Q: The path difference between two interfering waves at a point on the screen is \$ \lambda /6 \$. The ratio of intensity at the point and that the central bright fringe will be (Assume that internally due to each slit in same).

Given that, Path difference \$x=\dfrac{\lambda }{6}\$ We know that, \$ I={{I}_{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) \$ \$ \dfrac{I}{{{I}_{0}}}={{\cos }^{2}}\left( \dfrac{\phi }{2} \right)....(I) \$ Now, the phase difference is \$ \phi =\dfrac{2\pi }{\lambda }\times x \$ \$ \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} \$ \$ \phi =\dfrac{\pi }{3} \$ Now, put the value of \$\phi \$ in equation (I) \$ \dfrac{I}{{{I}_{0}}}={{\cos }^{2}}{{30}^{0}} \$ \$ \dfrac{I}{{{I}_{0}}}=\dfrac{3}{4} \$ \$ \dfrac{I}{{{I}_{0}}}=0.75 \$ Hence, the value of ratio of the intensity at the point is \$0.75\$

Q: Phase difference \$(\phi)\$ and path difference \$(\delta)\$ are related by :

Relation between phase difference and path difference is \$\phi = \dfrac{2\pi}{\lambda} \delta\$where \$\lambda\$ is the wavelength of the wave.For constructive interference : Path difference \$\delta = n\lambda\$ \$\implies \ \phi = 2n\pi\$where \$n = 0, \ 1,\ 2, \ ...\$For destructive interference : Path difference \$\delta = (n+\dfrac{1}{2})\lambda\$ \$\implies \ \phi = (2n+1)\pi\$

Q: Two waves having the intensities in the ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to

Let the intensity of the two waves be \$I_1\$ and \$I_2\$Given: \$I_2 : I_1 = 9 : 1 \implies I_2 = 9 I_1\$Now \$\dfrac{I_{max}}{I_{min}} = \dfrac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_2} - \sqrt{I_1})^2} = \dfrac{(\sqrt{I_1} + \sqrt{9 I_1})^2}{(\sqrt{9 I_1} - \sqrt{I_1})^2} = \dfrac{16 I_1}{4 I_1}\$ \$\implies I_{max} : I_{min} = 4 : 1\$

Q: A parallel beam of sodium light of wavelength 5890 \$\overset{0}{A}\$ is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is \$60^o\$. The smallest thickness of the plate which will make it dark by reflection:

Let the thickness of the Glass slab is \$t\$The condition for minimum thickness corresponding to a dark band is \$2\mu t \,cosr = \lambda\$\$ \therefore t= \dfrac{\lambda}{2 \mu cos \,r} = \dfrac{5890 \times 10^{-10}}{2 \times 1.5 \times cos 60^o} \$ \$t=3926 \times 10^{-10} m = 3926 {\overset{o}A} \$

Q: Two coherent sources of intensity ratio \$9:4\$ produce interference. The intensity ratio of maxima and minima of the interference pattern is:

Given : \$\dfrac{I_2}{I_1} = \dfrac{9}{4}\$Maximum intensity \$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\$Minimum intensity \$I_{min} = (\sqrt{I_1} -\sqrt{ I_2})^2\$Thus ratio of maximum to minimum intensities \$\dfrac{I_{max}}{I_{min}} = \dfrac{(1+\dfrac{\sqrt{I_2}}{\sqrt{I_1}})^2}{(1 - \dfrac{\sqrt{I_2}}{\sqrt{I_1}})^2}\$\$\therefore\$ \$\dfrac{I_{max}}{I_{min}} = \dfrac{(1+\dfrac{3}{2})^2}{(1 - \dfrac{3}{2})^2}\$Or \$\dfrac{I_{max}}{I_{min}} = \dfrac{(2+3)^2}{(2 - 3)^2} = \dfrac{25}{1}\$

Q: In Young's double slit experiment ,wavelength of light is \$6000 \mathring{A}\$ The the phase difference between the light waves reaching the third bright fringe from the central fringe will be:

Given that,Wave length \$\lambda =6000\,\overset{\circ }{\mathop{A}}\,\$ Phase difference =\$\phi\$Now, the phase difference is \$ \phi =\dfrac{2\pi }{\lambda }\times n\lambda \$ \$ \phi =2\pi \times 3 \$ \$ \phi =6\pi \$ Hence, the phase difference is \$6\pi \$

