}$

Substituting these values,

$ \mu=\frac{\left(20 \times 10^{-3}\right) \times\left(0.4 \times 10^{-3}\right)}{1 \times\left(20 \times 10^{-6}\right)}+1 $

$\Rightarrow $ $\mu=\frac{8 \times 10^{-6}}{20 \times 10^{-6}}+1$

$\Rightarrow $ $\mu=0.4+1=1.4=\frac{14}{10}$

$ \Rightarrow \frac{14}{10}=\frac{\alpha}{10} \Rightarrow \alpha=14 $

Therefore, the value of $\alpha$ is 14 .

Width of 20 maxima of double slit $=$ width of central maxima of single slit

$\begin{aligned} & \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm} \end{aligned}$

Value of $x$ is 15

Answer is 15

Calculate Maximum Intensity ($I_{\text{max}}$):

$ I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 $

Substituting the values, we have:

$ I_{\max} = \left(\sqrt{4I} + \sqrt{9I}\right)^2 $

$ I_{\max} = \left(2\sqrt{I} + 3\sqrt{I}\right)^2 = \left(5\sqrt{I}\right)^2 = 25I $

Calculate Minimum Intensity ($I_{\text{min}}$):

$ I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 $

Substituting the values, we have:

$ I_{\min} = \left(\sqrt{4I} - \sqrt{9I}\right)^2 $

$ I_{\min} = \left(2\sqrt{I} - 3\sqrt{I}\right)^2 = \left(-1\sqrt{I}\right)^2 = I $

Calculate the Difference Between Maximum and Minimum Intensities:

$ I_{\max} - I_{\min} = 25I - I = 24I $

Thus, the value of $x$, which represents the difference between the maximum and minimum intensities, is $24$.

$\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \theta_1+\theta_2=30^{\circ} \end{aligned}$

$\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\mathrm{a}}\right)+\sin ^{-1}\left(\frac{3 \lambda}{\mathrm{a}}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{2 \lambda}{\mathrm{a}} \sqrt{1-\left(\frac{3 \lambda}{\mathrm{a}}\right)^2}+\frac{3 \lambda}{\mathrm{a}} \sqrt{1+\left(\frac{2 \lambda}{\mathrm{a}}\right)^2}=\sin \frac{\pi}{6} \end{aligned}$

Here $\lambda=628 \mathrm{~nm}$

After solving

$\mathrm{A}=6.07 \mu \mathrm{~m}$

Approximate Method :

$\begin{aligned} & \theta=\theta_1+\theta_2 \\ & \Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{a}+\frac{3 \lambda}{a} \\ & \Rightarrow \frac{\pi}{6}=\frac{5}{a}(628 \mathrm{~nm}) \\ & \Rightarrow a=6 \mu \mathrm{~m} \end{aligned}$

Maxima condition

$2 \mu \mathrm{t}=\mathrm{n} \lambda \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{2 \mu}, \frac{2 \lambda}{2 \mu}, \ldots \ldots$

Minima condition $2 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2$

$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots . . \\ & \Delta \mathrm{t}=\frac{2 \lambda}{4 \mu} \end{aligned}$

Rate of evaporation $=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}$

In Young's double slit experiment, the distance of $n^{th}$ maxima from central maxima is given by,

$Y=\frac{n\lambda D}{d}$

Here, $n,D$ & $d$ are same for both the cases.

So, $Y \propto \lambda $

$ \Rightarrow {{{Y_1}} \over {{Y_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} \Rightarrow {{10} \over {{Y_2}}} = {{600} \over {660}} = {1 \over {11}}$

$ \Rightarrow {Y_2} = 11\,mm$.

To solve this problem, we need to understand how the interference pattern shifts due to the introduction of a thin glass plate in Young's double slit experiment. This shift occurs because the light passing through the glass experiences a different optical path length compared to the light passing through the other slit without glass.

### Step-by-Step Analysis:

1. Calculate the Path Difference Due to the Glass Plate:

• The path difference is influenced by the optical thickness of the glass, which is the product of the physical thickness $ t $ of the glass and the refractive index $ n $ minus the path it would have in air (i.e., $ n \times t - t $).


• The effective additional path difference in the medium of the glass plate is $ (n-1) \times t $.

1. Determine the Shift in Fringes:

• The shift of the central maximum to the position previously occupied by the fourth bright fringe indicates that the optical path difference created by the glass plate corresponds to four fringe spacings.


• Each fringe width corresponds to a change in path difference of one wavelength ($ \lambda $).

