Wave Optics

JEE Main level, numerical answer, medium

65 of 191

Question 65 of 191

As shown in the figure, in Young's double slit experiment, a thin plate of thickness t=10μmt=10\mu m and refractive index μ=1.2\mu =1.2 is inserted infront of slit S1{S}_{1}. The experiment is conducted in air (μ=1\mu =1) and uses a monochromatic light of wavelength λ=500nm\lambda =500nm. Due to the insertion of the plate, central maxima is shifted by a distance of xβ0x{\beta }_{0}. β0{\beta }_{0} is the fringe-width before the insertion of the plate. The value of the xx is ______.

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Answer: 4

Fringe shift due to slab in Young's double slit experiment is given by ∆x=D(μ−1)td∆x=\dfrac{D(\mu -1)t}{d}
Since, fringe width β0=Ddλ{\beta }_{0}=\dfrac{D}{d}\lambda
Then, ∆x=β0(μ−1)tλ∆x=\dfrac{{\beta }_{0}\left(\mu -1\right)t}{\lambda }
Putting the values, we have
=10×10−6(1.2−1)5×10−7β0=\dfrac{10\times {10}^{-6}(1.2-1)}{5\times {10}^{-7}}{\beta }_{0}
=10−5×0.25×10−7β0=4β0=\dfrac{{10}^{-5}\times 0.2}{5\times {10}^{-7}}{\beta }_{0} =4{\beta }_{0}
Hence, the value of x=4x=4.

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