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Chapter 4: Problem 111

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 \(1 \mathrm{mW} / \mathrm{cm}^{2}\)

Short Answer

Expert verified
The intensity levels of the first, second, and third order maxima are provided by the computations in step 3, using the formula for the intensity of light in diffraction patterns.

Step by step solution

01

Understand and Apply the Double Slit Interference Formula

The formula that gives the position of the bright fringes in double-slit interference is \( d \sin \theta = m \lambda \), where \( d \) is the separation between the slits, \( \theta \) is the angle to the bright fringe, \( \lambda \) is the wavelength of light, and \( m \) is an integer which is the interference order. Considering the first, second, and third order maxima, \( m \) takes values 1, 2, and 3. The aim is to compute the \(\theta \) for all three, which can be used to calculate relative intensity by comparing with the central maximum.
02

Calculation of Angles for Different Interference Orders

For \( m = 1 \) , \(\sin \theta_1 = (1) \cdot (500 \times 10^{-9} \mathrm{m}) / (1500 \times 10^{-9} \mathrm{m}) = 1/3\). Solving for \(\theta_1 = \arcsin(1/3)\). Similarly, for \( m = 2 \) and \( m = 3 \), the angles \(\theta _2\) and \(\theta_3\) are determined using the same formula.
03

Calculation of Relative Intensity

The intensity of light in the diffraction pattern can be calculated using the formula \(I = I_0 cos^2 (\frac{{\pi a \sin \theta}}{{\lambda}})\), where \( a \) is the slit width, \( I_0 \) is the intensity of the central peak (given as \(1 \mathrm{mW} / \mathrm{cm}^{2}\)) and \( \lambda \) is the wavelength of light. Compute \(I_1, I_2, I_3\) for \( \theta_1, \theta_2, \theta_3\) respectively.

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Most popular questions from this chapter

On a bright clear day, you are at the top of a mountain and looking at a city \(12 \mathrm{km}\) away. There are two tall towers \(20.0 \mathrm{m}\) apart in the city. Can your eye resolve the two towers if the diameter of the pupil is 4.0 mm? If not, what should be the minimum magnification power of the telescope needed to resolve the two towers? In your calculations use \(550 \mathrm{nm}\) for the wavelength of the light.

Microwaves of wavelength 10.0 mm fall normally on a metal plate that contains a slit \(25 \mathrm{mm}\) wide. (a) Where are the first minima of the diffraction pattern? (b) Would there be minima if the wavelength were \(30.0 \mathrm{mm}\) ?

(a) Calculate the angle at which a \(2.00-\mu \mathrm{m}\) -wide slit produces its first minimum for 410 -nm violet light. (b) Where is the first minimum for 700 -nm red light?

A source of light having two wavelengths \(550 \mathrm{nm}\) and \(600 \mathrm{nm}\) of equal intensity is incident on a slit of width \(1.8 \mu \mathrm{m} .\) Find the separation of the \(m=1\) bright spots of the two wavelengths on a screen \(30.0 \mathrm{cm}\) away.

Compare interference and diffraction.
See all solutions

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