If the first minima in a young's slit experiment occurs directly in front of one of the slits, (distance between slits and screen D = 12 cm and distance between slits d = 5 cm) then the wavelength of the radiation used can be:

2 cm

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4 cm

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\frac{2}{3} c m

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\frac{4}{3} c m

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Solution

The correct options are
A 2 cm
C $\frac{2}{3} c m$
Path difference = $\sqrt{D^{2} + d^{2}} - D = 1$ cm

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

$\Rightarrow= \frac{2 (1)}{2 n - 1}$

$λ = 2 c m, \frac{2}{3} c m, \frac{2}{5} c m$

Hence (A), (C)

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If the first minima in a young's slit experiment occurs directly in front of one of the slits, (distance between slits and screen D = 12 cm and distance between slits d = 5 cm) then the wavelength of the radiation used can be:

The correct options are A 2 cm C 23 cmPath difference = √D2+d2−D=1 cm Also ; (√D2+d2−D)=(2n−1)λ2 ⇒=2(1)2n−1 λ=2cm,23 cm,25 cm Hence (A), (C)

The correct options are
A 2 cm
C $\frac{2}{3} c m$
Path difference = $\sqrt{D^{2} + d^{2}} - D = 1$ cm

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

$\Rightarrow= \frac{2 (1)}{2 n - 1}$

$λ = 2 c m, \frac{2}{3} c m, \frac{2}{5} c m$

Hence (A), (C)