Q179 / 180 · COMEDK 2020 · Physics

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

  • A
    $3300 Å$
  • B
    $6600 Å$
  • C
    $2062 Å$
  • Correct answer
    D
    $5500 Å$

Solution

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, $\mu t=\frac{3 \lambda}{2}$ $$ \Rightarrow \quad t=\frac{3 \lambda}{2 \mu}=\frac{3 \times 5867}{2 \times 1.6}=5500 \hat{\lambda} $$

Hence, the correct answer is option: 'D'.

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