In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m the minimum distance between the slits $S_{1}$ and $S_{2}$ is ____ mm. [2024]

Question

Smart Achievers · Accepted Answer

(0.20)

Path difference for minima at P

$2 \sqrt{D^{2} + d^{2}} - 2 D = \frac{λ}{2}$

$∴$ $\sqrt{D^{2} + d^{2}} - D = \frac{λ}{4}$

$∴$ $\sqrt{D^{2} + d^{2}} = \frac{λ}{4} + D$

$\Rightarrow D^{2} + d^{2} = D^{2} + \frac{λ^{2}}{16} + \frac{D λ}{2}$

$\Rightarrow d^{2} = \frac{D λ}{2} + \frac{λ^{2}}{16}$

$= \frac{0.2 \times 400 \times 10^{- 9}}{2} + \frac{4 \times 10^{- 14}}{4} \Rightarrow d^{2} \approx 400 \times 10^{- 10}$

$∴$ $d = 20 \times 10^{- 5} \Rightarrow d = 0.20 mm$

Solution

Q.
In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m the minimum distance between the slits $S_{1}$ and $S_{2}$ is ____ mm. [2024]

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Solution

Q. In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m the minimum distance between the slits S1 and S2 is ____ mm. [2024]

Ans. (0.20) Path difference for minima at P 2D2+d2-2D=λ2 ∴ D2+d2-D=λ4 ∴ D2+d2=λ4+D ⇒D2+d2=D2+λ216+Dλ2 ⇒d2=Dλ2+λ216 =0.2×400×10-92+4×10-144⇒d2≈400×10-10 ∴ d=20×10-5⇒d=0.20 mm

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Q.
In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m the minimum distance between the slits $S_{1}$ and $S_{2}$ is ____ mm. [2024]