Physics 5 Marks Questions

Given that, t₁ = t₂ = 0.5mm = 0.5 × 10^-3m, $\mu_\text{m}=1.58$ and $\mu_\text{p}=1.55,$

$\lambda=590\text{nm}=590\times 10^{-9}\text{m}, \text{d}=0.12\text{cm},\\ \mu_{\text{m}}=1.58\text{ and }\mu_{\text{p}}=1.55,$

Fringe width $=\frac{\text{D}\lambda}{\text{d}}=\frac{1\times590\times10^{-9}}{12\times10^{-4}}=4.91\times10^{-4}\text{m}.$

When both the strips are fitted, the optical path changes by

$\Delta\text{x}=(\mu_\text{m}-1)\text{t}_1-(\mu_\text{p}-1)\text{t}_2=(\mu_\text{m}-\mu_\text{p})\text{t}$

$=(1.58-1.55)\times(0.5)(10^{-3})=0.015\times10^{-13}\text{m}$

So, No. of fringes shifted $=\frac{0.015\times10^{-3}}{590\times10^{-3}}=25.43$

⇒ There are 25 fringes and 0.43 th of a fringe.

⇒ There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.

So, position of first maximum on both sides will be given by,

$\therefore$ x = 0.43 × 4.91 × 10^-4 = 0.021cm

x' = (1 - 0.43) × 4.91 × 10^-4 = 0.028cm (since, fringe width = 4.91 × 10^-4m)

1. When intensity is half the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{2}$

$\Rightarrow\frac{4\text{a}^2\cos^2\big(\frac{\phi}{2}\big)}{4\text{a}^2}=\frac{1}{2}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{2}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{4}\Rightarrow\phi=\frac{\pi}{2}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{4}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{4\text{d}}$

2. When intensity is $\frac{1}{4}\text{th}$ of the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{4}$

$\Rightarrow\frac{4\text{a}^2\cos^2\Big(\frac{\phi}{2}\Big)}{4\text{a}^2}=\frac{1}{4}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{4}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{2}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{3}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{3\text{d}}$

$\therefore\text{S}_1\text{P}-\text{S}_2\text{P}=\text{x}=(2\text{n}+1)\frac{\lambda}{2}$

$\Rightarrow\sqrt{\text{X}^2+(2\lambda)^2}-\text{Z}=(2\text{n}+1)\frac{\lambda}{2}$

$\Rightarrow\text{Z}^2+4\lambda^2=\text{Z}^2+(2\text{n}+1)^2\frac{\lambda^2}{4}+\text{Z}(2\text{n}+1)\lambda$

$\Rightarrow\text{Z}=\frac{4\lambda^2-(2\text{n}+1)^2\Big(\lambda^2{4}\Big)}{(2\text{n}+1)\lambda}=\frac{16\lambda^2-(2\text{n}+1)^2\lambda^2}{4(2\text{n}+1)\lambda}\ ...(1)$

1. When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ at S₄, minimum intensity occurs

⇒ Amplitude = 0,

At S₃, path difference = 0

⇒ Maximum intensity occurs.

⇒ Amplitude = 2r.

So, on $\sum_2$ screen,

$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+0)^2}{(2\text{r}-0)^2}=1$

2. When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ At S₄, minimum intensity occurs.

⇒ Amplitude = 0.

At S₃, path difference = 0

⇒ Maximum intensity occurs.

⇒ Amplitude = 2r.

So, on $\sum_2$ screen,

$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+2\text{r})^2}{(2\text{r}-0)^2}=\infty$

3. When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$, At S₄, intensity $=\frac{\text{I}_\text{max}}{2}$

⇒ Amplitude $=\sqrt{2\text{r}}$

$\therefore$ At S₃, intensity is maximum

⇒ Amplitude = 2r

$\therefore\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+\sqrt{2\text{r}})^2}{(2\text{r}-\sqrt{2\text{r}})^2}=34.$

$\text{d}=1\times10^{-4}\text{cm}=10^{-6}\text{m},$ $\mu_{\text{oil}}=1.25\ \text{and}\ \mu_\text{x}=1.50$

$\lambda=\frac{2\mu\text{d}}{\Big(\text{n}+\frac{1}{2}\Big)}\frac{2\times10^{-6}\times1.25\times2}{2\text{n}+1}=\frac{5\times10^{-6}\text{m}}{2\text{n}+1}$

$\Rightarrow\lambda=\frac{5000\text{nm}}{2\text{n}+1}$

$\mu=1.6,$ t = 1.964 micron = 1.964 × 10^-6m

Given that, $\lambda= (400\text{nm to } 700\text{nm}),$ d = 0.5mm = 0.5 × 10^-3m,

D = 50cm = 0.5m and on the screen y_n = 1mm = 1 × 10^-3m

1. We know that for zero intensity (dark fringe)

$\text{y}_\text{n}=\Big(\frac{2\text{n}+1}{2}\Big)\frac{\lambda_\text{n}\text{D}}{\text{d}}$ where n = 0, 1, 2, ...

