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.