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Q104P

Expert-verifiedFound in: Page 1080

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig, monochromatic light of wavelength diffracts through narrow slit S in an otherwise opaque screen. On the other side, a plane mirror is perpendicular to the screen and a distance h from the slit. A viewing screen A is a distance much greater than h. (Because it sits in a plane through the focal point of the lens, screen A is effectively very distant. The lens plays no other role in the experiment and can otherwise be neglected.) Light travels from the slit directly to A interferes with light from the slit that reflects from the mirror to A. The reflection causes a half-wavelength phase shift. (a) Is the fringe that corresponds to a zero path length difference bright or dark? Find expressions (like Eqs. 35-14 and 35-16) that locate (b) the bright fringes and (c) the dark fringes in the interference pattern. (Hint: Consider the image of S produced by the minor as seen from a point on the viewing screen, and then consider Young’s two-slit interference.)**

- A zero path difference would correspond to a dark fringe.
- For bright fringes, $2h\mathrm{sin}\theta =\left(m+\frac{1}{2}\right)\lambda $
- For dark fringes, $2h\mathrm{sin}\theta =m\lambda $

Given that monochromatic light of wavelength diffracts through narrow slit *S* in an otherwise opaque screen.

On the other side, a plane mirror is perpendicular to the screen and a distance *h* from the slit. A viewing screen *A* is a distance much greater than h.

Light travels from the slit directly to *A* interferes with light from the slit that reflects from the mirror to *A*. The reflection causes a half-wavelength phase shift.

For bright fringes, the distance is either zero or an integer number of wavelengths $d\mathrm{sin}\theta =m\lambda $ for $m=0,\u200a1,\u200a2,...$

And for dark fringes, the distance is odd multiple of half of wavelength,$d\mathrm{sin}\theta =\left(m+\frac{1}{2}\right)\lambda $ for $m=0,\u200a1,\u200a2,...$

Here, d is distance, and $\lambda $is wavelength.

Also, reflected ray has a phase change of $\mathrm{\pi}$ that causes the location of dark and bright fringes to interchange.

(a)

The combination of the direct ray and reflected ray from the mirror will produce an interference pattern on the screen, like the double-slit experiment. However, here, the reflected ray has a phase change of $\mathrm{\pi}$, causing the locations of the dark and bright fringes to be interchanged. Thus, a zero path difference would correspond to a dark fringe.

(b)

Here, the distance is 2h and has a phase change of$\mathrm{\pi}$ occurs, So

Condition to locate bright fringe in interference pattern is $2h\mathrm{sin}\theta =\left(m+\frac{1}{2}\right)\lambda $

(c)

Here, the distance is 2h and has a phase change of $\mathrm{\pi}$occurs, So

Condition to locate dark fringe in interference pattern is $2h\mathrm{sin}\theta =m\lambda $

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