In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:
- A$2\lambda$
- B$\frac{2\lambda}{3}$
- C$\frac{\lambda}{3}$
- D$\lambda$
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In a Young's double-slit experiment,a glass slab of thickness $1.2 \, \mu m$ and refractive index $1.5$ is placed in front of one slit. Another slab of thickness $t$ and refractive index $2.5$ is placed in front of the other slit. If the position of the central fringe remains unchanged,then the thickness $t$ is equal to $...... \, \mu m$.
Medium
View Solution$A$ monochromatic beam of light falls on a $YDSE$ apparatus at an angle $\theta$ as shown in the figure. $A$ thin sheet of glass of thickness $t$ and refractive index $\mu$ is inserted in front of the lower slit $S_2$. The central bright fringe (path difference $= 0$) will be obtained:
Medium
View SolutionTo make the central fringe at the centre $O$,a mica sheet of refractive index $1.5$ is introduced. Choose the correct statement$(s)$.
In $YDSE$, a thin film $(\mu=1.6)$ of thickness $0.01 \,mm$ is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the $10^{\text{th}}$ bright fringe earlier. The wavelength of the wave is ......... $\mathring{A}$.
Easy
View SolutionIn a Young's double-slit experiment,a transparent plate of thickness $2.5 \times 10^{-5} \, m$ and refractive index $1.5$ is placed in the path of one of the beams. What will be the shift in the fringe pattern (in $, cm$)? The distance between the two slits $S_1$ and $S_2$ is $0.5 \, mm$ and the distance between the slits and the screen is $100 \, cm$.
Medium
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