1. Calculating the Thickness $ t $ of the Glass Plate:

• Since the central maximum shifts by four fringes, the path difference $ \Delta $ caused by the glass plate must be equal to $ 4 \lambda $.


• Therefore, $ (n-1) \times t = 4 \lambda $.

1. Insert Values and Solve for $ t $:

• Given:


• $ n = 1.5 $ (refractive index of the glass)


• $ \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} $ (wavelength of the light)


• $ (1.5 - 1) \times t = 4 \times 500 \times 10^{-9} \, \text{m} $


• $ 0.5 \times t = 2000 \times 10^{-9} \, \text{m} $


• $ t = \frac{2000 \times 10^{-9} \, \text{m}}{0.5} $


• $ t = 4000 \times 10^{-9} \, \text{m} $

Conversion to Micrometers:

• $ 4000 \times 10^{-9} \, \text{m} $ equals $ 4000 \, \text{nm} $


• Since $ 1 \, \mu\text{m} = 1000 \, \text{nm} $,


• $ t = 4 \, \mu\text{m} $.

Conclusion:

The thickness of the glass plate required to shift the central maximum to the position previously occupied by the fourth bright fringe is $ \mathbf{4 \, \mu\text{m}} $.

In a Young's double slit experiment, the intensity at a point can be expressed as a function of the phase difference between the light arriving from the two slits. The intensity at any point on the screen is given by:

$I = I_{\text{max}} \cos^2 \left( \frac{\delta}{2} \right)$

Where:

• $I_{\text{max}}$ is the maximum intensity.
• $\delta$ is the phase difference between the light waves from the two slits.

Given the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, we can write:

$\frac{I}{I_{\text{max}}} = \frac{1}{4}$

Substituting this into the intensity equation:

$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} \right)$

Taking the square root of both sides, we get:

$\cos \left( \frac{\delta}{2} \right) = \frac{1}{2}$

The possible solutions for $\delta$ are:

$\frac{\delta}{2} = \frac{\pi}{3}$ or $\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}\right)$ which gives $\delta = \frac{2\pi}{3}$ or $\delta = \frac{4\pi}{3}$.

Considering the smallest phase difference, $\delta = \frac{2\pi}{3}$, we use the relation for the phase difference due to path difference:

$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$

Thus, substituting the value of $\delta$, we have:

$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$

Solving for $\Delta x$, we get:

$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$

The minimum distance of the point from the central maximum on the screen can be found using the interference equation:

$y = \frac{\Delta x \cdot D}{d}$

Substituting the known values:

$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$

Simplifying this expression:

$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$

Hence, the minimum distance of the point from the central maximum is:

$200 \, \mu \mathrm{m}$

$\begin{aligned} & \text { Width central maxima }=10 \text { fringe width }=\frac{2 \lambda \theta}{a}=10 \beta \\ & \frac{2 \lambda \theta}{a}=10 \times \frac{\lambda D}{d} \\ & a=\frac{d}{5}=\frac{10^{-3}}{5}=2 \times 10^{-4} \end{aligned}$

$\begin{aligned} \theta & =(2)\left(\frac{2 \lambda}{a}\right) \\ & =\frac{4 \lambda}{a}=\frac{4 \times 600 \times 10^{-9}}{0.4 \times 10^{-3}} \\ & =6 \times 10^{-3} \end{aligned}$

When two coherent light beams interfere, the resulting intensity at a point depends on the principle of superposition and is related to their amplitudes. Let the amplitude of the first light beam be A, then its intensity, which is proportional to the square of its amplitude, is given as $I \propto A^2$. Since intensity is directly proportional to the square of amplitude, for the second beam with intensity $4I$, its amplitude would be $2A$, as $4I \propto (2A)^2$.

The maximum intensity ($I_{max}$) occurs when the two beams are in phase and their amplitudes add up constructively, which can be represented as:

$I_{max} \propto (A + 2A)^2 = (3A)^2 = 9A^2$

Given that $I \propto A^2$, substituting this relation to express $I_{max}$ in terms of $I$, we get:

$I_{max} = 9I$

The minimum intensity ($I_{min}$) occurs when the two beams are completely out of phase, leading their amplitudes to subtract destructively, thus

$I_{min} \propto (2A - A)^2 = A^2$

Again, using the fact that $I \propto A^2$, we find that:

$I_{min} = I$

Now, the difference between the maximum and minimum intensities in the resulting beam is:

$xI = I_{max} - I_{min} = 9I - I$

$xI = 8I$

Therefore, the value of $x$ is 8.

