JEE Advance - Physics (2014 - Paper 1 Offline - No. 9)
A light source, width emits two wavelengths $$\lambda$$1 = 400 nm and $$\lambda$$2 = 600 nm, is used in a Young's double-slit experiment. If recorded fringe widths for $$\lambda$$1 and $$\lambda$$2 are $$\beta$$1 and $$\beta$$2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then
$$\beta$$2 > $$\beta$$1
m1 > m2
from the central maximum, 3rd maximum of $$\lambda$$2 overlaps with 5th minimum of $$\lambda$$1
the angular separation of fringes of $$\lambda$$1 is greater than $$\lambda$$2
Explanation
Fringe width $$\beta = {{\lambda D} \over d}$$; $$\lambda$$1 = 400 nm; $$\lambda$$2 = 400 nm
Since $$\lambda$$2 > $$\lambda$$1, $$\beta$$2 > $$\beta$$1
Number of fringes within a distance y is given by
$${m_1} = {y \over {{\beta _1}}}$$
$${m_2} = {y \over {{\beta _2}}}$$
Since, $$\beta$$2 > $$\beta$$1, we get m2 < m1.
Position of 3rd maxima of $$\lambda$$2 :
$$y' = {D \over d}(3{\lambda _2}) = 1800{D \over d}$$
Position of 5th minima of $$\lambda$$1 :
$$y'' = {D \over d}\left( {{{9{\lambda _1}} \over 2}} \right) = 1800{D \over d}$$
Hence, y' = y''.
Angular fringe width is $$\theta = {\beta \over D} = {\lambda \over d}$$
Since, $$\beta$$2 > $$\beta$$1 $$\Rightarrow$$ $$\theta$$2 > $$\theta$$1
Since $$\lambda$$2 > $$\lambda$$1, $$\beta$$2 > $$\beta$$1
Number of fringes within a distance y is given by
$${m_1} = {y \over {{\beta _1}}}$$
$${m_2} = {y \over {{\beta _2}}}$$
Since, $$\beta$$2 > $$\beta$$1, we get m2 < m1.
Position of 3rd maxima of $$\lambda$$2 :
$$y' = {D \over d}(3{\lambda _2}) = 1800{D \over d}$$
Position of 5th minima of $$\lambda$$1 :
$$y'' = {D \over d}\left( {{{9{\lambda _1}} \over 2}} \right) = 1800{D \over d}$$
Hence, y' = y''.
Angular fringe width is $$\theta = {\beta \over D} = {\lambda \over d}$$
Since, $$\beta$$2 > $$\beta$$1 $$\Rightarrow$$ $$\theta$$2 > $$\theta$$1
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