Four light waves are represented by:
$(i)$ $y = a_1 \sin \omega t$
(ii) $y = a_2 \sin (\omega t + \phi)$
(iii) $y = a_1 \sin 2\omega t$
(iv) $y = a_2 \sin 2(\omega t + \phi)$
Interference fringes may be observed due to the superposition of:

The ratio of the amplitudes of two interfering waves is $3 : 4$. What is the ratio of their intensities?

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: The phase difference of two light waves changes if they travel through different media having the same thickness,but different indices of refraction.
Reason $R$: The wavelengths of waves are different in different media.
In the light of the above statements,choose the most appropriate answer from the options given below.

Two coherent sources of intensities,$I_1$ and $I_2$ produce an interference pattern. The maximum intensity in the interference pattern will be

In an interference experiment, the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda/6$. The ratio of intensities at points $A$ and $B$ is . . . . . . .

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_{1}$ through a medium of refractive index $n_{1}$ while the second wave travels a path of length $L_{2}$ through a medium of refractive index $n_{2}$. After this the phase difference between the two waves is: