Wave Optics

To find the rate of evaporation of the soap film, we need to determine the rate at which its thickness decreases over time.

• Wavelength of light: $\lambda = 580$ = $580 \times 10^{-9}$
• Refractive index of soap film: $\mu = 1.45$
• Minimum transmission occurs every $T = 12$
• Radius of the ring: $r = 3$ = $0.03$

A thin-film interference pattern changes when the optical path difference changes by half a wavelength in the medium:

$∆t=\frac{\lambda }{2\mu }$​

This means that the thickness decreases by:

$$\Delta t = \frac{\lambda}{2\mu}$$​

Substituting values:

$$\Delta t = \frac{580 \times 10^{-9}}{2 \times 1.45}$$

$$\Delta t = \frac{580 \times 10^{-9}}{2.9}$$

​ $$\Delta t = 200 \times 10^{-9} \text{ m} = 2 \times 10^{-7} \text{ m}$$

This thickness decreases every 12 seconds, so the rate of evaporation per second:

$\frac{∆t}{dt}=\frac{2\times 10^{-7}\text{ m}}{12}$​ $$= 1.67 \times 10^{-8} \text{ m/s}$$

The evaporating volume per second is:

$\text{Rate of evaporation}=\text{Surface area}\times \frac{∆t}{dt}$$=\pi r^2\times \frac{∆t}{dt}$​

Substituting values:

$$= \pi (0.03)^2 \times (1.67 \times 10^{-8})$$

$$= \pi (9 \times 10^{-4}) \times (1.67 \times 10^{-8})$$

$=15\pi \times 10^{-12}{\text{ m}}^3\mathrm{/}\text{s}$

From the given options, the correct answer : $15\pi \times 10^{-12}$

For the $6^{th}$Sv6Kpe[] maximum in air to coincide with the $n^{th}$Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[]

Sv6Kpe[]

Sv6Kpe[]

To achieve maximum transmission of green light, the thin film must cause destructive interference for the light reflected from its top and bottom surfaces. This is a standard case of thin-film interference, and the condition for minimum thickness for maximum transmission (destructive reflection) is:

$2 \mathrm{n} \mathrm{t} = \mathrm{m} \mathrm{\lambda }$

where: n is the refractive index of the film, t is the thickness of the film, $\mathrm{\lambda }$ is the wavelength of light in the medium of the film, m is the order of interference (minimum for the thinnest film).

Here, m = 1

So, $\mathrm{t} =\frac{\mathrm{m} \mathrm{\lambda }}{2 \mathrm{n}}=\frac{1\times 5000}{2 \times 2}=1250\stackrel{°}{\mathrm{A}}$

In Young’s Double Slit Experiment (YDSE), the position of the $n^{th}$ bright fringe is given by:

$$y_n = n \times \frac{\lambda D}{d}$$​

where:

• $y_n$​ = distance of the $n^{th}$ bright fringe from the central maximum,
• $\lambda$= wavelength of light used,
• $D$ = distance between slits and screen,
• $d$= separation between the two slits,
• $n$ = fringe order.

For $\lambda_1 = 600$, the 10th bright fringe is at 10 mm:

$$y_{10} = 10 \times \frac{600 D}{d} = 10 \text{ mm}$$

$$\frac{D}{d} = \frac{1}{600} \text{ mm/nm}$$

Now, using the same formula for $\lambda_2 = 660$:

$$y_{10} = 10 \times \frac{660 D}{d}$$​

Substituting $\frac{D}{d} = \frac{1}{600} \text{ mm/nm}$:

$$y_{10} = 10 \times \frac{660}{600}$$

​ $$y_{10} = 10 \times 1.1$$

$$y_{10} = 11 \text{ mm}$$

To find the rate of evaporation of the soap film, we need to determine the rate at which its thickness decreases over time.

• Wavelength of light: $\lambda = 580$ = $580 \times 10^{-9}$
• Refractive index of soap film: $\mu = 1.45$
• Minimum transmission occurs every $T = 12$
• Radius of the ring: $r = 3$ = $0.03$

A thin-film interference pattern changes when the optical path difference changes by half a wavelength in the medium:

$∆t=\frac{\lambda }{2\mu }$​

This means that the thickness decreases by:

$$\Delta t = \frac{\lambda}{2\mu}$$​

Substituting values:

$$\Delta t = \frac{580 \times 10^{-9}}{2 \times 1.45}$$

$$\Delta t = \frac{580 \times 10^{-9}}{2.9}$$

​ $$\Delta t = 200 \times 10^{-9} \text{ m} = 2 \times 10^{-7} \text{ m}$$

This thickness decreases every 12 seconds, so the rate of evaporation per second:

