Question

what is the wavelength λ2λ2lambda 2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if ddd = 0.350 mmmm ?

Answer 1

The distance separating two wave crests is known as the wavelength. The wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Double slit interference (constructive) can be described by the equation dsinθ=m λ

Here, the order is (m), the wavelength is (), and the slit distance is (d).

Information provided-

The laser's wavelength is d/8.

The slit separation is 0.500 mm in length.

Put the distance into consideration to determine the wavelength values as,

λ1=0.5 x 10⁻³/8

λ1=0.0625 x 10⁻³

Due to the second laser's wavelength, its second maximum would be at the same spot as the first laser's fourth minimum.

Therefore, we must first determine the fourth order minima as,

dsinθ= (4-1/2) λ1

dsinθ= (7/2) λ1

Using the second maxima, dsinθ= 2λ₂

Put the following value for in the equation above:

7/2λ1 = 2λ₂

7/2 x (0.0625 x 10⁻³) = 2λ₂

λ₂=0.000109375m.

Therefore, the wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Learn more about wavelength here-

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