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In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda$ ₁ = 12000 ${\AA}$ and $\lambda$ ₂ = 10000 $\AA$ . At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
Option: 1
3 mm

Option: 2
8 mm

Option: 3
6 mm

Option: 4
4 mm

Answers (1)

best_answer

As we learnt in

Fringe Width -

$eta = frac{lambda D}{d}$

- wherein

$eta = y_{n+1}-y_{n}$

$y_{n+1}=$ Distance of $left ( n+1 ight )^{th}$

Maxima $= left ( n+1 ight )frac{lambda D}{d}$

$y_{n}=$ Distance of $n^{th}$

maxima $= frac{nlambda D}{d}$

$\lambda _{1}=12000A^{\circ}$

$\lambda _{2}=10000A^{\circ}$

let m^th bright fringe of first will concide with n^th bright fringe of other

$\therefore m\beta _{1}=n.\beta _{2}$

$or\:m.\frac{\lambda _{1}D}{d}= n\frac{\lambda _{2}D}{d}$

$or\:m\lambda _{1}=n\lambda _{2}$

$or\:\frac{m}{n}=\frac{10000}{12000}=\frac{5}{6}$

$\therefore m=5,10,15,----$

$\therefore n=6,12,18,----$

minimum Value of m=5 & n= 6

$\therefore$ Minimum distance of such point from Central

Maxima $=\frac{5\lambda _{1}D}{d}$ $=\frac{5\times 12000\times 10^{-10}\times 2}{2\times 10^{-3}}$

$=6mm$

Posted by

Gaurav

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