$\begin{aligned} y & =\frac{n \lambda D}{a} \\ & =\frac{1 \times\left(550 \times 10^{-9}\right)(1)}{\left(0.2 \times 10^{-3}\right)} \\ & =275 \times 10^{-5} \mathrm{~m} \end{aligned}$

$\begin{aligned} x & =\frac{3 \lambda D}{d} \\ & =\frac{3 \times 5000 \times 10^{-10} \times 200 \times 10^{-2}}{0.3 \times 10^{-3}} \\ & =10 \mathrm{~mm} \end{aligned}$

In Young's double slit experiment, the condition for constructive interference (bright fringes) is given by:

$d \sin \theta = n \lambda$

where:

• $d$ is the distance between the slits
• $\theta$ is the angle of the fringe relative to the central maximum
• $n$ is the order of the fringe (an integer)
• $\lambda$ is the wavelength of the light

We are given two different wavelengths:

$\lambda_1 = 450 \, \text{nm}$

$\lambda_2 = 650 \, \text{nm}$

For the fringes produced by these two wavelengths to overlap, the path difference must be an integer multiple of both wavelengths. This means:

$d \sin \theta = m \lambda_1 = n \lambda_2$

where $m$ and $n$ are the orders of the fringes for $\lambda_1$ and $\lambda_2$, respectively.

To find the minimum order of fringe $n$ for $\lambda_2$ that coincides with a fringe for $\lambda_1$, we need to find the least common multiple (LCM) of these wavelengths in terms of their smallest integers. This can be formulated as:

$m \lambda_1 = n \lambda_2$

Dividing both sides by $\lambda_1$ and $\lambda_2$, we get:

$\frac{m}{\lambda_2} = \frac{n}{\lambda_1}$

Cross-multiplying, we get:

$m \lambda_1 = n \lambda_2$

Using the given wavelengths:

$m \times 450 = n \times 650$

Simplifying this equation, we get:

$\frac{m}{n} = \frac{650}{450}$

$\frac{m}{n} = \frac{13}{9}$

For the fringes to overlap, $m$ and $n$ must be integers. The smallest integers that satisfy this ratio are:

$m = 13$

$n = 9$

Therefore, the minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is:

$n = 9$

For incoherent wave $\mathrm{I}_1=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{I}_1=\mathrm{I}_0+9 \mathrm{I}_0$

$\mathrm{I}_1=10 \mathrm{I}_0$

For coherent wave $\mathrm{I_2=I_A+I_B+2 \sqrt{I_A I_B} \cos 60^{\circ}}$

$\begin{aligned} & \mathrm{I}_2=\mathrm{I}_0+9 \mathrm{I}_0+2 \sqrt{9 \mathrm{I}_0^2} \cdot \frac{1}{2}=13 \mathrm{I}_0 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{10}{13} \end{aligned}$

For $\mathrm{n}^{\text {th }}$ minima

$\mathrm{b} \sin \theta=\mathrm{n} \lambda$

($\lambda$ is small so $\sin \theta$ is small, hence $\sin \theta \simeq \tan \theta$)

$\mathrm{btan} \theta=\mathrm{n} \lambda$

$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$

$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(\text { Position of } \mathrm{n}^{\mathrm{th}}\right. \text { minima) }$

$\begin{aligned} & \mathrm{B} \rightarrow 1^{\text {st }} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rd}} \text { minima } \\ & \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \\ & \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \\ & 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \\ & \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \\ & \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \\ & x=2 \end{aligned}$

Path difference for minima at $\mathrm{P}$

$\begin{aligned} & 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \\ & \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \\ & \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \\ & \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\ & \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \\ & \therefore \mathrm{d}=20 \times 10^{-5} \\ & \Rightarrow \mathrm{d}=0.20 \mathrm{~mm} \end{aligned}$

For first minima

$\begin{aligned} & \operatorname{asin} \theta=\lambda \\ & \Rightarrow \sin \theta=\frac{\lambda}{a}=\frac{5000 \times 10^{-10}}{1 \times 10^{-6}}=\frac{1}{2} \\ & \Rightarrow \theta=30^{\circ} \end{aligned}$

When dealing with three polaroids, the intensity after the second polaroid will be given by Malus' law as $I_0 \cos^2 \theta$, where $\theta$ is the angle between the pass axes of the first two polaroids. However, as the third polaroid is orthogonal to the first, no light from the first polaroid passes through, only light from the second polaroid. So the final intensity is also modulated by a $\sin^2 \theta$ term (as the second and third polaroids are orthogonal).

