Question

In Young's double slit experiment, red light of wavelength $6000 \,\mathring{A}$ is used and the $n\, th$ bright fringe is obtained at a point $P$ on the screen. Keeping the same setting, the source of light is replaced by green light of wavelength $5000 \,\mathring{A}$ and now $(n+1)$ th bright fringe is obtained at the point $P$ on the screen. The value of $n$ is

• A. 4
• B. 5
• C. 6
• D. 3

Answer 1

Answer: (B)

Given, wavelength of red light
$=6000 \,\mathring{A}=6000 \times 10^{-10} \,m$
Wavelength of green light
$=5000\,\mathring{A}=5000 \times 10^{-10}\, m$
Fringe width, $\beta=\frac{\lambda D}{d}$
For $n$ number of fringes,
$\beta_{n}=n \frac{\lambda D}{d}$
For $(n+1)$ th number of fringes,
$\beta_{G}=(n+1) \frac{\lambda D}{d}$
According to the question,
$\beta_{R}=\beta_{G}$
$n \cdot \frac{\lambda_{R} D}{d}=(n+1) \frac{\lambda_{G} D}{d} $
$n .6000 \times 10^{-10} \frac{D}{d}=(n+1) 5000 \times 10^{-10} \frac{D}{d} $or ,
$6 n=(n+1) 5 $
or, $6 n=5 n+5$
or, $6 n-5 n=5$ or $n=5$