If yellow light in the Young's double slit experiment is replaced by red light,the fringe width will

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is:

In Young's double-slit experiment,the width of one slit is twice that of the other. What is the ratio of the maximum to the minimum intensity?

In Young's double slit experiment using sodium light $(\lambda_1 = 5898 \ \text{\AA})$, $92$ fringes are seen. If a light of wavelength $\lambda_2 = 5461 \ \text{\AA}$ is used instead, how many fringes will be seen in the same field of view?

Two parallel slits $0.6 \,mm$ apart are illuminated by a light source of wavelength $6000 \,\mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \,m$ away from the slits is: