
In the ideal double-slit experiment, when a glass plate (refractive index 1.5) of thickness $ t $ is introduced in the path of one of the interfering beams (wavelength $ \lambda $ ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is
$ (A)2\lambda $
$ (B)\dfrac{{2\lambda }}{3} $
$ (C)\dfrac{\lambda }{3} $
$ (D)\lambda $
Answer
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Hint: In modern physics, the double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles; moreover, it displays the fundamentally probabilistic nature of quantum mechanical phenomena.
Complete step by step solution:
Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used.
The quantity I is the intensity of the wave as a function of the phase difference of the two (identical) parent waves. If the two waves happen to be in phase, then the combined wave's intensity is Io when the two waves are in phase. In quantum mechanics the double-slit experiment demonstrated the inseparability of the wave and particle natures of light and other quantum particles. ... The results showed in many circumstances a pattern of interference, something which could only occur if wave patterns had been involved.
We know that in Young’s double slit experiment,
$ (\mu - 1)t = n\lambda $
Where $ \mu = $ refractive index
For minimum, we need to put $ n = 1 $
So, the required equation we get is,
$ (\mu - 1){t_{\min }} = \lambda $
By taking $ (\mu - 1) $ on the other side,
$ {t_{\min }} = \dfrac{\lambda }{{\mu - 1}} $
In this question we are given that $ \mu = 1.5 $ ,
$ {t_{\min }} = \dfrac{\lambda }{{1.5 - 1}} $
$ {t_{\min }} = \dfrac{\lambda }{{0.5}} $
On further solving, we get,
$ {t_{\min }} = 2\lambda $
So, the final answer is $ (A)\;2\lambda $ .
Note:
Coherent sources of light are those sources which emit a light wave having the same frequency, wavelength and in the same phase or they have a constant phase difference. A coherent source forms sustained interference patterns when superimposition of waves occurs and the positions of maxima and minima are fixed.
Complete step by step solution:
Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used.
The quantity I is the intensity of the wave as a function of the phase difference of the two (identical) parent waves. If the two waves happen to be in phase, then the combined wave's intensity is Io when the two waves are in phase. In quantum mechanics the double-slit experiment demonstrated the inseparability of the wave and particle natures of light and other quantum particles. ... The results showed in many circumstances a pattern of interference, something which could only occur if wave patterns had been involved.
We know that in Young’s double slit experiment,
$ (\mu - 1)t = n\lambda $
Where $ \mu = $ refractive index
For minimum, we need to put $ n = 1 $
So, the required equation we get is,
$ (\mu - 1){t_{\min }} = \lambda $
By taking $ (\mu - 1) $ on the other side,
$ {t_{\min }} = \dfrac{\lambda }{{\mu - 1}} $
In this question we are given that $ \mu = 1.5 $ ,
$ {t_{\min }} = \dfrac{\lambda }{{1.5 - 1}} $
$ {t_{\min }} = \dfrac{\lambda }{{0.5}} $
On further solving, we get,
$ {t_{\min }} = 2\lambda $
So, the final answer is $ (A)\;2\lambda $ .
Note:
Coherent sources of light are those sources which emit a light wave having the same frequency, wavelength and in the same phase or they have a constant phase difference. A coherent source forms sustained interference patterns when superimposition of waves occurs and the positions of maxima and minima are fixed.
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