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Young's double-slit experiment is carried with two thin sheets of thickness $$10.4\mu m$$ 4ach and refractive index $${ \mu  }_{ 1 }=1.52$$ and $${ \mu  }_{ 2 }=1.40$$ covering the slits $${S}_{1}$$ and $${S}_{2}$$, respectively. If white light of range $$400nm$$ to $$780nm$$ is used, then which wavelength will form maxima exactly at point $$O$$, the centre of the screen?

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A
416nm only
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B
624nm only
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C
416nm and 624nm only
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D
None of these
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Solution

The correct option is B $$416nm$$ and $$624nm$$ only
For maxima at center path difference should be an integer multiple of $$\lambda$$
$$(\mu_1-1)t-(\mu_2-1)t- \dfrac{yd}{D}= n\lambda$$ given y=0
$$(\mu_1-\mu_2)t= n\lambda$$ 
$$.12\times 10.4 \mu m=n\lambda$$ 
$$1248\; nm =n\lambda$$  
Checking option A $$\lambda=416 nm $$  

$$\Rightarrow n=\dfrac{1248 nm}{416 nm}=3 $$ hence, it will have maxima at center
Checking option B $$\lambda=624 nm $$ 

$$\Rightarrow n=\dfrac{1248 nm}{624nm}=2 $$ hence, it will have maxima at center
the correct answer is option C.

Physics
NCERT
Standard XII

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