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

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Wave Optics

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}

____

mm.

Answer: 0.2

Solution

For a dark fringe at P, the path difference from the two slits equals

λ /2

(measured via the extra geometric path in this two-source arrangement):

2 D^{2} + d^{2} - 2 D = \frac{λ}{2} ⟹ D^{2} + d^{2} = D + \frac{λ}{4}

Squaring and neglecting the tiny

λ^{2} /16

term:

D^{2} + d^{2} \approx D^{2} + \frac{D λ}{2} ⟹ d^{2} = \frac{D λ}{2}

With

D = 0.2 m

λ = 400 nm = 4 \times 1 0^{- 7} m

d^{2} = \frac{( 0.2 ) ( 4 \times 1 0 ^{- 7} )}{2} = 4 \times 1 0^{- 8} m^{2} ⟹ d = 2 \times 1 0^{- 4} m = 0.20 mm

Key confirmed: 0.20 mm.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Numerical answer

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