MHT CET · Physics · Wave Optics
A double slit experiment is immersed in water of refractive index 1.33. The slit separations \(1 \mathrm{~mm}\) and the 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 \(6.3 \times 10^{-4} \mathrm{~m}\)
- C \(8.6 \times 10^{-4} \mathrm{~m}\)
- D \(5.8 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(6.3 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{D}=1.33 \mathrm{~m} \\
& \lambda=6300 Å=6.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)
\(\lambda_{\mathrm{w}}=\) wavelength in water \(=\frac{6.3 \times 10^{-7}}{1.33} \mathrm{~m}\)
Fringe width,
\(
\mathrm{X}=\frac{\lambda_{\mathrm{w}} \mathrm{D}}{\mathrm{d}}=\frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}=6.3 \times 10^{-4} \mathrm{~m}
\)
\begin{aligned}
& \mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{D}=1.33 \mathrm{~m} \\
& \lambda=6300 Å=6.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)
\(\lambda_{\mathrm{w}}=\) wavelength in water \(=\frac{6.3 \times 10^{-7}}{1.33} \mathrm{~m}\)
Fringe width,
\(
\mathrm{X}=\frac{\lambda_{\mathrm{w}} \mathrm{D}}{\mathrm{d}}=\frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}=6.3 \times 10^{-4} \mathrm{~m}
\)
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