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Question

Two ideal slits $$S _ { 1 }$$ and $$S _ { 2 }$$ are at a distance $$d$$ apart and illuminated by light of wavelength $$\lambda$$ passing through an ideal source slit $$S$$ placed on the line through $$S _ { 2 }$$ as shown. The distance between the planes of slits and the source slit is $$D$$ . A screen is held at a distance $$D$$ from the plane of the slots. The minimum value of $$d$$ for which there is darkness at $$O$$ is
1378290_d77411035ee74765b4795b4f43890908.png


A
3λD2
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B
λD
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C
λD2
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D
3λD
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Solution

The correct option is C $$\sqrt { \dfrac { \lambda D } { 2 } }$$

Path difference between the waves reaching at $$P, \Delta=\Delta_{1}+\Delta_{2}$$

where $$\Delta_{1}=$$ initial path difference $$\Delta_{2}=$$ Path difference between the waves after emerging from slits.

$$\Delta_{1}=SS_{1}-SS_{2}=\sqrt{D^{2}+d^{2}}-D$$
and $$\Delta_{2}=S_{1}O-S_{2}O=\sqrt{D^{2}+d^{2}}-D$$

$$\therefore \Delta=2\left\{(D^{2}+d^{2})^{\dfrac{1}{2}}-D\right\}=2\left\{\left(D^{2}+\dfrac{d^{2}}{2D}\right)-D\right\}$$
$$=\dfrac{d^{2}}{D}$$   (From Binomial expansion)

For obtaining dark at $$O, \Delta$$ must be equals to $$(2n-1)\dfrac{\lambda}{2}$$ i.e.
$$\dfrac{d^{2}}{D}=(2n-1)\dfrac{\lambda}{2}\Rightarrow d\sqrt{\dfrac{(2n-1)\lambda D}{2}}$$
For minimum distance $$n-1$$ so $$d=\sqrt{\dfrac{\lambda D}{2}}$$

1760754_1378290_ans_774d6717e4f241e1907b3228030746ec.png

Physics

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