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  • In a two-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by <img alt="5 \times 10^{-2} \mathrm{~m}" src="ht

In a two-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 \times 10^{-2} \mathrm{~m} towards the slits, the change in fringe width is 3 \times 10^{-5} \mathrm{~m}. If the distance between the slits is 10^{-3} \mathrm{~m},The wave length of the light used is:

Option: 1

5000 \AA


Option: 2

5500 \AA


Option: 3

6000 \AA


Option: 4

6500 \AA


Answers (1)

best_answer

We know that fringe width is given by : \mathrm{\beta=D \lambda / 2 d}

In the given problem, wavelength \lambda  and separation between the slits 2 d is fixed. The fringe width changes due to a change of D (distance of screen from the slits).

\therefore \quad \Delta \beta=\Delta \mathrm{D} \cdot \frac{\lambda}{2 \mathrm{~d}} \text { or } \quad \lambda=\frac{\Delta \beta \cdot 2 \mathrm{~d}}{\Delta \mathrm{D}}
Given that \Delta \mathrm{D}=5 \times 10^{-2} \mathrm{~m}$ (decrease) and $\Delta \beta=3 \times 10^{-5} \mathrm{~m}$ and $2 \mathrm{~d}=10^{-3} \mathrm{~m}.

\therefore \quad \lambda=\frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}}=6 \times 10^{-7} \mathrm{~m}=6000 \AA.
 

Posted by

rishi.raj

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