Wave Optics

Here, $a=0.3 \mathrm{~cm}$

$ \begin{aligned} & =3 \times 10^{-3} \mathrm{~m} \\ \lambda & =6000 \mathop {\rm{A}}\limits^{\rm{o}}=6 \times 10^{-7} \mathrm{~m} \end{aligned} $

∵ Fresnel's distance

$ Z_F=\frac{a^2}{\lambda}=\frac{\left(3 \times 10^{-3}\right)^2}{6 \times 10^{-7}}=15 \mathrm{~m} $

• A) Steam engine: A steam engine operates by converting thermal energy (from steam) into mechanical work. This process is fundamentally governed by the Laws of thermodynamics, which describe how heat and work are related. Thus, A matches with II.

• B) Electron microscope: An electron microscope uses a beam of electrons instead of light to image a sample. The resolving power of an electron microscope is much higher than a light microscope because electrons, when accelerated, have a much shorter de Broglie wavelength than visible light. This relies on the wave nature of electrons, a concept from quantum mechanics (de Broglie hypothesis). Thus, B matches with III.

• C) Non-reflecting coatings: These coatings are thin films applied to optical surfaces to reduce reflection. They work by manipulating the phase of light waves. Light reflected from the top surface of the coating and light reflected from the bottom surface of the coating (at the coating-glass interface) interfere destructively, thereby reducing the overall reflected light. This phenomenon is Interference of light. Thus, C matches with IV.

• D) Tokamak: A Tokamak is a device that uses a strong magnetic field to confine plasma within a toroidal chamber for nuclear fusion research. The charged particles in the plasma are held in place by the magnetic field, preventing them from interacting with the reactor walls and maintaining the extremely high temperatures required for fusion. This principle is Magnetic confinement of plasma. Thus, D matches with I.

The highest intensity when two light waves mix is given by:

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

Case 1: Both waves have the same intensity

If both intensities are equal, let them be $ I_1 = I_2 = I_0 $.

$ I = \left(2 \sqrt{I_0}\right)^2 = 4 I_0 $

This means the maximum intensity is 4 times the intensity of one wave.

So, we can write: $ I_0=\frac{I}{4} $

Case 2: One wave is four times stronger

If the first wave is four times stronger, $ I_1 = 4I_0 $ and $ I_2 = I_0 $.

Maximum intensity becomes:

$ I' = (\sqrt{4I_0} + \sqrt{I_0})^2 $

$ I' = (2\sqrt{I_0} + \sqrt{I_0})^2 = (3\sqrt{I_0})^2 = 9I_0 $

Now, substitute $ I_0 = \frac{I}{4} $:

$ I' = 9 \times \frac{I}{4} = \frac{9I}{4} $

Step 1: Find the position equations for the fringes.

The position of the $n$th bright fringe is given by: $y_n = \frac{n D \lambda}{d}$

So, for the 5th bright fringe: $y_5 = \frac{5 D \lambda}{d}$

The position of the $n$th dark fringe is: $y_n = \frac{(2n - 1) D \lambda}{2d}$

For the 7th dark fringe: $y_7 = \frac{(2 \times 7 - 1) D \lambda}{2d} = \frac{13 D \lambda}{2d}$

Step 2: Use the given distance between 5th bright and 7th dark fringes.

Distance between the 7th dark and 5th bright fringe: $y_7 - y_5 = \frac{13 D \lambda}{2 d} - \frac{5 D \lambda}{d}$

First, make the denominators the same: $\frac{13 D \lambda}{2 d} - \frac{10 D \lambda}{2 d} = \frac{3 D \lambda}{2 d}$

The distance given in the question is $3~\mathrm{mm}$: $\frac{3 D \lambda}{2 d} = 3 \times 10^{-3}~\mathrm{m}$

Step 3: Find the value of $\frac{D \lambda}{d}$.

Divide both sides by $3$: $\frac{D \lambda}{2 d} = 1 \times 10^{-3}$ Now multiply both sides by $2$: $\frac{D \lambda}{d} = 2 \times 10^{-3}~\mathrm{m}$ So, $\frac{D \lambda}{d} = 2~\mathrm{mm} \quad \ldots (i)$

Step 4: Find the positions needed for the second distance.

Position of the 7th bright fringe: $y_7' = \frac{7 D \lambda}{d}$

Position of the 5th dark fringe: $y_5' = \frac{(2 \times 5 - 1) D \lambda}{2 d} = \frac{9 D \lambda}{2 d}$

Step 5: Calculate the distance between 5th dark and 7th bright fringes.

