Q. In a Young's double slit experiment, angular width of a fringe formed on a distant screen is $0.1^{\circ}$. If wavelength of light used is $6000 \,\mathring{A}$, then distance between the slits will be :

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A

$0.241 \,mm$

B

$0.344 \,mm$

C

$0.519 \,mm$

D

$0.413 \,mm$

Solution:

In YDSE setup angular width of fringes on screen is given as
$ \theta=\frac{\lambda}{d} $
$\Rightarrow d =\frac{6 \times 10^{-7} \times 180 \times 7}{0.1 \times 22}=0.344\, mm .$

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