MHT CET · Physics · Magnetic Properties of Matter
In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?
- A It remains same
- B Becomes \(1 / 10\)
- C Becomes \(1 / 20\)
- D Becomes \(1 / 90\)
Answer & Solution
Correct Answer
(C) Becomes \(1 / 20\)
Step-by-step Solution
Detailed explanation
Let \(\lambda\) be wavelength of monochromatic light, \(d\) the distance between coherent sources, and \(D\) the distance between screen and source, then fringe width is
\(\beta=\frac{D \lambda}{d}\)
Given, \(d_{1}=d, D_{1}=D, d_{2}=10 d, D_{2}=\frac{D}{2}\)
\(\therefore \beta_{2}=\frac{\frac{D}{2} \lambda}{10 d}=\frac{D \lambda}{20 d}\)
\(\Rightarrow \beta_{2}=\frac{W}{20}\)
\(\beta=\frac{D \lambda}{d}\)
Given, \(d_{1}=d, D_{1}=D, d_{2}=10 d, D_{2}=\frac{D}{2}\)
\(\therefore \beta_{2}=\frac{\frac{D}{2} \lambda}{10 d}=\frac{D \lambda}{20 d}\)
\(\Rightarrow \beta_{2}=\frac{W}{20}\)
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