Question

Monochromatic light of wavelength $400nm$ and $560nm$ are incident simultaneously and normally on double slits apparatus whose slit separation is $0.1mm$ and the screen distance is $1m$ . Distance between areas of total darkness will be:
(A) $4mm$
(B) $5.6mm$
(C) $14mm$
(D) $28mm$

Answer 1

Hint : Use the condition for minima and find the number of the dark bands for which both the lights have a minima to find the difference between the consecutive dark bands.
The minima condition is given by, path difference = $\left( {2m + 1} \right)\dfrac{\lambda }{2}$ where, $\lambda$ is the wavelength of the light and $m = 0,1,2,3...$ The separation between two minima bands ${n_1}$ and ${n_2}$ is given by, \Delta s = \dfrac{{D\lambda }}{d}\left\\{ {\dfrac{{\left( {2{n_2} + 1} \right) - \left( {2{n_1} + 1} \right)}}{2}} \right\\} where, $d$ is the separation of slits $D$ is the screen distance.

Complete Step By Step Answer:
Here we have two monochromatic sources. Now, both the sources will have minima pattern when their path difference is same for the minima condition, Hence, we can write, $\dfrac{{\left( {2n + 1} \right)}}{2}{\lambda _1} = \dfrac{{\left( {2m + 1} \right)}}{2}{\lambda _2}$ where, ${\lambda _1}$ is the wavelength of the first source and $n$ is the minima number ${\lambda _2}$ is the wavelength of the second source and $m$ is the minima number of its pattern.
Putting the ${\lambda _1} = 400nm = 400 \times {10^{ - 9}}$ and ${\lambda _2} = 560nm = 560 \times {10^{ - 9}}$ we get,
$\dfrac{{\left( {2n + 1} \right)}}{{\left( {2m + 1} \right)}} = \dfrac{{560}}{{400}} = \dfrac{7}{5}$
Therefore simplifying we get,
$10n = 14m + 2$
Now, we have to find the $n$ and $m$ values from this equation by observation,
So, we can see if we put, $n = 3$
$10 \times 3 = 14m + 2$
Or, $14m = 30 - 2 = 28$
Or, $m = 2$
Now, we have found the next overlapping minima for both the sources. Now, if we put,
$n = 10$
$10 \times 10 = 14m + 2$
Or, $14m = 100 - 2 = 98$
Or, $m = 7$
Now we know, the separation between two minima bands ${n_1}$ and ${n_2}$ is given by, \Delta s = \dfrac{{D\lambda }}{d}\left\\{ {\dfrac{{\left( {2{n_2} + 1} \right) - \left( {2{n_1} + 1} \right)}}{2}} \right\\} where, $d$ is the separation of slits $D$ is the screen distance
we found that for the first source at $n = 3$ and $n = 10$ we get over lapping dark bands. Hence putting the values of slit separation $d = 0.1 \times {10^{ - 3}}m$ , screen distance $D = 1m$ , consecutive minima numbers for which both the sources have minima condition ${n_2} = 10$ ${n_1} = 3$ in \Delta s = \dfrac{{D{\lambda _1}}}{d}\left\\{ {\dfrac{{\left( {2{n_2} + 1} \right) - \left( {2{n_1} + 1} \right)}}{2}} \right\\} we get,
\Delta s = \dfrac{{1 \times 400 \times {{10}^{ - 9}}}}{{0.1 \times {{10}^{ - 3}}}}\left\\{ {\dfrac{{\left( {2 \times 10 + 1} \right) - \left( {2 \times 3 + 1} \right)}}{2}} \right\\}
\Rightarrow \Delta s = \dfrac{{1 \times 400 \times {{10}^{ - 9}}}}{{0.1 \times {{10}^{ - 3}}}}\left\\{ {\dfrac{{\left( {21} \right) - \left( 7 \right)}}{2}} \right\\}
Calculating we get,
\Rightarrow \Delta s = 400 \times {10^{ - 5}}\left\\{ {\dfrac{{14}}{2}} \right\\}
$\Rightarrow \Delta s = 400 \times {10^{ - 5}} \times 7$
Hence,
$\Rightarrow \Delta s = 28 \times {10^{ - 3}}$
Therefore, the separation between complete darkness is $28 \times {10^{ - 3}}m$ or $28mm$ .
Hence, option (D) is correct.

Note :
We can also find the solution by finding the fringe width or separation between two consecutive dark bands for each of the sources using fringe width, $\beta = \dfrac{{\lambda D}}{d}$ and find the L.C.M of the values fringe width to find the separation between complete darkness.