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The Young's double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The lower slit $S_{2}$ is covered by a thin glass sheet of thickness 10.4 $μ m$ and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in the figure.
Now, if $600 n m$ light is replaced by white light of range $400$ to $700 n m$ , find the wavelengths of the light that form maxima exactly at point $O$ .

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576.8 nm

$433.33 n m$

453 nm

$650 n m$

answer is A, B.

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Detailed Solution

At $O$ , path difference is $∆ x = ∆ x_{2} = (\frac{μ_{g}}{μ_{m}} - 1) t$

For maximum intensity at $O$ , we have

$∆ x = n λ$ , where $n = 1, 2, 3 . . .$

$\Rightarrow λ = \frac{∆ x}{1}, \frac{∆ x}{2}, \frac{∆ x}{3}, . . .$ and so on

$\Rightarrow ∆ x = (\frac{1.5}{4 / 3} - 1) (10.4 \times 10^{- 6} m) \Rightarrow ∆ x = (\frac{1.5}{4 / 3} - 1) (10.4 \times 10^{3} n m) = 1300 n m$

So, maximum intensity will be corresponding to

$λ = 1300 n m, \frac{1300}{2} n m, \frac{1300}{3} n m, \frac{1300}{4} n m, \Rightarrow λ = 1300 n m, 650 n m, 433.33 n m, 325 n m$

The wavelength in the range $400 n m$ to $700 n m$ are $650 n m$ and $433.33 n m$ .

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