Question

In a Young's double slit experiment, (slit distance $d$) monochromatic light of wavelength $\lambda$ is used and the figure pattern observed at a distance $L$ from the slits. The angular position of the bright fringes is

• A. ${\sin }^{-1}\left(\begin{array}{c}\frac{N\lambda }{d}\end{array}\right)$
• B. ${\sin }^{-1}\left(\begin{array}{c}\frac{(N+\frac{1}{2})\lambda }{d}\end{array}\right)$
• C. ${\sin }^{-1}\left(\begin{array}{c}\frac{N\lambda }{L}\end{array}\right)$
• D. ${\sin }^{-1}\left(\begin{array}{c}\frac{(N+\frac{1}{2})\lambda }{L}\end{array}\right)$

Answer 1

Answer: (A)

${\sin }^{-1}\left(\begin{array}{c}\frac{N\lambda }{d}\end{array}\right)$

The path difference of light coming from two slits at height $y$ from the central maxima $=\frac{yd}{L}$

Hence for the bright fringe to form,

$\frac{yd}{L}=N\lambda$

Hence $\frac{y}{L}=\frac{N\lambda }{d}$

The angular postion of bright fringe is $\theta =si{n}^{-1}\frac{y}{L}$ $=si{n}^{-1}\left(\frac{N\lambda }{d}\right)$