KCET · Physics · Dual Nature of Matter
A fringe width of a certain interference pattern is \( \beta=0.002 \mathrm{~cm} \). What is the distance of \( 5^{\text {th }} \) dark
fringe from center?
- A \( 1 \times 10^{-2} \mathrm{~cm} \)
- B \( 11 \times 10^{-2} \mathrm{~cm} \)
- C \( 1.1 \times 10^{-2} \mathrm{~cm} \)
- D None of the above
Answer & Solution
Correct Answer
(D) None of the above
Step-by-step Solution
Detailed explanation
Given, fringe width \( \beta=0.002 \mathrm{~cm} \); distance of \( n^{\text {th }} \) fringe is given as
\( x_{n}=(2 n+1) \frac{\lambda D}{2 d} \)
We know fringe width is \( \beta=\frac{\lambda D}{2} \)
For \( 5 \) th dark fringe, \( n=4 \)
\( \alpha_{n}=(2 \times 4+1) \times \frac{0.002}{2}=9 \times 0.001=0.009 \mathrm{~cm} \)
\( x_{n}=(2 n+1) \frac{\lambda D}{2 d} \)
We know fringe width is \( \beta=\frac{\lambda D}{2} \)
For \( 5 \) th dark fringe, \( n=4 \)
\( \alpha_{n}=(2 \times 4+1) \times \frac{0.002}{2}=9 \times 0.001=0.009 \mathrm{~cm} \)
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