Question

# 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?

Solution

## 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, $$I=I_1+I_2+2\sqrt{I_1I_2}cos2\pi/3$$$$\Rightarrow I=I_1=K/4$$PhysicsNCERTStandard XII

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