$y_7' - y_5' = \frac{7 D \lambda}{d} - \frac{9 D \lambda}{2 d}$ Make denominators the same: $\frac{14 D \lambda}{2 d} - \frac{9 D \lambda}{2 d} = \frac{5 D \lambda}{2 d}$

Step 6: Find the final answer using the value from earlier.

From (i), $\frac{D \lambda}{d} = 2~\mathrm{mm}$, so: $\frac{5 D \lambda}{2 d} = \frac{5}{2} \times 2~\mathrm{mm} = 5~\mathrm{mm}$

Final Answer: Distance between 5th dark and 7th bright fringes is 5 mm.

$ \begin{aligned} &\text { According to Fresnel distance }\\ &\begin{aligned} Z_F & =\frac{a^2}{\lambda} \\ \lambda & =\frac{a^2}{Z_F}=\frac{\left(5 \times 10^{-3}\right)^2}{50} \\ & =0.5 \times 10^{-6} \mathrm{~m} \\ & =5000 \times 10^{-10} \mathrm{~m} \\ & =5000 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

Given, $\Delta x=\frac{\lambda}{3}$

$ \begin{aligned} & \theta=\pi \cdot \frac{\Delta x}{\lambda} \\ & \theta=\frac{\pi \cdot \lambda}{3 \cdot \lambda}=\frac{\pi}{3} \end{aligned} $

So, $I^{\prime}=I_{\text {max }} \cos ^2 \theta$

$ \begin{aligned} & I^{\prime}=I_{\max } \cdot \cos ^2\left(\frac{\pi}{3}\right) \\ & I^{\prime}=\frac{I_{\max }}{4} \quad \therefore I_{\max }=I \\ & I^{\prime}=\frac{I}{4} \end{aligned} $

We know that the resultant intensity in Young's double slit experiment is given as

$ I_{\mathrm{res}}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}} \cos \phi $

where $ \phi= $ phase difference,

$ \begin{aligned} I_{1} & =I_{2}=I_{0} \\\\ \phi & =\frac{2 \pi}{\lambda} \Delta x \end{aligned} $

For path difference is, $ \Delta x=\lambda $

$ \begin{aligned} \phi_{1} & =\frac{2 \pi}{\lambda} \times \lambda=2 \pi \\\\ I_{\text {res }} & =I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos (2 \pi) \\\\ I_{\text {res }} & =I=4 I_{0} \\\\ I_{0} & =\frac{I}{4} \end{aligned} $

For path difference, $ \Delta x=\frac{\lambda}{3} $

$ \begin{aligned} \phi_{2} & =\frac{2 \pi}{3} \\\\ \cos \frac{2 \pi}{3} & =\frac{-1}{2} \\\\ I^{\prime} & =I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos \left(\frac{2 \pi}{3}\right) \\\\ I^{\prime} & =2 I_{0}-I_{0} \end{aligned} $

where $ \rho $ is resistivity

(iii) $ \vdots $ Subsitute the value of $ I_{0} $ )

$ I^{\prime}=\frac{I}{4} $

Given, $d=36 \mathrm{~m}$

$ \lambda=540 \mathrm{~nm} $

From Rayleigh criterion, $\theta=1.22 \frac{\lambda}{D}$

$ \begin{aligned} & \quad=\frac{1.22 \times 540 \times 10^{-9}}{3.6} \\\\ & =183 \times 10^{-9} \mathrm{rad} \\\\ & =1.83 \times 10^{-7} \mathrm{rad} \end{aligned} $

In YDSE, the intensity of light at a point on the screen is

$ I=I_0 \cos ^2\left(\frac{\pi \Delta x}{\lambda}\right) $

Given, $\lambda=6000 \times 10^{-10} \mathrm{~m}$

Path difference,

$ \Delta x_1=2000 \times 10^{-10} \mathrm{~m} $

Path difference,

$ \Delta x_2=1000 \times 10^{-10} \mathrm{~m} $

For $I_1$,

$ \begin{aligned} & I_1=I_0 \cos ^2\left(\frac{\pi \times 2000 \times 10^{-10}}{6000 \times 10^{-10}}\right) \\\\ & I_1=I_0 \cos ^2\left(\frac{\pi}{3}\right) \\\\ & I_1=\frac{I_0}{4} \end{aligned} $

For $I_2$,

$ \begin{aligned} & I_2=I_0 \cos ^2\left(\frac{\pi \times 1000 \times 10^{-10}}{6000 \times 10^{-10}}\right) \\\\ & I_2=I_0 \cos ^2\left(\frac{\pi}{6}\right) \\\\ & I_2=\frac{3 I_0}{4} \end{aligned} $

Therefore, the ratio $I_1: I_2$ is

$ \begin{aligned} & I_1: I_2=\frac{I_0}{4}: \frac{3 I_0}{4} \\\\ & I_1: I_2=1: 3 \end{aligned} $

So, the ratio of the intensities is $1: 3$.

