Time taken by the sunlight to pass through a slab of $4 \ cm$ and refractive index $1.5$ is . . . . . . $s$.

$A$ convex lens with lateral magnification $2$ is used to image a point at the bottom of a tank. The image of the point is formed $60 \ cm$ above the lens. Now a liquid is filled into the tank to a height of $24 \ cm$. It is found that the distance of the image of the same point is now $120 \ cm$ above the lens. Find the refractive index of the liquid.

$A$ beam of light is converging towards a point $I$ on a screen. $A$ plane glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of the beam. The convergence point is shifted by:

In determining the refractive index of a glass slab using a travelling microscope,the following readings are tabulated:
$(a)$ Reading of travelling microscope for ink mark $= 5.123 \ cm$
$(b)$ Reading of travelling microscope for ink mark through glass slab $= 6.123 \ cm$
$(c)$ Reading of travelling microscope for chalk dust on glass slab $= 8.123 \ cm$
From the data,the refractive index of a glass slab is:

$A$ coin is placed at the bottom of a glass slab of refractive index $3$ and thickness $x$. Another glass slab of refractive index $\mu$ and thickness $x$ is placed on top of it. If the coin appears to be at the interface of the two slabs,then $\mu = $ . . . . . .

The figure shows a transparent slab of length $1 \, m$ placed in air,whose refractive index in the $x$-direction varies as $\mu = 1 + x^2$ for $0 \leq x \leq 1$. The optical path length of ray $R$ will be: