Two beams of light having intensities \$I\$ and \$4I\$ interfere to produce a fringe pattern on a screen.If the phase difference between the beams is \$\dfrac{\pi }{2}\$ at point A and \$\pi\$ at point B then the difference between the resultant intensities at A and B is

The resultant intensity is given by:\$I={I}_{1}+{I}_{2}+2\sqrt{{I}_{1}{I}_{2}}\cos{\phi}\$Thus difference is given by:\${I}_{A}-{I}_{B}=2\sqrt{{I}_{1}{I}_{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I\$

In the Young's double - slit experiment, the intensity of light at a point on the screen where the path difference is \$\lambda\$ is \$K, (\lambda\$ being the wave length of light used). The intensity at a point where the path difference is \$\lambda/4\$, will be

\$I=2I_0 (1+ \cos \theta)\$\$\phi =\dfrac{2 \pi}{\lambda}\$ (path as difference)\$k=2 z_0 (1+ \cos \left( \dfrac{2 z}{\lambda} \times \lambda\right))\$\$k=4I_0\$\$k^{'}=2I_0 (1+ \cos \left( \dfrac{2 z}{\lambda} \times \dfrac{\lambda}{4}\right))\$\$k^{'}=2I_0 (1+ \cos \left( \dfrac{2 z}{\lambda} \right)\$\$=2I_0\$so \$k^{'}=k/2\$

In young's double-slit experiment the intensity of light at a point on the screen where the path difference is \$\lambda\$ is \$\displaystyle I\$, \$\lambda\$ being the wavelength of light used. The intensity at a point where the path difference is \$\displaystyle\frac{\lambda}{4}\$ will be

For path difference \$\lambda\$, phasedifference = 2\$\pi\$ \$\left (Q = \displaystyle\frac{2\pi}{\lambda} x = \displaystyle\frac{2\pi}{\lambda}\lambda = 2\pi \right )\$\$\Rightarrow I_0 = I_0 + I_0 + 2I_0cos2\pi\$\$\Rightarrow I_0 = 4I_0\$ (\$\therefore cos2\pi\$ = 1 )For x = \$\displaystyle\frac{\lambda}{4}\$, phase difference = \$\displaystyle\frac{\pi}{2}\$ \$\therefore I'\$ =\$ I_1\$ + \$I_2\$ + 2\$\sqrt{I_1}\$ \$\sqrt{I_2}\$ cos \$\displaystyle\frac{\pi}{2}\$If \$I_1 \$= \$I_2 \$=\$ I_0\$ then \$I'= 2I_0\$ = 2.\$\displaystyle\frac{I}{4}\$ = \$\displaystyle\frac{I}{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).

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\$

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference '\$\lambda \$' is 'K' units. The intensity of light at a point where the path difference is \$\dfrac{\lambda}{3} \$ is (\$\lambda \$ being the wavelength of light used)

phase difference corresponding to path difference of \$\lambda /3\$ is\$\phi =\dfrac{2\pi }{3}\$So, \$I=k cos^{2}(\dfrac{2\pi /3}{2}) (\because I=I_{0} cos^{2}(\phi /2))\$ \$=k(+1/2)^{2} (\because cos \pi /3=1/2)\$ \$=\dfrac{k}{4}\$

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>>Class 12
>>Physics
>>Wave Optics
>>Superposition and Interference
>> $Introduction to Interference of Light$

Introduction to Interference of Light

Q: In Youngs double-slit experiment using monochromatic light of wavelength \$\lambda\$, the intensity of light at a point on the screen where path difference is \$\lambda\$, is K units. What is the intensity of light at a point where path difference is \$\lambda\$/3?

For monochromatic light, \$I_1+I_2\$ is the intensity.\$I^{'}=I_1+I_2+2\sqrt{I_1I_2}cos\theta\$Phase difference is given by:\$\phi=2\pi\times \dfrac{path\ difference}{\lambda}\$\$\phi=2\pi\times\dfrac{\lambda}{\lambda}\$\$=2\pi\$So, the new intensity can be obtained as: \$\Rightarrow I^{'}=4I_1\$given \$I^{'}=K,I_1=K/4,path\ difference=\lambda/3\$phase difference is \$2\pi\times 1/3\$Hence, \$I=I_1+I_2+2\sqrt{I_1I_2}cos2\pi/3\$\$\Rightarrow I=I_1=K/4\$

Q: Young's double slit experiment is first performed in air and then in a medium other than air. It is found that \$8^{th}\$ bright fringe in the medium lies where \$5^{th}\$ dark fringe lies in air. The refractive index of the medium is nearly:

In medium \$\lambda_{medium} = \dfrac{\lambda_{air}}{\mu}\$\$\lambda' =\dfrac{\lambda}{\mu}\$For Air, \$5^{th}\$ dark,\$y_s=(2m-1)\dfrac{\lambda D}{2d}\$\$m=5\$, \$y_s=\dfrac{9\lambda D}{2d}\$For bright in medium m\$y'=n\dfrac{\lambda' D}{d}\$, \$\lambda' = \dfrac{\lambda}{\mu}\$\$\therefore y' = \dfrac{8\lambda}{\mu} \dfrac{D}{d}\$\$y' = y\$\$\dfrac{9 \times D}{2d} = \dfrac{8\times D}{\mu d}\$ \$\therefore \mu = \dfrac{8 \times 2}{9}\$\$\mu = 1.78\$

Q: Two Polaroids \$P_1\$ and \$P_2\$ are placed with their axis perpendicular to each other. Unpolarised light \$I_0\$ is incident on \$P_1\$. A third polaroid \$P_3\$ is kept in between \$P_1\$ and \$P_2\$ such that its axis makes an angle \$45^o\$ with that of \$P_1\$. The intensity of transmitted light through \$P_2\$ is:

Initial intensity \$=I_0\$ After \$P_1\rightarrow I_1=\dfrac{I_0}{2}\$After \$P_3\rightarrow I_2=I_1cos^245^o\$\$=\dfrac{I_0}{2}\times \dfrac{1}{2}\$\$=\dfrac{I_0}{4}\$After \$P_2\rightarrow I_3=I_2cos^245^o\$\$=\dfrac{I_0}{4}\times \dfrac{a}{2}=\dfrac{I_0}{8}\$

Q: For constructive interference the phase difference between two waves must be _________.

Constructive interference between two waves occurs when there path difference is equal to the integral multiple of \$\lambda\$.For constructive interference, path difference \$\Delta x = n\lambda\$Phase difference \$\delta = \dfrac{2\pi}{\lambda} \Delta x \$\$\implies\$ \$\delta = \dfrac{2\pi}{\lambda} \times n\lambda = 2n\pi\$Hence for constructive interference, phase difference must be \$0,2\pi,4\pi,6\pi,........\$

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