Wave Optics Ans 12
Question 12 - 01 February - Shift 2
As shown in the figure, in Young’s double slit experiment, a thin plate of thickness $t=10 \mu m$ and refractive index $\mu=1.2$ is inserted infront of slit $S_1$. The experiment is conducted in air $(\mu=1)$ and uses a monochromatic light of wavelength $\lambda=500 nm$. Due to the insertion of the plate, central maxima is shifted by a distance of $x \beta_0 . \beta_0$ is the fringe-width before the insertion of the plate. The value of the $x$ is
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Answer: (4)
Solution:
Formula: YDSE
Fringe shift $=\frac{{ }^{\prime} \quad \prime}{\lambda} B$
$ \begin{aligned} & =\frac{10 \times 10^{-6}(1.2-1)}{5 \times 10^{-7}} B \\ & =\frac{10^{-5} \times 0.2}{5 \times 10^{-7}}=4 \end{aligned} $
