JEE Main 202125 JulyMorning Shift
- Exam
- JEE Main
- Year
- 2021
- Date
- 25 July
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Wave Optics
JEE Main 2021 · 25 July · Morning Shift · Physics · Wave Optics
In the Young's double slit experiment, the distance between the...
In the Young's double slit experiment, the distance between the slits varies in time as ; where and are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:
Options
- A\dfrac{2\lambdaD\left({d}_{0}\right)}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}
- B\dfrac{2\lambdaD{a}_{0}}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}
- C
- D
Correct answer
B. \dfrac{2\lambdaD{a}_{0}}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}
Step-by-step solution
Fringe width,
{\beta}_\max ⇒ {d}_\min and {\beta}_\min ⇒ {d}_\max
{d}_\max = {d}_{0} + {a}_{0} and {d}_\min = {d}_{0} - {a}_{0}
∴ {\beta}_\min = \dfrac{\lambdaD}{{d}_{0} + {a}_{0}} and ∴ {\beta}_\max = \dfrac{\lambdaD}{{d}_{0} - {a}_{0}}
{\beta}_\max - {\beta}_\min = \dfrac{\lambdaD}{{d}_{0} - {a}_{0}} - \dfrac{\lambdaD}{{d}_{0} + {a}_{0}} = \dfrac{2\lambdaD{a}_{0}}{{d}_{0}^{2} - {a}_{0}^{2}}