MHT CET · Physics · Wave Optics
In Young's double slit experiment, the distance between the slits is \(3 \mathrm{~mm}\) and the slits are \(2 \mathrm{~m}\) away from the screen. Two interference patterns can be obtained on
the screen due to light of wavelength \(480 \mathrm{~nm}\) and \(600 \mathrm{~nm}\) respectively. The
separation on the screen between the \(5^{\text {th }}\) order bright fringes on the two
interference patterns is
- A \(6 \times 10^{-4} \mathrm{~m}\)
- B \(8 \times 10^{-4} \mathrm{~m}\)
- C \(12 \times 10^{-4} \mathrm{~m}\)
- D \(4 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(4 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{d}=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\)
\(D=1 \mathrm{~m} \quad \lambda_{1}=480 \mathrm{~nm} \quad \lambda_{2}=600 \mathrm{~nm}\)
\({ }^{1} \mathrm{y}_{5}=5 \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}\)
\({ }^{2} \mathrm{y}_{5}=5 \frac{\lambda_{2} \mathrm{D}}{\mathrm{d}}\)
\({ }^{2} y_{5}-{ }^{1} y_{5}=\frac{5 D}{d}(600-480) n m\)
\(=\frac{5 \times 2}{3 \times 10^{-3}} \times 120\)
\(=10 \times 10^{-3+1} \times 4=4 \times 10^{5-9}\)
\(=4 \times 10^{-4}=0.4 \mathrm{~mm}\)
\(D=1 \mathrm{~m} \quad \lambda_{1}=480 \mathrm{~nm} \quad \lambda_{2}=600 \mathrm{~nm}\)
\({ }^{1} \mathrm{y}_{5}=5 \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}\)
\({ }^{2} \mathrm{y}_{5}=5 \frac{\lambda_{2} \mathrm{D}}{\mathrm{d}}\)
\({ }^{2} y_{5}-{ }^{1} y_{5}=\frac{5 D}{d}(600-480) n m\)
\(=\frac{5 \times 2}{3 \times 10^{-3}} \times 120\)
\(=10 \times 10^{-3+1} \times 4=4 \times 10^{5-9}\)
\(=4 \times 10^{-4}=0.4 \mathrm{~mm}\)
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