Q: Here, figure shows P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation PQ is 5.0 m and phase of Q is ahead of that of P by \$90^o\$. A, B and C are three distant points of observation equidistant from the mid-point of PQ. The intensity of radiations at A, B and C in ratio form will be :

At A, PA-QA =PQ = 5m\$=\dfrac{20}{4}=\dfrac{\lambda }{4}=\Delta x\$\$\Delta \phi '=\dfrac{2\pi}{\lambda }\Delta x =\dfrac{2\pi}{\lambda }\cdot \dfrac{\lambda }{4}=\dfrac{\pi}{2}\$\$\because \$ Phase of Q is greater by \$\dfrac{\pi}{2}\$ than P\$\therefore \$ Net phase difference \$\phi =\dfrac{\pi}{2}-\dfrac{\pi}{2}=0\$\$I_A=4I'\$At \$B, \phi =\dfrac{\pi}{2}, I_B=2I'\$At \$ C \,\,\,\Delta x=QA-PA=\dfrac{\lambda }{4}\$\$\phi '=\dfrac{\pi}{2}\$Net phase difference \$\phi =\dfrac{\pi}{2}+\dfrac{\pi}{2}=\pi\$\$I_c=0\$\$I_A:I_B:I_C=4I':2I':0\$\$=2: 1:0\$

Question 1

What is the path difference for destructive interference?

Accepted Answer

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Question 2

The path difference between two interfering waves at a point on the screen  is $ \lambda /6 $. The ratio of intensity at the point and that the central bright fringe will be (Assume that internally due to each slit in same).

Accepted Answer

Given&#10;that,&#10;&#10;Path difference $x=\dfrac{\lambda }{6}$ &#10;&#10;We know that,&#10;&#10;  $&#10;I={{I}_{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) $ &#10;&#10; $ \dfrac{I}{{{I}_{0}}}={{\cos&#10;}^{2}}\left( \dfrac{\phi }{2} \right)....(I) $ &#10;&#10;Now, the phase difference is &#10;&#10;  $ \phi&#10;=\dfrac{2\pi }{\lambda }\times x $ &#10;&#10; $ \phi&#10;=\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} $ &#10;&#10; $ \phi&#10;=\dfrac{\pi }{3} $ &#10;&#10;Now, put the value of $\phi $ in equation (I)&#10;&#10;  $ \dfrac{I}{{{I}_{0}}}={{\cos&#10;}^{2}}{{30}^{0}} $ &#10;&#10; $ \dfrac{I}{{{I}_{0}}}=\dfrac{3}{4}&#10;$ &#10;&#10; $ \dfrac{I}{{{I}_{0}}}=0.75&#10;$ &#10;&#10;Hence,&#10;the value of ratio of the intensity at the point is $0.75$

Question 3

Phase difference $(\phi)$ and path difference $(\delta)$ are related by :

Accepted Answer

Relation between phase difference and path difference is $\phi = \dfrac{2\pi}{\lambda} \delta$where $\lambda$ is the wavelength of the wave.For constructive interference :  Path difference  $\delta = n\lambda$  $\implies   \ \phi = 2n\pi$where $n = 0, \ 1,\  2, \ ...$For destructive interference :  Path difference  $\delta = (n+\dfrac{1}{2})\lambda$  $\implies   \ \phi = (2n+1)\pi$

Question 4

Two waves having the intensities in the ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to

Accepted Answer

Let the intensity of the two waves be $I_1$  and  $I_2$Given:    $I_2  :  I_1  =  9  :  1        \implies  I_2   =  9  I_1$Now       $\dfrac{I_{max}}{I_{min}}  =  \dfrac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_2} - \sqrt{I_1})^2} = \dfrac{(\sqrt{I_1} + \sqrt{9  I_1})^2}{(\sqrt{9  I_1} - \sqrt{I_1})^2} = \dfrac{16  I_1}{4  I_1}$ $\implies    I_{max}  :  I_{min}  =  4  : 1$

Question 5

A parallel beam of sodium light of wavelength 5890 $\overset{0}{A}$ is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is $60^o$. The smallest thickness of the plate which will make it dark by reflection:

