In a double-slit experiment,the angular width of the fringes is $0.20^o$ for sodium light $(\lambda = 5890 \ \mathring{A})$. In order to increase the angular width of the fringes by $10\%$,the necessary change in the wavelength is:

The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen $1 \ m$ away. The wavelength of light used is $600 \ nm$. What is the change in the angular width of the fringe if the entire measurement system is immersed in water (in $^{\circ}$)? (Use refractive index of water as $\mu = \frac{4}{3}$)

The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

In Young's double-slit experiment,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)