Question

In a Young's double slit experiment, the slits are $2mm$ apart and are illuminated with a mixture of two wavelength ${\lambda }_0=750nm$ and $\lambda =900nm$ . The minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

• A. $1.5mm$
• B. $3mm$
• C. $4.5mm$
• D. $6mm$

Answer 1

Answer: (C)

$4.5mm$

From the given data, note that the fringe width $\left({\beta }_1\right)$ for ${\lambda }_{1}900nm$ is greater than fringe width $\left({\beta }_2\right)$ for ${\lambda }_2=750nm$. This means that at though the central maxima of the two coincide, but first maximum for ${\lambda }_1=900nm$ will be further away from the first maxima for ${\lambda }_{2}750nm$ and so on. A stage may come when this mismatch equals ${\beta }_2$, then again maxima of ${\lambda }_1=900nm$ will coincide with a maxima of ${\lambda }_2=750nm$ let this correspond to nth order fringe for ${\lambda }_1$.Then it will correspond to $(n+1{)}^{th}$ order fringe for ${\lambda }_2$.
Therefore $\frac{n{\lambda }_{1}D}{d}=\frac{(n+1){\lambda }_{2}D}{d}$
$\Rightarrow n\times 900\times {10}^{-9}=(n+)750\times {10}^{-9}\Rightarrow n=5$
Minimum distance from
Central maxima $=\frac{n{\lambda }_{1}D}{d}=\frac{5\times 900\times {10}^{-9}\times 2}{2\times {10}^{-3}}$
$=45\times {10}^{-4}m=4.5mm$