The inverse square law for illuminance is valid for:

Two plane mirrors of length $L$ are separated by a distance $L$,and a man $M_2$ is standing at a distance $L$ from the connecting line of the mirrors,as shown in the figure. $A$ man $M_1$ is walking in a straight line at a distance $2L$ parallel to the mirrors at a speed $u$. Then,the man $M_2$ at $O$ will be able to see the image of $M_1$ for a total time of:

An electric bulb illuminates a plane surface. The intensity of illumination on the surface at a point $2 \ m$ away from the bulb is $5 \times 10^{-4} \ phot$ $(lumen/cm^2)$. The line joining the bulb to the point makes an angle of $60^{\circ}$ with the normal to the surface. The intensity of the bulb in candela is

$A$ lamp is hanging at a height of $4\,m$ above a table. The lamp is lowered by $1\,m$. The percentage increase in illuminance will be.....$\%$

$A$ beaker of radius $r$ is filled with water (refractive index $\frac{4}{3}$) up to a height $H$ as shown in the figure. The beaker is kept on a horizontal table rotating with angular speed $\omega$. This makes the water surface curved so that the difference in the height of the water level at the center and at the circumference of the beaker is $h$ $(h \ll H, h \ll r)$,as shown in the figure. Take this surface to be approximately spherical with a radius of curvature $R$. Which of the following is/are correct? ($g$ is the acceleration due to gravity)
$(A)$ $R=\frac{h^2+r^2}{2 h}$
$(B)$ $R=\frac{r^2}{2 h}$
$(C)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H}{4}\left(1+\frac{\omega^2 H}{4 g}\right)^{-1}$
$(D)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H}{2}\left(1+\frac{\omega^2 H}{2 g}\right)^{-1}$

$A$ light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in $List-I$ with their values in $List-II$ and choose the correct option.

$List-I$	$List-II$
$(P)$ If $n=2$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be	$(1)$ $30^{\circ}$ and $0^{\circ}$
$(Q)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be	$(2)$ $60^{\circ}$ and $0^{\circ}$
$(R)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\phi_0$ will be	$(3)$ $45^{\circ}$ and $0^{\circ}$
$(S)$ If $n=\sqrt{2}$ and $\theta_0=45^{\circ}$,then all the possible values of $\alpha$ will be	$(4)$ $150^{\circ}$
	$(5)$ $0^{\circ}$