JEE Advance - Physics (2015 - Paper 1 Offline - No. 7)
A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x2 = p2m2$$\lambda$$2 $$-$$ d2, where $$\lambda$$ is the wavelength of light in air (refractive index = 1). 2d is the separation between the slits and m is an integer. The value of p is
Answer
3
Explanation
$$\mu ({S_2}P) - {S_1}P = m\lambda $$
$$ \Rightarrow \mu \sqrt {{d^2} + {x^2}} - \sqrt {{d^2} + {x^2}} = m\lambda $$
$$ \Rightarrow (\mu - 1)\sqrt {{d^2} + {x^2}} = m\lambda $$
$$ \Rightarrow \left( {{4 \over 3} - 1} \right)\sqrt {{d^2} + {x^2}} = m\lambda $$
or, $$\sqrt {{d^2} + {x^2}} = 3m\lambda $$
Squaring this equation we get,
$${x^2} = 9{m^2}{\lambda ^2} - {d^2}$$
$$ \Rightarrow {p^2} = 9$$ or $$p = 3$$
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