Physics M.C.Q (1 Marks)

Therefore, other fringes will be coloured.

$\theta_w$ independent of $D$ but depends on $\lambda$

$n _{1} \lambda_{1}= n _{2} \lambda_{2}$

$(8)$ $(600 nm )= n _{2}(400)$

$n _{2}=12$

When Screen is moving away from slits then $D$ increases, so that fringe width increases.

$I_2=I_1 \cos ^2 \theta$

Where $I_1$ is incident Polarised light

$I _2= I \cos ^2 30= I \left(\frac{\sqrt{3}}{2}\right)^2$

$I _2=\frac{3 I }{4}$

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

$D ^{\prime}=2 D , d ^{\prime}=\frac{ d }{2}$

$\beta^{\prime}=\frac{\lambda \times 2 D}{d / 2}=\frac{4 \lambda D}{d}$

$\beta^{\prime}=4 \beta$

Fringe with becomes $4$ times

$\mu_{2}>1$

$\therefore \tan i _{ b }>1$

$\therefore 90^{\circ}> i _{ b }>45^{\circ}$

$\because 0=\frac{2 \pi}{\lambda} \Delta x$

where $\Delta x$ is path difference and $\phi$ is phase difference.

$=\frac{1.22 \times 6 \times 10^{-7}}{2}$

$=3.66 \times 10^{-7} rad$

$\therefore \theta^{\prime}=\frac{0.2^{\circ}}{4 / 3}=0.15^{\circ}$

$\Rightarrow \frac{\theta_{0}}{0.7 \theta_{0}}=\frac{\frac{6000 {\mathring A}}{\mathrm{d}}}{\frac{\lambda}{\mathrm{d}}} $$\Rightarrow \lambda=4200 \mathring A$

For fifth minima $(n=5)=\frac{9 \lambda}{2}$

${0.20^\circ } = \frac{\lambda }{{2\,{\text{mm}}}}$  .... $(i)$ and  ${0.21^\circ } = \frac{\lambda }{d}$   ..... $(ii)$

Dividing we get, $\frac{{0.20}}{{0.21}} = \frac{d}{{2\,{\text{mm}}}}$

$\therefore $ $d=1.9 \mathrm{mm}$.

Also, $tan\,\,i=\,\,\mu $ (Brewster angle)

$I_{\max }=\left(A_{1}+A_{2}\right)^{2}, I_{\min }=\left(A_{1}-A_{2}\right)^{2}$ where

$A=$amplitude

$=\frac{4 A_{1} A_{2}}{2\left(A_{1}^{2}+A_{2}^{2}\right)}=\frac{2 A_{1} A_{2}}{A_{1}^{2}+A_{2}^{2}}$

Now, dividing the numerator and denominator by $A _{1} A _{2}$, we get

$\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2}{\left[\frac{A_{1}}{A_{2}}+\frac{A_{2}}{A_{1}}\right]}=\frac{2}{\left[\sqrt{\alpha}+\frac{1}{\sqrt{\alpha}}\right]}=\frac{2 \sqrt{\alpha}}{(\alpha+1)}$

$x=\sqrt{2} \cdot \frac{\sqrt{3}}{2}$

$x=\frac{\sqrt{3}}{\sqrt{2}}$

$x: y=\sqrt{3}: \sqrt{2}$

$\frac{0.1}{6000}=\frac{v}{3 \times 10^{8}}$

$v=5\; km / s$

$\mathrm{RP}=\frac{2 \mu \sin \theta}{\lambda}$

For wavelength ${\lambda _1} = 4000\,\mathop {\text{A}}\limits^o $, resolving power will be

$\mathrm{RP}_{1}=\frac{2 \mu \sin \theta}{4000}$  ..... $(i)$

For wavelength ${\lambda _2} = 6000\,\mathop {\text{A}}\limits^o $, resolving power will be

$\mathrm{RP}_{2}=\frac{2 \mu \sin \theta}{6000}$  ..... $(ii)$

On dividing eqn. $(i)$ by eqn. $(ii)$

$=\frac{\mathrm{RP}_{1}}{\mathrm{RP}_{2}}=\frac{6000}{4000}=\frac{3}{2}$

$I_{1}=\frac{I_{0}}{2}$

The intensity of transmitted light through $P_{3}$

${I_2} = {I_1}{\cos ^2}{45^o} = $ $\frac{{{I_0}}}{2}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$ $ = \frac{{{I_0}}}{2} \cdot \frac{1}{2} = \frac{{{I_0}}}{4}$

