Optics 6 Question 45
47. In a Young’s double slit experiment, two wavelengths of $500 nm$ and $700 nm$ were used. What is the minimum distance from the central maximum where their maximas coincide again? Take $D / d=10^{3}$. Symbols have their usual meanings.
(2004, 4M)
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Answer:
Correct Answer: 47. $3.5 mm$
Solution:
- Let $n _1$ bright fringe corresponding to wavelength $\lambda _1=500 nm$ coincides with $n _2$ bright fringe corresponding to wavelength $\lambda _2=700 nm$.
$$ \begin{aligned} \therefore & & n _1 \frac{\lambda _1 D}{d} & =n _2 \frac{\lambda _2 D}{d} \\ \text { or } & & \frac{n _1}{n _2} & =\frac{\lambda _2}{\lambda _1}=\frac{7}{5} \end{aligned} $$
This implies that 7th maxima of $\lambda _1$ coincides with 5th maxima of $\lambda _2$. Similarly 14 th maxima of $\lambda _1$ will coincide with 10th maxima of $\lambda _2$ and so on.
$\therefore$ Minimum distance $=\frac{n _1 \lambda _1 D}{d}=7 \times 5 \times 10^{-7} \times 10^{3}$
$$ =3.5 \times 10^{-3} m=3.5 mm $$
