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Question: If a transparent medium of refractive index \(1.5\) and thickness \(t=2.5\times {{10}^{-5}}m\) is in...

If a transparent medium of refractive index 1.51.5 and thickness t=2.5×10−5mt=2.5\times {{10}^{-5}}m is inserted in front of the slit’s of Young’s double slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5mm0.5mm and that between slits and screen is 100cm100cm:
(A). 5cm5cm
(B). 2.5cm2.5cm
(C). 0.25cm0.25cm
(D). 0.1cm0.1cm

Explanation

Solution

In Young’s double slit experiment, two slits are made and light from coherent sources are made to pass through it and observed on a screen. An interference pattern is obtained with equally spaced bright and dark fringes. When a medium is introduced in front of the slits, the interference pattern shifts and the shift depends on the distance between slit and screen, distance between slits, thickness and refractive index of the medium. Substituting given values in the relation, we can calculate the shift.
Formulas used:
s=D(μ−1)tds=\dfrac{D(\mu -1)t}{d}

Complete step-by-step solution:
In Young’s double slit experiment, light from coherent sources and constant phase difference is made to pass through very small slits and a pattern of continuous bright and dark fringes is obtained on the screen. When a medium with definite refractive index is introduced in front of the slits, the shift that takes place in the interference pattern is given by-
s=D(μ−1)tds=\dfrac{D(\mu -1)t}{d}
Here, ss is the shift in interference pattern
DD is the distance between the screen and slit
μ\mu is the refractive index of the medium
tt is the thickness
dd is the distance between the slits
Given, μ=1.5\mu =1.5, t=2.5×10−5mt=2.5\times {{10}^{-5}}m, d=0.5mmd=0.5mm, D=100cmD=100cm
In the above equation, substituting given values, we get,
s=100×10−2(1.5−1)×2.5×10−50.5×10−3 ⇒s=2.5cm=0.025m \begin{aligned} & s=\dfrac{100\times {{10}^{-2}}(1.5-1)\times 2.5\times {{10}^{-5}}}{0.5\times {{10}^{-3}}} \\\ & \Rightarrow s=2.5cm=0.025m \\\ \end{aligned}
Therefore, the shift that takes places in the interference pattern is 2.5cm2.5cm. Hence, the correct option is (B).

Note: The interference pattern contains alternate bright and dark fringes. The intensity of bright fringes is maximum while the intensity of dark fringes is minimum. All the bright and dark fringes in the interference pattern have equal size throughout. Sources with constant frequency are called coherent sources. No interference pattern is observed if non coherent sources are used.