Q85 / 120 · VITEEE 2010 · Physics

In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 Å$. If the angular width of a fringe formed on a distance screen is $1^{\circ}$, then value of $d$. is

  • A
    $1 \mathrm{~mm}$
  • B
    $0.05 \mathrm{~mm}$
  • Correct answer
    C
    $0.03 \mathrm{~mm}$
  • D
    $0.01 \mathrm{~mm}$

Solution

Here, $\sin \theta=\left(\frac{Y}{D}\right) \quad \therefore \Delta \theta=\frac{\Delta Y}{D}$ Angular fringe width $\theta_{0}=\Delta \theta$ $$ \text { (width } \Delta \mathrm{Y}=\beta \text { ) } $$ $$ \begin{array}{l} \theta_{0}=1^{\circ}=\frac{\pi}{180} \mathrm{rad} \text { and } \lambda=6 \times 10^{-7} \mathrm{~m} \\ d=\frac{\lambda}{\theta_{0}}=\frac{180}{\pi} \times 6 \times 10^{-7}=0.03 \mathrm{~mm} \end{array} $$

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

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