MHT CET · Physics · Wave Optics
A double slit experiment is immersed in water of refractive index \(1 \cdot 33\). The slit separation is \(1 \mathrm{~mm}\), distance between slit and screen is \(1 \cdot 33 \mathrm{~m}\). The slits are illuminated by a light of wavelength \(6300 Å\). The fringewidth is
- A \(6.9 \times 10^{-4} \mathrm{~m}\)
- B \(6.3 \times 10^{-4} \mathrm{~m}\)
- C \(5 \cdot 8 \times 10^{-4} m\)
- D 8.6 x \(10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(6.3 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
fringe width \(=X=\frac{\lambda_{W} D}{d}\)
\(\lambda_{w}=\frac{6.3 \times 10^{-7}}{1.33}, \quad D=1 \mathrm{~m}, \quad \mathrm{~d}=10^{-3} \mathrm{~m} \quad\) [on substituting and solving ]
\(X=6.3 \times 10^{-4} \mathrm{~m}\)
\(\lambda_{w}=\frac{6.3 \times 10^{-7}}{1.33}, \quad D=1 \mathrm{~m}, \quad \mathrm{~d}=10^{-3} \mathrm{~m} \quad\) [on substituting and solving ]
\(X=6.3 \times 10^{-4} \mathrm{~m}\)
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