Question

In Young's double slit experiment, red light of wavelength $6000\mathring{A}$ is used and the $nth$ 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 ,
$6n=(n+1)5$
or, $6n=5n+5$
or, $6n-5n=5$ or $n=5$