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Chapter 32: Problem 14

The 546 -nm green line of gaseous mercury falls on a double-slit apparatus. If the fifth dark fringe is at \(0.113^{\circ}\) from the centerline, what's the slit separation?

Short Answer

Expert verified
The slit separation, \(d\), for the double-slit experiment, given these conditions, would be found through the given steps.

Step by step solution

01

Find the Wavelength in Meter

First, the wavelength of light has been given in nanometers, but for the calculation it's needed in meters. Therefore, convert the wavelength \(546\) nm into meters by multiplying the value by \(1.0 x 10^{-9}\) m/nm. So, \( \lambda = 546 \times 10^{-9}\) m.
02

Convert the Angle to Radians

Next, convert the given angle from degrees to radians because trigonometric functions in the formula use radians. Use the conversion factor that \(\pi\) rad = \(180^{\circ}\). Apply the conversion factor to \(0.113^{\circ}\) as followed: \( \theta = 0.113 \times (\pi / 180) \) rad.
03

Find the Slit Separation, \(d\)

Now, substitute \(m\), \(\lambda\), and \(\theta\) into the formula for a double-slit experiment, and then solve for the slit separation \(d\). In this instance, \(d= \frac{m \lambda}{\sin(\theta)} \). Since \(m=5\) ( as it's the fifth fringe), \(\lambda= 546 \times 10^{-9}\) m and \(\theta= 0.113 \times (\pi / 180)\) rad.

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Most popular questions from this chapter

In deriving the intensity in double-slit interference, why can't you simply add the intensities from the two slits?

You know that it's safe to microwave plastic containers, since their molecules don't respond significantly to 2.4 -GHz microwaves. But since you've learned about thin-film interference, you worry about enhanced microwave intensity due to multiple reflections in plastic cookware. You calculate the minimum thickness for a plastic tray with refractive index 1.3 that will cause enhanced reflection of microwaves incident normal to the plate. Are your worries assuaged?

Monochromatic light shines on a glass wedge with refractive index \(1.65,\) and enhanced reflection occurs where the wedge is \(450 \mathrm{nm}\) thick. Find all possible values for the wavelength in the visible range.

While driving at night, your eyes' irises dilate to 3.1 -mm diameter. If your vision were diffraction limited, what would be the greatest distance at which you could see as distinct the two headlights of an oncoming car, spaced \(1.5 \mathrm{m}\) apart? Take \(\lambda=550 \mathrm{nm}\)

The CIA wants your help identifying individual terrorists in a photo of a training camp taken from a spy satellite at \(100-\mathrm{km}\) altitude. You ask for details of the optical system used, but they're classified. However, they do tell you that the optics are diffraction limited and can resolve facial features as small as \(5 \mathrm{cm} .\) Assuming a typical optical wavelength of \(550 \mathrm{nm},\) what do you conclude about the size of the mirror or lens in the satellite camera?
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