MHT CET · Physics · Wave Optics
A double slit experiment is immersed in water of refractive index 1.33. The slit separation is \(1 \mathrm{~mm}\), distance between slit and screen is \(1.33 \mathrm{~m}\) The slits are illuminated by a light of wavelength \(6300 Å\). The fringe width is
- A \(4.9 \times 10^{-4} \mathrm{~m}\)
- B \(5.8 \times 10^{-4} \mathrm{~m}\)
- C \(6.3 \times 10^{-4} \mathrm{~m}\)
- D \(8.6 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(6.3 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \lambda_{\text {liquid }}=\frac{\lambda_{\text {sir }}}{\mu} \\
& \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33}
\end{aligned}
\)
Fringe width,
\(
\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}
\)
\begin{aligned}
& \lambda_{\text {liquid }}=\frac{\lambda_{\text {sir }}}{\mu} \\
& \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33}
\end{aligned}
\)
Fringe width,
\(
\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}
\)
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