Explanation 1) Concept: Here, we need to use the concept of interference of waves from thin films. Here, as the reflection causes the phase shift of half wavelength, the position of dark and bright fringes will change. 2) Formula: i) P a t h l e n g t h d i f f e r e n c e = o d d n u m b e r 2 × λ 3) Given: i) The light of wavelength λ diffracts from the slit. ii) The distance of slit is ‘h’ from the principle axis. 4) Calculation: a) As the reflection causes the phase shift of half wavelength, the rays coming at the viewing screen are out of phase. Hence, at the center, it will be a dark fringe. b) As the waves are already out of phase due to reflection from the mirror, for the bright fringes, the waves need further phase shift of half wavelength. So, P a t h l e n g t h d i f f e r n c e = o d d n u m b e r 2 × w a v e l e n g t h where, p a t h l e n g t h d i f f e r e n c e = 2 h s i n θ Here, h is the distance between the slit and the principle axis of lens. So, 2 h s i n θ = m + 1 2 λ , f o r m = 0 , 1 , 2 , …

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Dispersion of White Light

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Chapter 35, Problem 104P

To determine

To find:

a. Is the fringe that corresponds to a zero path length difference bright or dark?

b. The expression for the bright fringes

c. The expression for the dark fringes

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Chapter 35, Problem 103P

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Chapter 35 Solutions

Fundamentals of Physics Extended

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a. Is the fringe that corresponds to a zero path length difference bright or dark? b. The expression for the bright fringes c. The expression for the dark fringes

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To find: a. Is the fringe that corresponds to a zero path length difference bright or dark? b. The expression for the bright fringes c. The expression for the dark fringes

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Chapter 35 Solutions

To find:

a. Is the fringe that corresponds to a zero path length difference bright or dark?

b. The expression for the bright fringes

c. The expression for the dark fringes