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Chapter 4: Problem 30

A single slit of width \(3.0 \mu \mathrm{m}\) is illuminated by a sodium yellow light of wavelength 589 nm. Find the intensity at a \(15^{\circ}\) angle to the axis in terms of the intensity of the central maximum.

Short Answer

Expert verified
The intensity at a \(15^{\circ}\) angle to the axis is approximately \(0.0957I_0\), which means it is about 9.57% of the intensity of the central maximum.

Step by step solution

01

Convert Units

Before we proceed with any formulas, it is crucial to make sure all the variables have consistent units. The width of the slit is given in micrometers (μm), we will convert it to nanometers (nm) to match the wavelength unit, nm. $$ 1 \mu m = 1000\ nm $$ Slit width: \(3.0 \mu m = 3.0 * 1000\ nm = 3000\ nm\) Now the given variables are: - Slit width: 3000 nm - Wavelength: 589 nm - Angle: 15°
02

Apply the Single-slit Diffraction Formula

The single-slit diffraction formula is as follows: $$ I(\theta) = I_0 \left(\frac{\sin(\frac{\pi a}{\lambda}\sin\theta)}{(\frac{\pi a}{\lambda}\sin\theta)}\right)^2 $$ Where: \(I(\theta)\) is the intensity at angle \(\theta\), \(I_0\) is the intensity of the central maximum, \(a\) is the width of the slit, \(\lambda\) is the wavelength of the light, and \(\theta\) is the angle. We will use this formula to find \(I(15^\circ)\) in terms of \(I_0\).
03

Obtain Final Intensity Value

Plugging the given values into the single-slit diffraction formula and using \( \theta\) = 15°: $$ I(15^\circ) = I_0 \left(\frac{\sin(\frac{(\pi)(3000)}{(589)}\sin(15^\circ))}{(\frac{(\pi)(3000)}{(589)}\sin(15^\circ))}\right)^2 $$ Now it's just a matter of calculating the value of the fraction inside the brackets and squaring it. Remember that we only need to find the intensity in terms of the central maximum intensity, \(I_0\). After evaluating: $$ I(15^\circ) \approx 0.0957 I_0 $$ This means that the intensity at 15° relative to the axis is approximately 9.57% of the intensity of the central maximum.

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Most popular questions from this chapter

Can an astronaut orbiting Earth in a satellite at a distance of \(180 \mathrm{km}\) from the surface distinguish two skyscrapers that are \(20 \mathrm{m}\) apart? Assume that the pupils of the astronaut's eyes have a diameter of \(5.0 \mathrm{mm}\) and that most of the light is centered around \(500 \mathrm{nm}\).

Determine the intensities of three interference peaks other than the central peak in the central maximum of the diffraction, if possible, when a light of wavelength \(500 \mathrm{nm}\) is incident normally on a double slit of width \(1000 \mathrm{nm}\) and separation \(1500 \mathrm{nm} .\) Use the intensity of the central spot to be \(1 \mathrm{mW} / \mathrm{cm}^{2}\)

Two lamps producing light of wavelength 589 nm are fixed \(1.0 \mathrm{m}\) apart on a wooden plank. What is the maximum distance an observer can be and still resolve the lamps as two separate sources of light, if the resolution is affected solely by the diffraction of light entering the eye? Assume light enters the eye through a pupil of diameter \(4.5 \mathrm{mm}\).

Suppose that the central peak of a single-slit diffraction pattem is so wide that the first minima can be assumed to occur at angular positions of \(\pm 90^{\circ} .\) For this case, what is the ratio of the slit width to the wavelength of the light?

At what angle does a diffraction grating produce a second-onder maximum for light having a first-order maximum at \(20.0^{\circ}\) ?
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