Wave OpticsMedium
Question
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n( = 1.5)$, the central fringe shifts by 0.2 cm . The value of t is $\_\_\_\_$ cm .
Options
A.$8 \times 10^{- 4}$
B.$6.0 \times 10^{- 3}$
C.$5.6 \times 10^{- 4}$
D.$5.0 \times 10^{- 3}$
Solution
$dsin\theta = (\mu - 1)t$
$$d\left\lbrack \frac{x}{D} \right\rbrack = (\mu - 1)t$$
$$\begin{matrix} t & \ = \frac{xd}{D(\mu - 1)} \\ & \ = \frac{(0.2)(0.1)}{50(1.5 - 1)} \end{matrix}$$
$$t = 8 \times 10^{- 4}\text{ }cm$$
Create a free account to view solution
View Solution FreeMore Wave Optics Questions
Properly arrange the effects with their reasons Effects Reasons(i) Rainbow formation (a) Reflection(ii) Red colour of da...For Fraunhoffer single slit diffraction :...M1 and are plane mirrors and kept parallel to each other. At point O there will be a maxima for wavelength λ. Light...Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. ...In interference experiment the path difference between two interfering waves at a point A on the screen is $\lambda/3$, ...