Saved Bookmarks
| 1. |
In young’s double slit experiment, the fringe width is 12mm. If the entire arrangement is placed in water of refractive index 4/3, then the fringe width becomes (in mm)(A) 16(B) 9(C) 48(D) 12 |
|
Answer» Correct option is (B) 9 For a given light wavelength corresponding a medium of refractive index µ \(\lambda_{med} = \frac{\lambda_{vacuum}}\mu\) and we know that fringe width \(\beta = \frac {\lambda D} d\) Therefore, \(\beta_{med} = \frac{\beta_{vacuum}}\mu= \frac{12}{\frac 43} = 9 mm\) |
|