$\Rightarrow\lambda_\text{n}=\frac{2}{(2\text{n}+1)}\frac{\lambda_\text{n}\text{d}}{\text{D}}=\frac{2}{2\text{n}+1}\times\frac{10^{-3}\times0.5\times10^{-3}}{0.5}$

$\Rightarrow\frac{2}{(2\text{n}+1)}\times10^{-6}\text{m}=\frac{2}{(2\text{n}+1)}\times10^3\text{nm}$

If $\text{n}=1,\lambda_1=\Big(\frac{2}{3}\Big)\times1000=667\text{nm}$

If $\text{n}=1,\lambda_2=\Big(\frac{2}{5}\Big)\times1000=400\text{nm}$

So, the light waves of wavelengths 400nm and 667nm will be absent from the out coming light.

2. For strong intensity (bright fringes) at the hole

$\text{y}_\text{n}=\frac{\text{n}\lambda_\text{n}\text{D}}{\text{d}}\Rightarrow\lambda_\text{n}=\frac{\text{y}_\text{n}\text{d}}{\text{nD}}$

When, $\text{n}=1,\lambda_1=\frac{\text{y}_\text{n}\text{d}}{\text{D}}$

$=\frac{10^{-3}\times0.5\times10^{-3}}{0.5}=10^{-6}\text{m}=1000\text{nm}.$

1000nm is not present in the range 400nm - 700nm

Again, where $\text{n}=2,\lambda_2=\frac{\text{y}_\text{n}\text{d}}{2\text{D}}=500\text{nm}$

So, the only wavelength which will have strong intensity is 500nm.

1. When, $\text{z}=\frac{\text{D}\lambda}{\text{d}}$

So, $\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{2\text{d}}$

⇒ Dark fringe at S₃ and S₄.

⇒ At S₃, intensity at S₃ = 0 ⇒ I₁ = 0

At S₄, intensity at S₄ = 0 ⇒ I₂ = 0

At P, path difference = 0 ⇒ Phase difference = 0.

$\Rightarrow\text{I}=\text{I}_1+\text{I}_2+\sqrt{\text{I}_1\text{I}_2}\cos0^\circ=0+0+0=0$

⇒ Intensity at P = 0.

2. Given that, when $\text{z}=\frac{\text{D}\lambda}{2\text{d}}$, intensity at P = I

Here, $\text{OS}_3=\text{OS}_4=\text{y}=\frac{\text{D}\lambda}{4\text{d}}$

$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{\pi}{2}.$

$\Big[$Since, x = path difference $=\frac{\text{yd}}{\text{D}}\Big]$

Let, intensity at S₃ and S₄ = I'

$\therefore$ At P, phase difference = 0

So, $\text{I}'+\text{I}'+2\text{I}'\cos0^\circ=\text{I}$

$\Rightarrow4\text{I}'=\text{I}\Rightarrow\text{I}'=\frac{1}{4}$

When, $\text{z}=\frac{3\text{D}\lambda}{2\text{d}},\Rightarrow\text{y}=\frac{3\text{D}\lambda}{4\text{d}}$

$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{3\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{3\pi}{2}$

Let, I'' be the intensity at S₃ and S₄ when, $\phi=\frac{3\pi}{2}$

Now comparing,

$\frac{\text{I}"}{\text{I}}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos\Big(\frac{3\pi}{2}\Big)}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{2\text{a}^2}{2\text{a}^2}=1$ $\Rightarrow\text{I}"=\text{I}'=\frac{\text{I}}{4}$

$\therefore$ Intensity at $\text{P}=\frac{\text{I}}{4}+\frac{\text{I}}{4}+2\times\Big(\frac{\text{I}}{4}\Big)\cos0^\circ=\frac{\text{I}}{2}+\frac{\text{I}}{2}=1$

3. When $\text{z}=\frac{2\text{D}\lambda}{\text{d}}$

$\Rightarrow\text{y}=\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{\text{d} }$

$\therefore\phi=\frac{2\pi\text{X}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{\text{d}}\times\frac{\text{d}}{\text{D}}=2\pi.$

Let, I"' = intensity at S₃ and S₄ when, $\phi=2\pi.$

$\frac{\text{I}'''}{\text{I}'}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos2\pi}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{4\text{a}^2}{2\text{a}^2}=2$

$\Rightarrow\text{I}'''=2\text{I}'=2\Big(\frac{\text{I}}{4}\Big)=\frac{\text{I}}{2}$

At P, I_resultant $=\frac{\text{I}}{2}+\frac{\text{I}}{2}+2\Big(\frac{\text{I}}{2}\Big)\cos0^\circ=\text{I}+\text{I}=2\text{I}$

So, the resultant intensity at P will be 2I.