Angle between polaroids $P_{3}$ and $P_{2}=\left(90^{\circ}-45^{\circ}\right)$ $=45^{\circ}$

$\therefore \quad$ Intensity of transmitted light through $P_{2}$

${I_3} = {I_2}{\cos ^2}{45^o} = $ $\frac{{{I_o}}}{4}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{{{I_o}}}{8}$

distance $=\frac{2 \lambda D}{d}$

$x=\frac{8 \lambda_{m} D}{d}$

Position of $5^{\text {th }}$ dark fringe in air,

${x^{\prime}=\frac{\left(5-\frac{1}{2}\right) \lambda_{\operatorname{air}} D}{d}}$

${x^{\prime}=\frac{4.5 \lambda_{\operatorname{air}} D}{d}}$

Given $x=x^{\prime}$

$\therefore \,\,\frac{{8{\lambda _m}D}}{d} = \frac{{4.5{\lambda _{{\text{air}}}}D}}{d}$

${\mu _m} = \frac{{{\lambda _{{\text{air}}}}}}{{{\lambda _m}}} = \frac{8}{{4.5}} = 1.78$

$\lambda=5 \times 10^{-5} \,\mathrm{cm}=5 \times 10^{-7}\, \mathrm{m}$

$D=60\, \mathrm{cm}=0.6\, \mathrm{m}$

Position of first minima on the diffraction pattern

${y_1} = \frac{{D\lambda }}{a} = $ $\frac{{0.6 \times 5 \times {{10}^{ - 7}}}}{{2 \times {{10}^{ - 4}}}}$ $ = 15 \times {10^{ - 4}}{\text{m}} = 0.15\,\,{\text{cm}}$

Resultant Intensity at $y=\frac{d}{2}, I_{y}=?$

The path difference between two waves at $y=\frac{d}{2}$

$\Delta x = d\tan \theta  = $ $d \times \frac{y}{D} = $ $\frac{{d \times \frac{d}{2}}}{{10d}} = $ $\frac{d}{{20}} = \frac{{5\lambda }}{{20}} = \frac{\lambda }{4}$

Corresponding phase difference, $\phi=\frac{2 \pi}{\lambda} \Delta x=\frac{\pi}{2}$

Now, maximum intensity in Young's double slit experiment,

${I_{\max }} = {I_1} + {I_2} + 2{I_1}{I_2}$

${I_0} = 4I$ $(\because \,\,{I_1}\, = \,{I_2}\, = \,I)$

$\therefore I=\frac{I_{0}}{4}$

Required intensity,

${I_y} = {I_1} + {I_2} + 2{I_1}{I_2}\cos \frac{\pi }{2}$ $ = 2I = \frac{{{I_0}}}{2}$

$a \sin \theta=\lambda$

Here $\theta=30^{\circ} \Rightarrow \sin \theta=\frac{1}{2}$

$\therefore \quad a=2 \lambda$    ..... $(i)$

For first secondary maximum, the path difference between extreme waves

$a \sin \theta=\frac{3}{2} \lambda$ or $(2 \lambda) \sin \theta^{\prime}=\frac{3}{2} \lambda$  [Using eqn $(i)$]

or $\sin \theta^{\prime}=\frac{3}{4} \quad \therefore \theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)$

$\phi_{e}=\phi_{f}$

on subtracts

$\phi_{c}-\phi_{e}=\phi_{d}-\phi_{f}$

                                                  $\left\{ {\because  \phi  = \frac{{2\pi }}{\lambda }\Delta } \right\}$

==> $\frac{{{I_1}}}{{{I_2}}} = \frac{{{{\cos }^2}\left( {\frac{{\pi {\Delta _1}}}{\lambda }} \right)}}{{{{\cos }^2}\left( {\frac{{\pi {\Delta _2}}}{\lambda }} \right)}} = \frac{{{{\cos }^2}\left( {\frac{{\pi .\frac{\lambda }{4}}}{\lambda }} \right)}}{{{{\cos }^2}(0)}} = \frac{1}{2}$

For that Phase difference Should be constant, and Pattern will not be constant

As there is no change, it should be conerent So it should have Same frequency.

When Amplitude is same, the minime that is formed is completely dark but it is not necessary Condition.we know I $\alpha(\text { Amplitade })^2$