MHT CET · Physics · Wave Optics
If fringe width is \(0.4 \mathrm{~mm}\), the distance between fifth bright and third dark band on same side is
- A \(1 \mathrm{~mm}\)
- B \(2 \mathrm{~mm}\)
- C \(3 \mathrm{~mm}\)
- D \(4 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(A) \(1 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Position of \(n\) th bright fringe from central
\(
\begin{array}{l}
\text { maxima } x_{r_{1}}=\frac{n_{1} \lambda D}{d} \text { here } n_{1}=5 \\
\therefore \quad x_{n_{4}}=\frac{5 \lambda D}{d}
\end{array}
\)
Position of \(n\) th dark fringe from central maxima
\(
\begin{aligned}
x_{n} &=\frac{(2 n-1) \lambda D}{2 d}, \text { Here } n=3 \\
x_{n} &=\frac{5}{2} \frac{\lambda D}{d} \\
x_{r_{4}}-x_{n} &=\frac{2.5 \lambda D}{d}=2.5 \beta
\end{aligned}
\)
Given \(\quad \beta=0.4 \mathrm{~mm}\)
\(
\Rightarrow \quad x_{n_{1}}-x_{n}=1 \mathrm{~mm}
\)
\(
\begin{array}{l}
\text { maxima } x_{r_{1}}=\frac{n_{1} \lambda D}{d} \text { here } n_{1}=5 \\
\therefore \quad x_{n_{4}}=\frac{5 \lambda D}{d}
\end{array}
\)
Position of \(n\) th dark fringe from central maxima
\(
\begin{aligned}
x_{n} &=\frac{(2 n-1) \lambda D}{2 d}, \text { Here } n=3 \\
x_{n} &=\frac{5}{2} \frac{\lambda D}{d} \\
x_{r_{4}}-x_{n} &=\frac{2.5 \lambda D}{d}=2.5 \beta
\end{aligned}
\)
Given \(\quad \beta=0.4 \mathrm{~mm}\)
\(
\Rightarrow \quad x_{n_{1}}-x_{n}=1 \mathrm{~mm}
\)
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