If a torch is used in place of monochromatic light in Young's experiment,what will happen?

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.

Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

In Young's double slit experiment,the slits are horizontal. The intensity at a point $P$ on the screen is $\frac{I_0}{4}$,where $I_0$ is the maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $d = 2 \lambda$,then the value of $\theta$ is:

In a Young's double slit experiment,two slits are separated by $2 \, mm$ and the screen is placed $1 \, m$ away. When light of wavelength $500 \, nm$ is used,the fringe separation will be ........ $mm$.

Light consisting of plane waves of wavelengths $\lambda_1 = 8 \times 10^{-5} \ cm$ and $\lambda_2 = 6 \times 10^{-5} \ cm$ generates an interference pattern in Young's double-slit experiment. If $n_1$ denotes the $n_1^{\text{th}}$ dark fringe due to light of wavelength $\lambda_1$ which coincides with the $n_2^{\text{th}}$ bright fringe due to light of wavelength $\lambda_2$, then: