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Light from a helium-neon laser \((\lambda=633 \mathrm{nm})\) passes through a circular aperture and is observed on a screen \(4.0 \mathrm{m}\) behind the aperture. The width of the central maximum is \(2.5 \mathrm{cm} .\) What is the diameter (in \(\mathrm{mm}\) ) of the hole?

Short Answer

Expert verified
The diameter of the hole is approximately 2.01 mm.

Step by step solution

01

Identify the Relevant Formula

The relation between the distance from the hole to the screen (L), the width of the central maximum (w), the wavelength of the light (\( \lambda \)), and the diameter of the hole (D) is given by \( w = \frac{2 \lambda L}{D} \). We are asked to find D.
02

Rearrangement of the Formula

We can isolate D by multiplying both sides of the equation by D and then dividing by w. The rearranged formula is \( D = \frac{2 \lambda L}{w} \).
03

Insert Known Values

Now, we can substitute the known values into the equation. The distance L is 4.0 m, the width w is 2.5 cm, and the wavelength \(\lambda\) is 633 nm. Don't forget to convert the units to the same system - in this case, we should use meters: \( D = \frac{2 * 633 * 10^{-9} m * 4.0 m}{2.5 * 10^{-2} m} \).
04

Evaluate the Expression

After substituting the values, simplify the equation to find the diameter of the hole. This leads to a result of approximately 0.00201 m. Converting this result to millimeters gives us a diameter of 2.01 mm.

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