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 . What will be the intensity of light in front of one of the slits on a screen at a distance $$D=10d$$?

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 . What will be the intensity of light in front of one of the slits on a screen at a distance $$D=10d$$?

Infinity Learn · Accepted Answer

Path difference, Δ$$x=ydD$$ Here, $$y=d2=5$$λ$$2And$$ $$D=10d=50$$λ (as$$  $$d=5$$λ$$)So$$, Δ$$x=(5$$λ$$2)(5$$λ$$50$$λ$$)=$$λ$$4Corresponding$$ phase difference will beϕ$$=(2$$πλ$$)($$Δ$$x)=(2$$πλ$$)($$π$$4)=$$π$$2Or$$ ϕ$$2=$$π$$4$$∴$$I=I0cos2($$ϕ$$2)=I0cos2($$π$$4) $$=I02$$

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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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 . What will be the intensity of light in front of one of the slits on a screen at a distance D=10d?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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