Accepted Answer

Let the thickness of the Glass slab is $t$The condition for minimum thickness corresponding to a dark band is $2\mu t \,cosr = \lambda$$ \therefore t= \dfrac{\lambda}{2 \mu cos \,r} = \dfrac{5890 \times 10^{-10}}{2 \times 1.5 \times cos 60^o} $    $t=3926 \times 10^{-10} m  = 3926 {\overset{o}A} $

Question 6

Two coherent sources of intensity ratio $9:4$ produce interference. The intensity ratio of maxima and minima of the interference pattern is:

Accepted Answer

Given :  $\dfrac{I_2}{I_1} = \dfrac{9}{4}$Maximum intensity   $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$Minimum intensity   $I_{min} = (\sqrt{I_1} -\sqrt{ I_2})^2$Thus ratio of maximum to minimum intensities   $\dfrac{I_{max}}{I_{min}} = \dfrac{(1+\dfrac{\sqrt{I_2}}{\sqrt{I_1}})^2}{(1 - \dfrac{\sqrt{I_2}}{\sqrt{I_1}})^2}$$\therefore$   $\dfrac{I_{max}}{I_{min}} = \dfrac{(1+\dfrac{3}{2})^2}{(1 - \dfrac{3}{2})^2}$Or   $\dfrac{I_{max}}{I_{min}} = \dfrac{(2+3)^2}{(2 - 3)^2} = \dfrac{25}{1}$

Question 7

In Young's double slit experiment ,wavelength of light is $6000 \mathring{A}$ The the phase difference between the light waves reaching the third bright fringe from the central fringe will be:

Accepted Answer

Given that,Wave length $\lambda =6000\,\overset{\circ }{\mathop{A}}\,$&#10;Phase difference =$\phi$Now, the phase difference is&#10;&#10;  $ \phi =\dfrac{2\pi&#10;}{\lambda }\times n\lambda  $ &#10;&#10; $ \phi =2\pi \times&#10;3 $ &#10;&#10; $ \phi =6\pi  $ &#10;&#10;Hence, the phase difference is $6\pi $

Question 8

Here, figure shows P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation PQ is 5.0 m and phase of Q is ahead of that of P by $9 0^{o}$ . A, B and C are three distant points of observation equidistant from the mid-point of PQ. The intensity of radiations at A, B and C in ratio form will be :

Accepted Answer

At A, PA-QA =PQ = 5m$=\dfrac{20}{4}=\dfrac{\lambda }{4}=\Delta x$$\Delta \phi '=\dfrac{2\pi}{\lambda }\Delta x =\dfrac{2\pi}{\lambda }\cdot \dfrac{\lambda }{4}=\dfrac{\pi}{2}$$\because $ Phase of Q is greater by $\dfrac{\pi}{2}$ than P$\therefore $ Net phase difference $\phi =\dfrac{\pi}{2}-\dfrac{\pi}{2}=0$$I_A=4I'$At   $B, \phi =\dfrac{\pi}{2}, I_B=2I'$At   $ C   \,\,\,\Delta x=QA-PA=\dfrac{\lambda }{4}$$\phi '=\dfrac{\pi}{2}$Net phase difference $\phi =\dfrac{\pi}{2}+\dfrac{\pi}{2}=\pi$$I_c=0$$I_A:I_B:I_C=4I':2I':0$$=2: 1:0$

Question 9

Two waves of intensities $I$ and $4I$ superimpose. The minimum and maximum intensities will respectively be

Accepted Answer

The Intensity of the wave is directly proportional to the square of its amplitude.$I \propto A^{2}$;$I = cA$$^2$;   '$c$' is an arbitrary constant.So, if a wave with Intensity $I$ has an amplitude of $A(A{_{1}}$)A wave with an Intensity of $4I$ would respectively have an amplitude of $2A$($A{_{2}}$)If 2 waves with amplitudes $A{_{1}}$,$A{_{2}}$ are superimposed, the resultants would beMaximum of $A{_{1}}$ + $A{_{2}} = 3A$ (Constructive Interference)Minimum of $|A{_{1}}$ - $A{_{2}} |   = A $ (Destructive Interference)The wave of amplitude $3A$ would have an Intensity of $9I$The wave of amplitude $A$ would have an Intensity of $I$

Question 10

For constructive interference, the path difference between two waves must be

Accepted Answer

For constructive interference, the superimposing waves must be in phase or the phase difference should be integer multiple of $2\pi$. So  the path difference between two waves must be  integer multiple of $\lambda$. Ans: D

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