Given,

Wavelength $d=6000\mathop {\rm{A}}\limits^{\rm{o}}$

Decrease in angular width $=30 \%$

Let original angular width $=\beta$

New width $=\beta-30 \% \beta=0.7 \beta$

Fringe width is given by

$ \begin{aligned} & \beta=\frac{\lambda D}{d} \\ & \beta \propto \lambda\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

$ \beta_1 \propto \lambda_1 $

From Eqs. (i) and (ii)

$ \begin{aligned} & \frac{\beta_1}{\beta}=\frac{\lambda_1}{\lambda} \\ & \lambda_1=\frac{\beta_1 \lambda}{\beta} \Rightarrow \frac{\beta \times 0.7 \times 6000}{\beta} \\ & \lambda_1=4200 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Hence, the correct option is (b).

The point source emits the spherical wavefront as the light diverge from a point source in all direction. Hence, incident wave front will be spherical in nature.

When the rays got reflected by concave mirror the reflected wavefront will be planet.

$ \text { Hence, option (b) is the correct option. } $

$ \begin{aligned}Given,\,\, d & =1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m} \\ \lambda & =6.5 \times 10^{-7} \mathrm{~m} \\ D & =1 \mathrm{~m} \end{aligned} $

Position of third dark fringes,

$ \begin{aligned} x_3 & =(2 n-1) \frac{D \lambda}{2 d} \\ & =(2 \times 3-1) \frac{D \lambda}{2 d}=\frac{5 D \lambda}{2 d} \end{aligned} $

Position of fifth bright fringe,

$ \begin{aligned} & x_5=n \frac{D \lambda}{d}=5 \frac{D \lambda}{d} \\ \therefore \quad & x_5-x_3=\frac{5 D \lambda}{d}-\frac{5 D \lambda}{2 d} \\ & =\frac{5 D \lambda}{2 d} \\ & =\frac{5 \times 1 \times 6.5 \times 10^{-7}}{2 \times 1 \times 10^{-3}} \\ & =1.625 \times 10^{-3} \mathrm{~m} \\ & =1.625 \mathrm{~mm} \end{aligned} $

Given, $d=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

$ \begin{aligned} & D=1 \mathrm{~m} \\ & \lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m} \end{aligned} $

Fringe width,

$ \begin{aligned} \beta & =\frac{D \lambda}{d}=\frac{1 \times 5 \times 10^{-7}}{10^{-3}} \\ & =5 \times 10^{-4} \mathrm{~m}=0.5 \times 10^{-3} \mathrm{~m} \\ & =0.5 \mathrm{~mm} \end{aligned} $

Given:

Slit width $ a = 2 \, \mathrm{mm} = 2 \times 10^{-3} \, \mathrm{m} $

Wavelength of light $ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \mathrm{m} $

To find the Fresnel distance $ f_z $, use the formula:

$ f_z = \frac{a^2}{\lambda} $

Substitute the values into the formula:

$ f_z = \frac{(2 \times 10^{-3} \, \mathrm{m})^2}{4000 \times 10^{-10} \, \mathrm{m}} $

Calculate:

$ f_z = \frac{4 \times 10^{-6} \, \mathrm{m}^2}{4 \times 10^{-7} \, \mathrm{m}} = 10 \, \mathrm{m} $

Therefore, the Fresnel distance is approximately 10 meters.

Given:

Central maxima in air: $ Y = 2 \text{ cm} $

10th maxima in air: $ Y = 5 \text{ cm} $

Refractive index of liquid (µ): $ 1.5 = \frac{3}{2} $

When the apparatus is immersed in a liquid, the wavelength changes due to the refractive index. The relationship between wavelengths and refractive index is:

$ \mu = \frac{c}{v} = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}} = \frac{\lambda_{\text{vacuum}}}{\lambda_{\text{medium}}^{\prime}} $

From this, the new wavelength in the medium is:

$ \lambda_{\text{medium}}^{\prime} = \frac{2}{3} \lambda_{\text{vacuum}} $

Considering path difference along the X-axis:

The path difference for maxima is given by:

$ \Delta x = 0, \lambda, 2\lambda, \ldots, 10\lambda $

For maxima along the Y-axis, the relationship is:

$ \Delta Y = \frac{\Delta x D}{d} = \frac{n \lambda D}{d} \Rightarrow \Delta Y \propto \lambda $

