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In a young's double slit experiment with slit sepration 0.1mm, one observes a bright fringe at angle $\frac{1}{40}$ rad by using light of wavelength $\lambda _{1}$ . When the light of wavelength $\lambda _{2}$ is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda _{1}$ and $\lambda _{2}$ are in visible range (380nm to 740nm), their values are:

Option 1)
625 nm, 500nm
Option 2)
380 nm, 525 nm
Option 3)
380 nm, 500 nm
Option 4)
400 nm, 500 nm

Answers (1)

best_answer

Young Double Slit Experiment -

- wherein

$y=\Delta x\cdot \left ( \frac{D}{d} \right )$

$y=$ Distance of a point on screen from central maxima

$\Delta x=$ Path difference at that point

$path \: \:difference = d\: \: sin\theta \approx d\times \Theta$

$d=0.1, \: \: \Theta =\frac{1}{40}$ $so\: \: path\: \: difference =0.1 \times \frac{1}{40}=2500nm$

$for\: \: bright\: \: fringe$

$path\: \: difference\: \: is\: \: multiple \: of \: \lambda$

$for \: \lambda _{1}$

$2500=n\lambda _{1}$

$for\: \: \lambda _{2}$

$2500=m\lambda _{2}$

$\Rightarrow 2500=n\lambda _{1}=m\lambda _{2}$

$for\: n=4,\lambda _{1}=625nm$

$for \: \: m=5, \lambda _{2}=500nm$

Option 1)
625 nm, 500nm
Option 2)

380 nm, 525 nm

Option 3)

380 nm, 500 nm

Option 4)

400 nm, 500 nm

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