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Question: Answered & Verified by Expert
Two slits separated by a distance of $1 \mathrm{~mm}$ are illuminated with light of wavelength $6.5 \times 10^{-7} \mathrm{~m}$. The interference fringes are observed on a screen placed at $1 \mathrm{~m}$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to
PhysicsWave OpticsTS EAMCETTS EAMCET 2023 (13 May Shift 2)
Options:
  • A $0.655 \mathrm{~mm}$
  • B $1.625 \mathrm{~mm}$
  • C $3.125 \mathrm{~mm}$
  • D $4.785 \mathrm{~mm}$
Solution:
1602 Upvotes Verified Answer
The correct answer is: $1.625 \mathrm{~mm}$
Distance between two slit, $\mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$ Light of wavelength, $\lambda=6.5 \times 10^{-7} \mathrm{~m}$ Distance of screen from the slit, $D=1 \mathrm{~m}$ For distance of dark fringe,
$$
\mathrm{y}_{\mathrm{n}}=\left(\mathrm{n}-\frac{1}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{d}}
$$
take $\mathrm{n}=3$
$$
\begin{aligned}
& y_3=\left(3-\frac{1}{2}\right) \frac{6.5 \times 10^{-7} \times 1}{10^{-3}} \\
& =\frac{5}{2} \times 6.5 \times 10^{-4} \\
& =16.25 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
For distance of bright fringe, $y_n=\frac{n \lambda D}{d}$ take, $\mathrm{n}=5$
$$
\begin{aligned}
& y_5=\frac{5 \times 6.5 \times 10^{-7} \times 1}{10^{-3}} \\
& =32.5 \times 10^{-4} \mathrm{~m} \\
& y_5-y_3=32.5 \times 10^{-4}-16.25 \times 10^{-4} \\
& =16.25 \times 10^{-4} \mathrm{~m} \\
& =1.625 \mathrm{~mm} .
\end{aligned}
$$

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