$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.
- AThe intensity at $O$ is $4I_o$.
- BThe intensity at $O$ is zero.
- CThe intensity at a point on the screen $4 \ mm$ from $O$ is $4I_o$.
- DThe intensity at a point on the screen $4 \ mm$ from $O$ is zero.
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In an ideal Young's double-slit experiment,a glass plate of thickness $t$ and refractive index $\mu = 1.5$ is placed in the path of one of the interfering beams. If the central bright fringe shifts to the position originally occupied by the first bright fringe (corresponding to wavelength $\lambda$),then the minimum thickness $t$ of the glass plate is:
Medium
View SolutionIn a double slit experiment,when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits,the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is ($\lambda$ is the wavelength of the light used).
In Young's double slit experiment, two slits $S_1$ and $S_2$ are $d$ distance apart and the separation from slits to screen is $D$ (as shown in figure). Now, if two transparent slabs of equal thickness $0.1 \, mm$ but refractive indices $1.51$ and $1.55$ are introduced in the path of the beam $(\lambda = 4000 \, \mathring{A})$ from $S_1$ and $S_2$ respectively, the central bright fringe spot will shift by $..........$ number of fringes.
DifficultJEE MAIN 2023
View SolutionIn a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.
Difficult
View SolutionIn 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 . . . . . . $cm$.
DifficultJEE MAIN 2026
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