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  • One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the cen

One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness t and wavelength of light 480 nm. Then, the value of t is:

Option: 1

\mathrm{2.4 \mu \mathrm{m}}


Option: 2

\mathrm{4.8 \mu \mathrm{m}}


Option: 3

\mathrm{8 \mu \mathrm{m}}


Option: 4

16 \mathrm{~m}


Answers (1)

best_answer

Five fringes have to be shifted

\mathrm{ \because \quad 5=\frac{\left(\mu_1-1\right) \mathrm{t}-\left(\mu_2-1\right) \mathrm{t}}{\lambda}=\frac{\left(\mu_1-\mu_2\right) \mathrm{t}}{\lambda} }

\mathrm{\therefore \quad \text { Thickness, } \mathrm{t}=\frac{5 \lambda}{\left(\mu_1-\mu_2\right)}=\frac{5 \times 480 \times 10^{-9}}{1.7-1.4}=8 \times 10^{-6} \mathrm{~m}}

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