Question

The distance of a point on the screen from two slits in biprism experiment is $1.8\times 10^{-5}m$ and $1.23\times 10^{-5}m$. If wavelength of light used is $6000\mathring{A}$, the fringe formed at that point is

• A. $10^{th}$ bright
• B. $10^{th}$ dark
• C. $9^{th}$ bright
• D. $9^{th}$ dark

Answer 1

Answer: (B)

Path difference between light rays coming from two slits,
$\Delta x=1.8\times 10^{-5}-1.23\times 10^{-5}$
$=10^{-5}(1.8-1.23)=0.57\times 10^{-5}=57\times 10^{-7}$
$=57000\times 10^{-10}=57000\mathring{A}$
It is not an integral multiple of wavelength. Thus, only dark fringes are possible. For $n$ th dark fringe, $\Delta x=(2n-1)\frac{\lambda }{2}$
$\Rightarrow 57000\mathring{A}=(2n-1)\times \frac{6000\mathring{A}}{2}$
$\Rightarrow 2n-1=2\times \frac{57}{6}=\frac{57}{3}=19$
$\Rightarrow 2n-1=19$
or $2n=20$ or $n=10$
Thus, $10th$ dark fringe is formed.