In Young's double slit experiment, red light of wavelength $6000 Å$ is used and the $n$th bright fringe is…
In Young's double slit experiment, red light of wavelength $6000 Å$ 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 $5000 Å$ and now $(n+1)$ th bright fringe is obtained at the point $P$ on the screen. The value of $n$ is
$4$
$5$
$6$
$3$
Solution
Fringe width, $\beta_G=n \frac{\lambda D}{d}$
And for $n$ number of fringes,
$\beta_G=n \frac{\lambda D}{d} \text { or, } \beta_R=n \frac{\lambda D}{d}$
For $(n+1)$ th number of fringes,
$\begin{aligned}
& \beta_G=(n+1) \frac{\lambda D}{d} \\
& \because \beta_{\mathrm{R}}=\beta_{\mathrm{G}} \\
& \text { or, } n \frac{\lambda_R D}{d}=(n+1) \frac{\lambda_R D}{d} \\
& n .6000 \times 10^{-10} \frac{D}{d}=(n+1) 5000 \times 10^{-10} \frac{D}{d} \text { or, } \\
& \Rightarrow 6 n=(n+1) 5 \\
& \Rightarrow 6 n=5 n+5 \text { or, } 6 n-5 n=5 \\
& \therefore \quad n=5
\end{aligned}$
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