Question
In Young’s double $-$ slit experiment using monochromatic light of wavelength $\lambda, $ the intensities of two sources is $I.$ What is the intensity of light at a point where path difference between wave front is $\lambda / 4$ ?
✓
Answer
$\lambda_1=4 \times 10^{-7} m \lambda_2=6 \times 10^{-7} m$
Distance at which dark fringe is observed $x=\left(n+\frac{1}{2}\right) \frac{\lambda D}{d}$
First Dark fringe for $\lambda_1 d_1=\frac{1}{2} \frac{4 \times 10^{-7}}{10^{-2}} m=2 \times 10^{-5} m$
First Dark fringe for $\lambda_2 d_2=\frac{1}{2} \frac{6 \times 10^{-7}}{10^{-2}} m=3 \times 10^{-5} m$
First dark fringe will be the distance where both dark fringes will coincide
i.e $\text {LCM}$ of $d_1 \ d_1$ i.e.
$ 2 \times 10^{-5} m \times 3 \times 10^{-5} m$
$=6 \times 10^{-5} m$
Distance at which dark fringe is observed $x=\left(n+\frac{1}{2}\right) \frac{\lambda D}{d}$
First Dark fringe for $\lambda_1 d_1=\frac{1}{2} \frac{4 \times 10^{-7}}{10^{-2}} m=2 \times 10^{-5} m$
First Dark fringe for $\lambda_2 d_2=\frac{1}{2} \frac{6 \times 10^{-7}}{10^{-2}} m=3 \times 10^{-5} m$
First dark fringe will be the distance where both dark fringes will coincide
i.e $\text {LCM}$ of $d_1 \ d_1$ i.e.
$ 2 \times 10^{-5} m \times 3 \times 10^{-5} m$
$=6 \times 10^{-5} m$
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