So if we set up the equation for the final intensity $I_{\text{net}}$:

$I_{\text{net}} = I_0 \cos^2 \theta \sin^2 \theta$

And we substitute the given values $I_{\text{net}} = 3 \, \text{W/m}^2$ and $I_0 = \frac{32 \, \text{W/m}^2}{2} = 16 \, \text{W/m}^2$:

$3 = 16 \cos^2 \theta \sin^2 \theta$

This simplifies to:

$\frac{3}{16} = \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2$

Taking the square root of both sides gives:

$\frac{\sqrt{3}}{2} = \left|\sin 2\theta\right|$

The solutions for this are $\theta = 30^\circ$ and $\theta = 60^\circ$.

So, the angle between the pass axes of the first two polaroids is either $30^\circ$ or $60^\circ$.

In Young's double slit experiment, we have two slits separated by a distance $d$, and a screen placed at a distance $L$ from the slits. When light with a single wavelength $\lambda$ passes through the slits, an interference pattern is formed on the screen with bright and dark fringes.

In this problem, we have a beam of light consisting of two wavelengths $\lambda_1 = 7000~\mathop A\limits^o$ and $\lambda_2 = 5500~\mathop A\limits^o$. We are given the distance between the slits $d = 2.5 \mathrm{~mm}$ and the distance between the plane of the slits and the screen $L = 150 \mathrm{~cm}$. Our goal is to find the least distance from the central fringe where the bright fringes due to both wavelengths coincide.

First, let's convert the wavelengths and distances to meters:

$\lambda_1 = 7 \times 10^{-7} \mathrm{~m}$ $\lambda_2 = 5.5 \times 10^{-7} \mathrm{~m}$ $d = 2.5 \times 10^{-3} \mathrm{~m}$ $L = 1.5 \mathrm{~m}$

The fringe width $\beta$ for a single wavelength is given by:

$\beta = \frac{\lambda D}{d}$

Now, let's consider the condition for the bright fringes due to both wavelengths to coincide. Let the $n^{\text{th}}$ bright fringe of $\lambda_1$ match with the $m^{\text{th}}$ bright fringe of $\lambda_2$. In this case, we have:

$n \beta_1 = m \beta_2$

Substituting the expression for fringe width, we get:

$n \frac{\lambda_1 L}{d} = m \frac{\lambda_2 L}{d}$

Simplifying, we get:

$\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{11}{14}$

The least values of $n$ and $m$ that satisfy this condition are $n = 11$ and $m = 14$.

Now, let's find the position $y$ of the coincident bright fringe on the screen:

$y = n \beta_1 = n \frac{\lambda_1 L}{d} = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m}}$

$y = k \times 10^{-5} \mathrm{~m}$

Calculating the value of $k$:

$k = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m} \times 10^{-5} \mathrm{~m}} = 462$

So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $462 \times 10^{-5} \mathrm{~m}$.

$I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)$

so ${I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)$

${I_1} = 2{I_0}$

& ${I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)$

${I_2} = {I_0}$

So ${{{I_1} + {I_2}} \over {{I_0}}} = 3$

Path difference introduced by two slabs $=(\mu_2-\mu_1)t$

$\Rightarrow$ Number of shifts $ = {{({\mu _2} - {\mu _1})t} \over \lambda }$

$ = {{0.04 \times 0.1\,\mathrm{mm}} \over {4000\,\mathop A\limits^o }}$

$ = {{4 \times {{10}^{ - 2}} \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 7}}}} = 10$

We know that

$\tan {\theta _0} = {{{\mu _2}} \over {{\mu _1}}}$

$\tan {\theta _0} = \sqrt {{{6.8} \over {2.8}}} = \sqrt {{{17} \over 7}} $

${\theta _0} = {\tan ^{ - 1}}\sqrt {1 + {{10} \over 7}} \Rightarrow \theta = 7$

${I_A} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} = 25I$

${I_B} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} = I$

So, ${I_A} - {I_B} = 24I$

$\lambda = 560 \times {10^{ - 9}}$

${B_1} = 7.2 \times {10^{ - 3}}$

${B_2} = 8.1 \times {10^{ - 3}}$

${{{B_1}} \over {{B_2}}} = {{{\lambda _1}} \over {{\lambda _2}}}$

$ \Rightarrow {\lambda _2} = {{560 \times {{10}^{ - 9}} \times 8.1 \times {{10}^{ - 3}}} \over {7.2 \times {{10}^{ - 3}}}}$

$ = 6.3 \times {10^{ - 7}}$ m

$ = 630$ nm