Central maxima remains unchanged because it corresponds to $ n = 0 $:

$ Y' = 2 \text{ cm} $

For the 10th maxima:

We know:

$ \frac{\Delta Y}{\Delta Y^{\prime}} = \frac{\lambda}{\lambda_{\text{medium}}^{\prime}} $

By substituting the given values:

$ \frac{5 - 2}{Y' - 2} = \frac{3}{2} $

Solving for $ Y' $:

$ 3 = \frac{3}{2} (Y' - 2) $

$ Y' - 2 = 2 $

$ Y' = 4 \text{ cm} $

Thus, the new coordinates when the apparatus is immersed in the liquid are $ 2 \text{ cm} $ and $ 4 \text{ cm} $.

Distance between the two maxima is equal to fringe width which is given as

$ \begin{gathered} \beta=\frac{D \lambda}{d} \Rightarrow \beta \propto \frac{\lambda}{d} \\ \Rightarrow \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}=\frac{4 \lambda_1}{\lambda_1} \times \frac{d_1}{d_{1 / 2}}=4 \times 2=8 \\ \end{gathered} $

$ \Rightarrow \quad \beta_2=8 \beta_1 $

In Young's double slit experiment,

$ \begin{aligned} & n_1=12, \lambda_1=6000 \mathop {\rm{A}}\limits^{\rm{o}}=6 \times 10^{-7} \mathrm{~m} \\ & \lambda_2=4800 \mathop {\rm{A}}\limits^{\rm{o}}=4.8 \times 10^{-7} \mathrm{~m} \end{aligned} $

Then, $n_2=$ ?

In Young's double slit experiment, position of $n$th fringe,

$ \begin{aligned} & & x_n & =\frac{n D \lambda}{d} \\ \Rightarrow & & x_n & =\frac{n_1 D \lambda_1}{d}=\frac{n_2 D \lambda_2}{d} \\ \Rightarrow & & n_1 \lambda_1 & =n_2 \lambda_2 \\ \Rightarrow & & n_2 & =\frac{n_1 \lambda_1}{\lambda_2}=\frac{12 \times 6 \times 10^{-7}}{4.8 \times 10^{-7}}=15 \end{aligned} $

In Young's double slit experiment.

Angular fringe width, $\theta=0.15^{\circ}$

$ D=80 \mathrm{~cm}=0.8 \mathrm{~m} $

Wavelength of light used,

$ \lambda=490 \mathrm{~nm}=4.9 \times 10^{-7} \mathrm{~m} $

Angular fringe width when entire apparatus is immersed in a medium of refractive index $\mu=5 / 3$, is given as,

$ \theta^{\prime}=\frac{\theta}{\mu} $

$ =\frac{0.15}{5 / 3}=\frac{0.15 \times 3}{5}=0.09^{\circ} $

Given, wavelength of red light,

$ \lambda_R=780 \mathrm{~nm}=7.8 \times 10^{-7} \mathrm{~m} $

Wavelength of blue light,

$ \lambda_B=520 \mathrm{~nm}=5.2 \times 10^{-7} \mathrm{~m} $

According to question,

$ \begin{aligned} & & x_R & =x_B \\ \Rightarrow & & \frac{n \lambda_R D}{d} & =\frac{(n+1) \lambda_B D}{d} \end{aligned} $

$ \Rightarrow \quad n \lambda_R=(n+1) \lambda_B \Rightarrow \frac{\lambda_R}{\lambda_B}=\frac{n+1}{n} $

$ \begin{aligned} \Rightarrow & & \frac{7.8 \times 10^{-7}}{52 \times 10^{-7}} & =\frac{n+1}{n} \Rightarrow \frac{39}{26}=\frac{n+1}{n} \\ \Rightarrow & & 39 n & =26 n+26 \\ \Rightarrow & & 39 n-26 n & =26 \\ \Rightarrow & & 13 n & =26 \Rightarrow n=2 \end{aligned} $

Given, when young's double slit experiment is performed in air, the angular fringe width is $\theta=0.2$

$ =0.2 \times \frac{\pi}{180} \mathrm{rad} $

But we know that,

$ \theta=\frac{\lambda}{d} $

$ \begin{aligned} \Rightarrow \quad d & =\frac{\lambda}{\theta}=\frac{600 \times 10^{-9}}{0.2 \times \frac{\pi}{180}} \\ & {\left[\because \lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\right] } \\ & =\frac{600 \times 180 \times 10^{-9}}{0.2 \times \pi} \\ & =\frac{1.08 \times 10^{-4}}{0.2 \pi} \mathrm{~m} \end{aligned} $

When the entire measurement system is immersed in water of refractive index $\mu=\frac{4}{3}$, then new value of angular fringe width.

