JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2023
- Hard
- Numerical
In Young's double slit experiment, two slits \(S_1\) and \(S_2\) are ' \(d\) ' distance apart and the separation from slits to screen is \(D\) (as shown in figure). Now if two transparent slabs of equal thickness \(0.1\,mm\) but refractive index \(1.51\) and \(1.55\) are introduced in the path of beam \((\lambda=4000 \mathring A)\) from \(S_1\) and \(S_2\) respectively. The central bright fringe spot will shift by \(..........\) number of fringes.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
Path difference at \(P\) be \(\Delta x\)
\(\Delta x =\left(\mu_2-\mu_1\right) t\)
\(=(1.55-1.51) 0.1\,mm\)
\(=0.04 \times 10^{-4}\)
\(\Delta x =4 \times 10^{-6}=4\,\mu m\)
\(y =\frac{\Delta xD }{ d }=4 \times 10^{-6} \frac{ D }{ d }\)
\(\{ y\) is the distance of central maxima from geometric center \(\}\)
fringe width \((\beta)=\frac{\lambda D}{d}=4 \times 10^{-6} m \frac{ D }{ d }=4\,\mu m \frac{ D }{ d }\)
Central bright fringe spot will shift by ' \(x\) '
Number of shift \(=\frac{y}{\beta}\)
\(=\frac{4 \times 10^{-6} D / d }{4 \times 10^{-7} D / d }=10\,Ans\)
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