$\frac{∆t}{dt}=\frac{2\times 10^{-7}\text{ m}}{12}$​ $$= 1.67 \times 10^{-8} \text{ m/s}$$

The evaporating volume per second is:

$\text{Rate of evaporation}=\text{Surface area}\times \frac{∆t}{dt}$$=\pi r^2\times \frac{∆t}{dt}$​

Substituting values:

$$= \pi (0.03)^2 \times (1.67 \times 10^{-8})$$

$$= \pi (9 \times 10^{-4}) \times (1.67 \times 10^{-8})$$

$=15\pi \times 10^{-12}{\text{ m}}^3\mathrm{/}\text{s}$

From the given options, the correct answer : $15\pi \times 10^{-12}$

$I\propto {\left(\text{ width }\right)}^2$

${\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2=\frac{9}{4}$

$\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1-\sqrt{I_2}}}=\frac{3}{2}$

$\frac{(x+1)\text{d}}{(\text{x}-1)\text{d}}=\frac{3}{2}$

$\Rightarrow 3\text{x}-3=2\text{x}+2$

$\text{x}=5$

$I\propto {\left(\text{ width }\right)}^2$

${\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2=\frac{9}{4}$

$\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1-\sqrt{I_2}}}=\frac{3}{2}$

$\frac{(x+1)\text{d}}{(\text{x}-1)\text{d}}=\frac{3}{2}$

$\Rightarrow 3\text{x}-3=2\text{x}+2$

$\text{x}=5$

The radius of refractive wavefrant will be twice the ford length =2f=R

• Fringe separation (β) in YDSE = λD / d.
• β = (600 × 10⁻⁹ m) × (1 m) / (1.5 × 10⁻³ m) = (600 / 1.5) × 10⁻⁶ m = 400 × 10⁻⁶ m = 4 × 10⁻⁴ m.

  (NEW NCERT 12^th Page No. 265, 266)

• In YDSE with white light, the central fringe (path difference = 0) is white because all colours interfere constructively. Fringes on either side are coloured due to different path differences for different wavelengths.
• Central bright white fringe surrounded by coloured fringes.

  (NEW NCERT 12^th Page No. 265, 266)

$$\begin{gathered} \frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d} \\ n480=m600 \\ n_{\min }=5 \end{gathered}$$

• The convex lens focuses light rays that were initially diverging from the point source.

• The light rays become nearly parallel after passing through the lens.

• The wavefronts become approximately flat and parallel to each other.

To achieve maximum transmission of green light, the thin film must cause destructive interference for the light reflected from its top and bottom surfaces. This is a standard case of thin-film interference, and the condition for minimum thickness for maximum transmission (destructive reflection) is:

$2 \mathrm{n} \mathrm{t} = \mathrm{m} \mathrm{\lambda }$

where: n is the refractive index of the film, t is the thickness of the film, $\mathrm{\lambda }$ is the wavelength of light in the medium of the film, m is the order of interference (minimum for the thinnest film).

Here, m = 1

So, $\mathrm{t} =\frac{\mathrm{m} \mathrm{\lambda }}{2 \mathrm{n}}=\frac{1\times 5000}{2 \times 2}=1250\stackrel{°}{\mathrm{A}}$

$$\begin{gathered} \frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d} \\ n480=m600 \\ n_{\min }=5 \end{gathered}$$

For the $6^{th}$Sv6Kpe[] maximum in air to coincide with the $n^{th}$Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[]

Sv6Kpe[]

Sv6Kpe[]

$$\begin{gathered} \text{Fringe width is given by } \\ \frac{\lambda D}{\mu d}=\frac{690\times {10}^{-9}\times 0.72}{1.44\times 1.5\times {10}^{-3}}=0.23mm \end{gathered}$$

For the $6^{th}$Sv6Kpe[] maximum in air to coincide with the $n^{th}$Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[]

Sv6Kpe[]

Sv6Kpe[]

According to Brewster's Law, when reflected light is completely polarized, the angle of incidence \((i)\) and the angle of
refraction \((r)\) sum up to \(90^\circ\).

\(i + r = 90^\circ\)

\(60^\circ + r = 90^\circ\) \(\Longrightarrow\) \(r = 30^\circ\).

For the $6^{th}$Sv6Kpe[] maximum in air to coincide with the $n^{th}$Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[]

Sv6Kpe[]

Sv6Kpe[]