Q89 / 180 · AIIMS 2015 · Physics

In a certain double slit experimental arrangement, interference fringes of width $1 \mathrm{~mm}$ each are observed when light of wavelength $5000 Å$ is used. Keeping the setup unaltered, if the source is replaced by another of wavelength $6000 Å$, the fringe width will be

  • Correct answer
    A
    $1.2 \mathrm{~mm}$
  • B
    $0.5 \mathrm{~mm}$
  • C
    $1 \mathrm{~mm}$
  • D
    $1.5 \mathrm{~mm}$

Solution

Fringe width, $\beta=\frac{\lambda D}{d}$ Since $D$ and $d$ are unaltered, $\beta \propto \lambda$. $$ \begin{aligned} & \therefore \frac{\beta^{\prime}}{\beta}=\frac{\lambda^{\prime}}{\lambda} \\ & \text { or } \beta^{\prime}=\beta \times \frac{\lambda^{\prime}}{\lambda}=1 \times \frac{6000}{5000}=1.2 \mathrm{~mm} \end{aligned} $$

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

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