A Young's double slit interference arrangement with slits and is immersed in water (refractive index ) as shown in the figure. The positions of maximum on the surface of water are given by where is the wavelength of light in air (refractive index ),

Question

A Young's double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $= \frac{4}{3}$ ) as shown in the figure.

The positions of maximum on the surface of water are given by $x^{2} = p^{2} m^{2} λ^{2} - d^{2},$ where $λ$ is the wavelength of light in air (refractive index $= 1$ ), $2 d$ is the separation between the slits and $m$ is an integer. The value of $p$ is [2015]

Smart Achievers · Accepted Answer

(3)

For maxima, Path difference $= m λ$

$∴$ $S_{2} A - S_{1} A = m λ$

$∴$ $[(n - 1) \sqrt{d^{2} + x^{2}} + \sqrt{d^{2} + x^{2}}] - \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $(n - 1) \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $(\frac{4}{3} - 1) \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $\sqrt{d^{2} + x^{2}} = 3 m λ$

$∴$ $d^{2} + x^{2} = 9 m^{2} λ^{2}$

$∴$ $x^{2} = 9 m^{2} λ^{2} - d^{2}$

Comparing this equation with the given equation $x^{2} = p^{2} m^{2} λ^{2} - d^{2},$

we get $p^{2} = 9$

$∴$ $p = 3$

Solution

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Solution

Ans. (3) For maxima, Path difference =mλ ∴ S2A-S1A=mλ ∴ [(n-1)d2+x2+d2+x2]-d2+x2=mλ ∴ (n-1)d2+x2=mλ ∴ (43-1)d2+x2=mλ ∴ d2+x2=3mλ ∴ d2+x2=9m2λ2 ∴ x2=9m2λ2-d2 Comparing this equation with the given equation x2=p2m2λ2-d2, we get p2=9 ∴ p=3

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