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  • In Young's double slit experiment, the fringe width is  If the entire arrangement is placed in water of refract

In Young's double slit experiment, the fringe width is 12 \mathrm{~mm}. If the entire arrangement is placed in water of refractive index \frac{4}{3}, then the fringe width becomes (in mm) :

Option: 1

16


Option: 2

9


Option: 3

48


Option: 4

12


Answers (1)

best_answer

\mathrm{\beta=\frac{\lambda D}{d}=12 \times 10^{-3} \mathrm{~m}}
Fringe width in water \mathrm{=\beta^{\prime}=\frac{\beta}{\mu}}

\mathrm{\beta^{\prime}=\frac{12 \times 10^{-3}}{4 / 3} =9 \times 10^{-3} \mathrm{~m} }

                               \mathrm{=9 \mathrm{~mm} }

Hence (2) option is correct

Posted by

shivangi.bhatnagar

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