Q177 / 180 · KCET 2009 · Physics

Young's double slit experiment gives interference fringes of width $0.3 \mathrm{~mm}$. A thin glass plate made of material of refractive index $1.5$ is kept in the path of light from one of the slits, then the fringe width becomes

  • A
    zero
  • Correct answer
    B
    $0.3 \mathrm{~mm}$
  • C
    $0.45 \mathrm{~mm}$
  • D
    $0.15 \mathrm{~mm}$

Solution

When a this glass plate of thickness $t$ and refractive index $\mu$ is introduced in one of the paths of the interfering waves then the path increases by $(\mu-1) \mathrm{t}$ and the whole pattern shifts by $$ \mathrm{y}_{0}=\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t} $$ Shifting is towards the side in which the plate is introduced without any change in fringe width. Therefore, when a glass plate of refractive index $1.5$ is kept in the path of light from one of the slits, only the fringes get shifted but the fringe width remains unchanged.

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

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