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If the distance between the first and tenth minima of a double-slit pattern is 18.0 mm and the slits are separated by 0.150 mm with the screen 50.0 cm from the slits, what is the wavelength of the light used?

Short Answer

Expert verified

Thus, the wavelength of light is 600nm.

Step by step solution

01

The separation of slits.

The condition for a minimum in the two-slit interference pattern is»å²õ¾±²Ôθ=(m+12)λ,where dis the slit separation, λis the wavelength,mis an integer, and θis the angle made by the interfering rays with the forward direction. If θis small, ²õ¾±²Ôθmay be approximately byθin radians.

Then, , θ=m+12λdand the distance from the minimum to the central fringe is:

y=Dtanθ≈Dsinθ≈Dθ=m+12Dλd

Where Dis the distance from the slits to the screen. For the first minimumm=0 for the tenth onem=9 . The separation is:

Δy=9+12Dλd-12Dλd=9Dλd

02

Calculate the wavelength of light. 

Solve for the wavelength as follows:

λ=dΔy9D=0.15×10-1m18×10-1m950×10-2m=6.0×10-7m=600nm

Hence, the wavelength of light is 600nm.

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