In YDSE, d=1.5 mm and D=1.2 m. If white light is incident on the slits ( wavelength range 4000A0 TO 8000A0) then the wavelength in A0 that absent in front of one of the slit is

Path difference in front of the slit is D2+d2-D=2n-1λ2 to have minimum thereD1+d22D2-D=2n-1λ2d22D=λ22n-1λ=1.5×10-321.22n-1

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In YDSE, d=1.5 mm and D=1.2 m. If white light is incident on the slits ( wavelength range $4000 \overset{0}{A}$ TO $8000 \overset{0}{A}$ ) then the wavelength in $\overset{0}{A}$ that absent in front of one of the slit is

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Detailed Solution

$Path difference in front of the slit is \sqrt{D^{2} + d^{2}} - D = (2 n - 1) \frac{λ}{2} to have minimum there$

$D (1 + \frac{d^{2}}{2 D^{2}}) - D = (2 n - 1) \frac{λ}{2}$

$\frac{d^{2}}{2 D} = \frac{λ}{2} (2 n - 1)$

$λ = \frac{{(1.5 \times 10^{- 3})}^{2}}{1.2 (2 n - 1)}$

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