Question

In a Young's double slit experiment the slit is illuminated by a source having two wavelengths of $400\text{nm}$ and $600\text{nm}.$ If distance between slits, $d=1\text{nm},$ and distance between the plane of the slit and screen, $D=10\text{m}$ the smallest distance from the central maximum where there is complete darkness is-

• A. $2\text{mm}$
• B. $3\text{mm}$
• C. $12\text{mm}$
• D. There is no such point

Answer 1

The correct option is D There is no such point

Given,

${\lambda }_1=400\text{nm};{\lambda }_2=600nmd=1\text{mm};D=10\text{m}$

For destructive interference of both wavelengths, $n^{th}$ minima of ${\lambda }_1$ coinside with $m^{th}$ minima of ${\lambda }_2.$

$n\times 400=m\times 600$

$n=\frac{3m}{2}$

$Y=\frac{(2n+1)\lambda D}{2d}$

For both to be coinciding,

$(2n+1){\lambda }_1=(2m+1){\lambda }_2$

$(2n+1)400=(2m+1)600$

$8n+4=12m+6$

$8n=12m+2$

$n=\frac{3}{2}m+\frac{1}{4}$

Here we can see for any values of $m$ and $n$ there is no. integer. Hence such points can't be found.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.