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A plastic sheet (refractive index $=1.6$ ) covers one slit of a double slit arrangement for the Young's experiment. When the double slit is illuminated by monochromatic light (wavelength $=5867 Å$ ), the centre of the screen appears dark rather than bright. The minimum thickness of the plastic sheet to be used for this to happen is
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The correct answer is:
$5500 Å$
Given, wavelength of monochromatic light, $\lambda=5867 \hat{A}$
Refractive index, $\mu=1.6$
Let $t$ be the thickness of the plastic.
Since, the centre of the screen appears dark when plastic sheet is introduced, hence path difference,
$$
\mu t=\frac{\lambda}{2} \text { or } \frac{3 \lambda}{2}
$$
when, $\mu t=\frac{3 \lambda}{2}$
$$
\Rightarrow \quad t=\frac{3 \lambda}{2 \mu}=\frac{3 \times 5867}{2 \times 1.6}=5500 \hat{\lambda}
$$
Refractive index, $\mu=1.6$
Let $t$ be the thickness of the plastic.
Since, the centre of the screen appears dark when plastic sheet is introduced, hence path difference,
$$
\mu t=\frac{\lambda}{2} \text { or } \frac{3 \lambda}{2}
$$
when, $\mu t=\frac{3 \lambda}{2}$
$$
\Rightarrow \quad t=\frac{3 \lambda}{2 \mu}=\frac{3 \times 5867}{2 \times 1.6}=5500 \hat{\lambda}
$$
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