MCQ
In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength fo the light used is $6280 \;\mathring A$. $...........\,mm$ is the distance between the two coherent sources.
  • $0.036$
  • B
    $0.12$
  • C
    $6$
  • D
    $4$

Answer

Correct option: A.
$0.036$
a
(a)

The angular fringe width is given by $\alpha=\frac{\lambda}{ d }$ where $\lambda$ is wavelength and $d$ is the distance between two coherent sources. Thus $d =\frac{\lambda}{\alpha}$

Given, $\lambda=6280 \mathring A, \alpha=1^{\circ}=\frac{\pi}{180}$ radian .

$d =\frac{6280 \times 10^{-10}}{3.14} \times 180$

$=3.6 \times 10^{-5}\,m =0.036\,mm$

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