SATHEE: Optics 6 Question 45

Optics 6 Question 45

47. In a Young’s double slit experiment, two wavelengths of $500 n m$ and $700 n m$ 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. $3.5 m m$

Solution:

Let $n_{1}$ bright fringe corresponding to wavelength $λ_{1} = 500 n m$ coincides with $n_{2}$ bright fringe corresponding to wavelength $λ_{2} = 700 n m$ .

$\begin{aligned} ∴ & n_{1} \frac{λ_{1} D}{d} & = n_{2} \frac{λ_{2} D}{d} \\ or & \frac{n_{1}}{n_{2}} & = \frac{λ_{2}}{λ_{1}} = \frac{7}{5} \end{aligned}$

This implies that 7th maxima of $λ_{1}$ coincides with 5th maxima of $λ_{2}$ . Similarly 14 th maxima of $λ_{1}$ will coincide with 10th maxima of $λ_{2}$ and so on.

$∴$ Minimum distance $= \frac{n_{1} λ_{1} D}{d} = 7 \times 5 \times 10^{- 7} \times 10^{3}$

$= 3.5 \times 10^{- 3} m = 3.5 m m$

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Optics 6 Question 45

47. In a Young’s double slit experiment, two wavelengths of 500nm and 700nm were used. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d=103. Symbols have their usual meanings.

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47. In a Young’s double slit experiment, two wavelengths of $500 n m$ and $700 n m$ 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.