(A) The shift in the central bright fringe due to a plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{d}(\mu - 1)t =

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(A) The shift in the central bright fringe due to a plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{d}(\mu - 1)t = \frac{\beta}{\lambda}(\mu - 1)t$.When both slits are covered,the net shift is $\Delta x_{net} = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$.Given that the central fringe shifts to the position of the $5^{th}$ bright fringe,the net shift is $5\beta$.Therefore,$5\beta = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$.$5 = \frac{1}{\lambda}(\mu_2 - \mu_1)t$.$t = \frac{5\lambda}{\mu_2 - \mu_1} = \frac{5 \times 4800 \times 10^{-10} \, m}{1.8 - 1.5} = \frac{24000 \times 10^{-10}}{0.3} \, m$.$t = 80000 \times 10^{-10} \, m = 8 \times 10^{-6} \, m = 8 \, \mu m$.

Class 12 Physics Wave Optics Variations in YDSE (Young's Double Slit Experiment)

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In a Young's double-slit experiment,light of wavelength $4800 \, \mathring A$ is used. If both slits are covered by transparent plates of the same thickness $t$ having refractive indices $\mu_1 = 1.5$ and $\mu_2 = 1.8$,and the central bright fringe shifts to the position of the fifth bright fringe,then the thickness of the plates is ....... $\mu m$.

A
$8$
B
$80$
C
$0.8$
D
None of these

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Variations in YDSE (Young's Double Slit Experiment)

In a double slit experiment,when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits,the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is ($\lambda$ is the wavelength of the light used).

MediumJEE MAIN 2019

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If a transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is inserted in front of one of the slits of Young's Double Slit experiment,how much will be the shift in the interference pattern? The distance between the slits is $0.5 \, mm$ and that between slits and screen is $100 \, cm$. (Answer in $cm$)

MediumAIIMS 1999

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When a thin plastic film of refractive index $\mu = 1.45$ is placed in the path of one of the interfering waves,the central fringe shifts by a distance equal to $5$ fringes. If the wavelength of light used is $5890 \, \mathring{A}$,find the thickness of the film.

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In a Young's double-slit experiment,a glass slab of thickness $1.2 \, \mu m$ and refractive index $1.5$ is placed in front of one slit. Another slab of thickness $t$ and refractive index $2.5$ is placed in front of the other slit. If the position of the central fringe remains unchanged,then the thickness $t$ is equal to $...... \, \mu m$.

Medium

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$A$ transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is placed in front of one of the slits in a Young's double-slit experiment. By what distance (in $cm$) will the interference pattern shift? The distance between the two slits is $d = 0.5 \, mm$ and the distance between the screen and the slits is $D = 100 \, cm$.

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In a double slit experiment,when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits,the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is ($\lambda$ is the wavelength of the light used).

When a thin plastic film of refractive index $\mu = 1.45$ is placed in the path of one of the interfering waves,the central fringe shifts by a distance equal to $5$ fringes. If the wavelength of light used is $5890 \, \mathring{A}$,find the thickness of the film.

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