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In Young's double slit experiment, the distance between the slits is $3 \mathrm{~mm}$ and the slits are $2 \mathrm{~m}$ away from the screen. Two interference patterns can be obtained on
the screen due to light of wavelength $480 \mathrm{~nm}$ and $600 \mathrm{~nm}$ respectively. The
separation on the screen between the $5^{\text {th }}$ order bright fringes on the two
interference patterns is
Options:
the screen due to light of wavelength $480 \mathrm{~nm}$ and $600 \mathrm{~nm}$ respectively. The
separation on the screen between the $5^{\text {th }}$ order bright fringes on the two
interference patterns is
Solution:
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Verified Answer
The correct answer is:
$4 \times 10^{-4} \mathrm{~m}$
$\mathrm{d}=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$
$D=1 \mathrm{~m} \quad \lambda_{1}=480 \mathrm{~nm} \quad \lambda_{2}=600 \mathrm{~nm}$
${ }^{1} \mathrm{y}_{5}=5 \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}$
${ }^{2} \mathrm{y}_{5}=5 \frac{\lambda_{2} \mathrm{D}}{\mathrm{d}}$
${ }^{2} y_{5}-{ }^{1} y_{5}=\frac{5 D}{d}(600-480) n m$
$=\frac{5 \times 2}{3 \times 10^{-3}} \times 120$
$=10 \times 10^{-3+1} \times 4=4 \times 10^{5-9}$
$=4 \times 10^{-4}=0.4 \mathrm{~mm}$
$D=1 \mathrm{~m} \quad \lambda_{1}=480 \mathrm{~nm} \quad \lambda_{2}=600 \mathrm{~nm}$
${ }^{1} \mathrm{y}_{5}=5 \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}$
${ }^{2} \mathrm{y}_{5}=5 \frac{\lambda_{2} \mathrm{D}}{\mathrm{d}}$
${ }^{2} y_{5}-{ }^{1} y_{5}=\frac{5 D}{d}(600-480) n m$
$=\frac{5 \times 2}{3 \times 10^{-3}} \times 120$
$=10 \times 10^{-3+1} \times 4=4 \times 10^{5-9}$
$=4 \times 10^{-4}=0.4 \mathrm{~mm}$
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