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Updated on : 2022-09-05

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The wavelength of the light is $λ_{1}=650nm$

The wavelength of seond light,$λ_{2}=520nm$

Distance between the slit and the screen is $1.4m$.

Distance between the slits is $0.28mm$.

$(a)$

The relation between the $n_{th}$ bright fringe and the width of fringe is:

$x=nλ_{1}dD $

For third bright fringe, $n=3$

$x=3×6500.28×10_{−3}1.4 =1950×5×10_{3}nm$

$x=9.75×10_{−3}m$

$=9.75mm$

$(b)$

We can consider that $n_{th}$ bright frind of $λ_{2}$ and the $(n−1)_{th}$ bright fringe of wavelength $λ_{1}$ coincide with each other.

$nλ_{2}=(n−1)λ_{1}$

$520n=650n−650$

$650=130n$

$n=5$

therefore, the least distance from the central maximum can be obtained as:

$x_{′}=nλ_{2}dD $

$x_{′}=5×520dD =26000.28×10_{−3}1.4 nm$

$x_{′}=1.30×10_{−2}m=1.3cm$

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