The maximum intensity in Young's double slit experiment is $$I0$$. Distance between the slits is d $$=$$ $$5$$λ, where λ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D $$=$$ $$10d$$ is

Question

The maximum intensity in Young's double slit experiment is $$I0$$. Distance between the slits is d $$=$$ $$5$$λ, where  λ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D $$=$$ $$10d$$ is

Infinity Learn · Accepted Answer

Suppose P is a point in front of one slit at which intensity is $$I0$$. From figure, it is clear that $$y=d2$$. Path difference between the waves reaching at P isΔ$$x=ydD=d2d10d=d20=5$$λ$$20=$$λ$$4Hence$$ corresponding phase difference is given by         ϕ$$=2$$πλ×λ$$4=$$π$$2Resultant$$ intensity at P is    $$I=Imaxcos2$$⁡ϕ$$2=I0cos2$$⁡π$$4=I02$$

The maximum intensity in Young's double slit experiment is I₀. Distance between the slits is d = 5 $λ$ , where $λ$ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D = 10d is

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The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d = 5λ, where λ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D = 10d is

Unlock the full solution & master the concept.

Ready to Test Your Skills?

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The maximum intensity in Young's double slit experiment is I₀. Distance between the slits is d = 5 $λ$ , where $λ$ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D = 10d is