In a double slit experiment,when light of wavelength $400 \; nm$ was used,the angular width of the first minima formed on a screen placed $1 \; m$ away was found to be $0.2^{\circ}$. What will be the angular width of the first minima if the entire experimental apparatus is immersed in water (in $^{\circ}$)? $(\mu_{water} = 4/3)$

If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

$6300 \mathring A$ wavelength light shines on two narrow slits separated by a distance of $1.0 \ mm$ and illuminates a screen at a distance of $1.5 \ m$ away. When one slit is covered by a thin glass plate of refractive index $1.8$ and the other slit by a thin glass plate of refractive index $\mu$,the central maxima shifts by $6^o$. Both plates have the same thickness of $0.5 \ mm$. The value of the refractive index $\mu$ of the plate is:

$A$ double slit experiment is performed with light of wavelength $500\, nm$. $A$ thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

Young's double slit experiment is carried out with two thin sheets of thickness $t = 10.4 \, \mu m$ each and refractive indices $\mu_1 = 1.52$ and $\mu_2 = 1.40$ covering the slits $S_1$ and $S_2$,respectively. If white light of range $400 \, nm$ to $780 \, nm$ is used,then which wavelength will form maxima exactly at point $O$,the centre of the screen?

Light of wavelength $500 \, nm$ is used to form an interference pattern in Young's double-slit experiment. $A$ uniform glass plate of refractive index $1.5$ and thickness $0.1 \, mm$ is introduced in the path of one of the interfering beams. The number of fringes by which the central maximum shifts is