Question

In a Young's double slit experiment, $d=0.5\, mm$ and $D=100 \,cm $ It is found that $9^{\text {th }}$ bright fringe is at a distance of $7.5 mm$ from the second dark fringe of fringe pattern. The wavelength of light used is (in $\mathring{A})$

• A. $\frac{2500}{7}$
• B. 2500
• C. 5000
• D. $\frac{5000}{7}$

Answer 1

Answer: (C)

For bright fringes, $x=\frac{n \lambda D}{d}$
where $n=0,1,2,3, \dots$
For dark fringes, $x=\frac{(2 n-1) \lambda D}{2 d}$
where $n=1,2,3, \dots$
As per question, $\frac{9 \lambda D}{d}-\frac{3 \lambda D}{2 d}=7.5 \times 10^{-3} m$
or $\frac{15 \lambda D}{2 d}=7.5 \times 10^{-3} m$ or $\lambda=\frac{2 \times 7.5 \times 10^{-3} \times d}{15 D}$
Substituting the given values, we get
$\lambda=\frac{2 \times 7.5 \times 10^{-3} m \times 0.5 \times 10^{-3} m }{15 \times 1 m }$
$=0.5 \times 10^{-6} m =5000 \, \mathring{A}$