Question

A narrow slit S transmitting light of wavelength

λis placed a distance d above a large plane mirror, as shown in the figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?

Answer 1

Answer:

a) the intensity at a point just above the mirror, i.e. just above O is zero.

b)y= λD/4d

Explanation:

A) Since there is a phase difference of π between direct light and reflecting light  the intensity just above the mirror will be zero.

B)separation between two slits is  2d.

Wavelength of the light is  λ.

Distance of the screen from the slit is  D.

Consider that the bright fringe is formed at position y. Then,

path difference,

∆x=y×2d/D=nλ

But as there is a phase reversal of λ/2

yx2d/D= λ/2= nλ

yx2d/D=nλ

=λ/2 (n-1)/2 =λ/2

so for first order put n=0 then

y= λD/4d

Answer 2

Answer:

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