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 , …

11th Edition

ISBN: 9781119459163

Author: Halliday

Publisher: WILEY

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Dispersion of White Light

Snell’s law is a formula that gives a relation between the angle of incidence and angle of refraction and is given as:

Question

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

Chapter 35, Problem 105P

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Chapter 35 Solutions

WILEY PLUS ACCESS CODE

Ch. 35 - Does the spacing between fringes in a two-slit...Ch. 35 - a If you move from one bright fringe in a two-slit...Ch. 35 - Figure 35-22 shows two light rays that are...Ch. 35 - In Fig. 35-23, three pulses of lighta, b, and cof...Ch. 35 - Is there an interference maximum, a minimum, an...Ch. 35 - Figure 35-24a gives intensity I verus position x...Ch. 35 - Figure 35-25 shows two sources S1 and S2 that emit...Ch. 35 - Figure 35-26 shows two rays of light, of...Ch. 35 - Light travels along the length of a 1500-nm-long...Ch. 35 - Figure 35-27a shows the cross section of a...

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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

Want to see the full answer?

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