$ \begin{aligned} \theta^{\prime} & =\frac{\lambda}{d \mu}=\frac{600 \times 10^{-9}}{\frac{1.08 \times 10^{-4}}{0.2 \pi} \times \frac{4}{3}} \\ & =\frac{3 \times 0.2 \pi \times 600 \times 10^{-9}}{1.08 \times 4 \times 10^{-4}} \\ & =2.62 \times 10^{-3} \mathrm{rad} \\ & =2.62 \times 10^{-3} \times \frac{180}{\pi} \text { degree } \\ & =015 \text { degree } \end{aligned} $

∴ Change in angular width $=\theta-\theta^{\prime}$

$ =0.2^{\circ}-0.15^{\circ}=0.05^{\circ} $

In Young's double slit experiment,

$ \begin{aligned} & d=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m} \\ & D=60 \mathrm{~cm}=0.6 \mathrm{~m} \\ & \lambda=550 \mathrm{~nm}=5.5 \times 10^{-7} \mathrm{~m} \\ & \mu=\frac{11}{9} \end{aligned} $

∴ Fringe width when whole apparatus is immersed in liquid,

$ \begin{aligned} \beta & =\frac{D \lambda}{d \mu}=\frac{0.6 \times 5.5 \times 10^{-7}}{2 \times 10^{-4} \times \frac{11}{9}} \\ & =\frac{9 \times 0.6 \times 5.5 \times 10^{-3}}{2 \times 11} \\ & =1.35 \times 10^{-3} \mathrm{~m}=1.35 \mathrm{~mm} \end{aligned} $

As we know that, Resolving power of optical instrument,

$ \mathrm{RP}=\frac{2 \mu \sin \theta}{1.22 \lambda} \Rightarrow \mathrm{RP} \text { or } R \propto \frac{1}{\lambda} $

Hence, ratio of resolving power of two instruments is

$ \begin{aligned} & \frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1} \Rightarrow \frac{R_1}{R_2}=\frac{5000}{4000} \\ & {\left[\because \lambda_1=4000 \mathop {\rm{A}}\limits^{\rm{o}} \text { and } \lambda_2=5000 \mathop {\rm{A}}\limits^{\rm{o}}\right]} \\ & \frac{R_1}{R_2}=\frac{5}{4} \text { or } 5: 4 \end{aligned} $

Fringe width in Young's double slit experiment is given as

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

where, $D=$ distance between source and screen and $d=$ slit width.

$ \begin{array}{ll} \therefore & \frac{\beta_2}{\beta_1}=\frac{d_1}{d_2} \times \frac{D_2}{D_1} \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{d_1}{2 d_1} \times \frac{D_1 / 2}{D_1} \\ & \quad\left[\because d_2=2 d_1 \text { and } D_2=\frac{D_1}{2}\right] \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{1}{4} \Rightarrow B_2=\frac{\beta_1}{4} \end{array} $

Limit of resolution or angular resolution for a telescope is

$ \alpha_{\min }=\frac{1.22 \lambda}{D} $

Here, $\alpha_{\text {min }}=3 \times 10^{-7} \mathrm{rad}$,

$ \lambda=525 \mathrm{~nm}=525 \times 10^{-9} \mathrm{~m} $

∴ Diameter of objective lens is

$ \begin{aligned} D & =\frac{1.22 \lambda}{\alpha_{\min }} \\ & =\frac{1.22 \times 525 \times 10^{-9}}{3 \times 10^{-7}} \\ & =213.5 \times 10^{-2}=2.1 \mathrm{~m} \end{aligned} $

Given, $\lambda_1=6600 \mathop {\rm{A}}\limits^{\rm{o}}, \lambda_2=4400 \mathop {\rm{A}}\limits^{\rm{o}}$,

$ n_1=60 $

For same set up of YDSE,

$ \begin{aligned} n_1 \lambda_1 & =n_2 \lambda_2 \\ \Rightarrow n_2 & =\frac{n_1 \lambda_1}{\lambda_2}=\frac{60 \times 6600}{4400} \\ & =\frac{60 \times 6}{4}=15 \times 6=90 \text { fringes } \end{aligned} $

$ \begin{aligned} &\text { Fringe width in YDSE is given by }\\ &\begin{aligned} & \beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \lambda \\ & \text { As, } \lambda_R>\lambda_G>\lambda_B \\ & \Rightarrow \beta_R>\beta_G>\beta_B \end{aligned} \end{aligned} $