Search any question & find its solution
Question:
Answered & Verified by Expert
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?
Options:
Solution:
1825 Upvotes
Verified Answer
The correct answer is:
Becomes $1 / 20$
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
$\quad \beta=\frac{D \lambda}{d}$
Given, $d_{1}=d, D_{1}=D, d_{2}=10 d, D_{2}=\frac{D}{2}$
$\therefore \quad \quad \beta_{2}=\frac{\frac{D}{2} \lambda}{10 d}=\frac{D \lambda}{20 d}$
$\Rightarrow \quad \beta_{2}=\frac{W}{20}$
$\quad \beta=\frac{D \lambda}{d}$
Given, $d_{1}=d, D_{1}=D, d_{2}=10 d, D_{2}=\frac{D}{2}$
$\therefore \quad \quad \beta_{2}=\frac{\frac{D}{2} \lambda}{10 d}=\frac{D \lambda}{20 d}$
$\Rightarrow \quad \beta_{2}=\frac{W}{20}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.