Question

When one of the slits in Young's experiment is covered with a transparent sheet of thickness $3.6\times {10}^{-3}$ cm the central fringe shifts to a position originally occupied by the 30th brigth fringe. If $\lambda =6000A,$ the refractive index of the sheet is

• A. 1.5
• B. 1.55
• C. 1.60
• D. 1.65

Answer 1

The correct option is A 1.5

The position of the 30th bright fringe is given by
$y_{30}=30\frac{\lambda D}{d}Hence,theshiftofthecentralfringeis{y}_0=30\frac{\lambda D}{d}But{y}_0=\frac{D}{d}(\mu -1)t\therefore 30\frac{\lambda D}{d}=\frac{D}{d}(\mu -1)tor(\mu -1)=\frac{30\lambda }{t}=\frac{30\times (6000\times {10}^{-10})}{(3.6\times {10}^{-5})}=0.5\therefore \mu =1.5$