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  • In a Young's double slit experiment, the slits are apart and are illuminated with a m

In a Young's double slit experiment, the slits are 2 \mathrm{~mm} apart and are illuminated with a mixture of two wavelength \mathrm{\lambda_0=750 \mathrm{~nm}} and \mathrm{ \lambda=900 \mathrm{~nm}. }The minimum distance from the common central bright fringe on a screen \mathrm{ 2 m} from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

Option: 1

1.5 mm


Option: 2

3 mm


Option: 3

4.5 mm


Option: 4

6 mm


Answers (1)

best_answer

From the given data, note that the fringe width \left(\beta_1\right) for \mathrm{\lambda_1=900 \mathrm{~nm}} is greater than fringe width \mathrm{\left(\beta_2\right)} for

\mathrm{\lambda_2=750 \mathrm{~nm}}. This means that at though the central maxima of the two coincide, but first maximum for \mathrm{\lambda_1=900 \mathrm{~nm}}

will be further away from the first maxima for \mathrm{\lambda_2=750 \mathrm{~nm},} and so on. A stage may come when this mismatch equals \mathrm{\beta_2},

then again maxima of \mathrm{\lambda_1=900 \mathrm{~nm}}, will coincide with a maxima of \mathrm{\lambda_2=750 \mathrm{~nm}}, let this correspond to \mathrm{n^{\text {th }}} order

fringe for \mathrm{\lambda_1.} Then it will correspond to \mathrm{(n+1)^{\text {th }}} order fringe for  \mathrm{\lambda_2.}

Therefore \mathrm{\frac{n \lambda_1 D}{d}=\frac{(n+1) \lambda_2 D}{d} \Rightarrow n \times 900 \times 10^{-9}=(n+1) 750 \times 10^{-9} \Rightarrow n=5}

Minimum distance from

Central maxima \mathrm{=\frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}=45 \times 10^{-4} \mathrm{~m}=4.5 \mathrm{~mm}}

Posted by

Gautam harsolia

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