Concept explainers
(a)
The angle at which the first maximum of sound intensity located.
(a)
Answer to Problem 12P
The angle at which the first maximum of sound intensity located is
Explanation of Solution
Write the expression to find the wavelength.
Here,
Write the expression for location of maxima.
Here,
Conclusion:
Substitute
Write the expression to find the angle of maxima from equation (II).
Substitute
Therefore, the angle at which the first maximum of sound intensity located is
(b)
The slit separation if the sound wave is replaced by microwave.
(b)
Answer to Problem 12P
The slit separation if the sound wave is replaced by microwave is
Explanation of Solution
Write the expression for location of maxima.
Here,
Conclusion:
Substitute
Therefore, the slit separation if the sound wave is replaced by microwave is
(c)
The
(c)
Answer to Problem 12P
The frequency of light that gives the same angle to the first maximum intensity is
Explanation of Solution
Write the expression for location of maxima.
Here,
Conclusion:
Substitute
Therefore, the frequency of light that gives the same angle to the first maximum intensity is
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Chapter 37 Solutions
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
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- When a monochromatic light of wavelength 430 nm incident on a double slit of slit separation 5 m, there are 11 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.8 nm for the same double slit?arrow_forwardFor 600-nm wavelength light and a slit separation of 0.12 mm, what are the angular positions of the first and third maxima in the double slit interference pattern?arrow_forwardShow that the distribution of intensity in a double-slit pattern is given by Equation 36.9. Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes E1=E0sintE2=E0sin(t+) The phase angle in in E2 is due to the extra path length traveled by the lower beam in Figure 36.4. Recall from Equation 33.27 that the intensity of light is proportional to the square of the amplitude of the electric field. In addition, the apparent intensity of the pattern is the time-averaged intensity of the electromagnetic wave. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You will also need the trigonometric identity sinA+sinB=2sin(A+B2)cos(AB2)arrow_forward
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