D. Suppose that a thin piece of glass were placed in nt of the lower slit in a double-slit apparatus so that the plitude for a photon of wavelength λ to reach that slit ers in phase by 180° with the amplitude to reach the slit. (a) Describe in detail the interference pattern on screen. At what angles will there be bright fringes? What is the minimum thickness of glass required, ming the index of refraction for the glass is n?

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1.30. Suppose that a thin piece of glass were placed in
front of the lower slit in a double-slit apparatus so that the
amplitude for a photon of wavelength λ to reach that slit
differs in phase by 180° with the amplitude to reach the
top slit. (a) Describe in detail the interference pattern on
the screen. At what angles will there be bright fringes?
(b) What is the minimum thickness of glass required,
assuming the index of refraction for the glass is n?
Transcribed Image Text:1.30. Suppose that a thin piece of glass were placed in front of the lower slit in a double-slit apparatus so that the amplitude for a photon of wavelength λ to reach that slit differs in phase by 180° with the amplitude to reach the top slit. (a) Describe in detail the interference pattern on the screen. At what angles will there be bright fringes? (b) What is the minimum thickness of glass required, assuming the index of refraction for the glass is n?
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(a) When a thin piece of glass is placed in front of the lower slit in a double-slit apparatus, the amplitude for a photon of wavelength 2 to reach that slit differs in phase by 180° with the amplitude to reach the top slit. This means that the two waves will interfere destructively at certain angles and constructively at others.

The interference pattern on the screen will have a series of bright and dark fringes, similar to the pattern obtained in a regular double-slit experiment. However, in this case, the bright fringes will be shifted downwards from their normal positions, since the waves from the top and bottom slits will interfere constructively at a lower point on the screen.

To find the angles at which there will be bright fringes, we can use the equation for constructive interference in a double-slit experiment:

d sinθ = mλ

where d is the distance between the slits, θ is the angle between the line connecting the slits and the line from the slits to the point on the screen, m is an integer representing the order of the bright fringe, and λ is the wavelength of the light.

Since the wavelength is given as 2, we have:

d sinθ = m(2)

The distance between the slits, d, is assumed to be much smaller than the distance to the screen, so we can make the approximation sinθ ≈ tanθ ≈ y/L, where y is the distance from the central maximum to the bright fringe and L is the distance from the slits to the screen. Substituting this approximation into the equation above, we get:

y = mLλ/2d

Since the waves from the top and bottom slits will interfere constructively at a lower point on the screen, we need to subtract the thickness of the glass, t, from the distance between the slits, d, in the equation above. Therefore, the equation becomes:

y = mLλ/[2(d-t)]

The angles at which there will be bright fringes are given by:

θ = tan^-1(y/L)

where L is the distance from the slits to the screen.

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