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, $\Delta x=(2 n-1) \frac{\lambda}{2}$
$\Rightarrow \,\,\, 57000 \mathring{A}=(2 n-1) \times \frac{6000 \mathring{A}}{2}$
$\Rightarrow \,\,\, 2 n-1=2 \times \frac{57}{6}=\frac{57}{3}=19$
$\Rightarrow \,\,\, 2 n-1=19$
or $2 n=20$ or $n=10$
Thus, $10\,th$ dark fringe is formed.