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

Question

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

Infinity Learn · Accepted Answer

Here, $$d=5$$λ,$$D=10d$$,$$y=d2$$ Resultant Intensity at $$y=d2$$,$$Iy=$$? The path difference between two waves at $$y=d2$$ isΔ$$x=dtan$$θ$$=d$$×$$yD=d$$×$$d210d=d20=5$$λ$$20=$$λ$$4$$ Corresponding phase difference, ϕ$$=2$$πλΔ$$x=$$π$$2$$ .  Now, maximum intensity in Young's double slit experiment,  $$Imax=I1+I2+2I1I2$$  $$I0=4I$$∵$$I1=I2=I$$∴$$I=I04$$ intensity at y is  $$Iy=I1+I2+2I1I2cos$$$$π$$$$2=2I=I02$$

The intensity at the maximum in a Young’s double slit experiment is $I_{0}$ . Distance between two slits is $d = 5 λ$ , where $λ$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10 d ?$

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The intensity at the maximum in a Young’s double slit experiment is I0. Distance between two slits is d=5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10d?

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Ready to Test Your Skills?

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The intensity at the maximum in a Young’s double slit experiment is $I_{0}$ . Distance between two slits is $d = 5 λ$ , where $λ$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10 d ?$