$A$ vessel of depth $d$ is half filled with oil of refractive index $n_1$ and the other half is filled with water of refractive index $n_2$. The apparent depth of this vessel when viewed from above will be

$A$ parallel-sided block of glass with a refractive index of $1.5$ and a thickness of $36 \, mm$ rests on the floor of a tank filled with water (refractive index $= 4/3$). The difference between the apparent depth of the floor at $A$ (where the glass block is present) and $B$ (where only water is present) when seen from vertically above is equal to $....... \, mm$.

$A$ fish rising vertically upward with a uniform velocity of $8\,ms^{-1}$ observes that a bird is diving vertically downward towards the fish with a velocity of $12\,ms^{-1}$. If the refractive index of water is $\frac{4}{3}$,then the actual velocity of the diving bird to pick the fish will be $.......\,ms^{-1}$.

$A$ bird is looking at a fish underwater from the air. $h_1$ is the height of the bird above the water surface and $h_2$ is the depth of the fish below the water surface. If $\mu$ is the refractive index of water with respect to air,then the distance of the fish as observed by the bird is:

Consider three media $P$, $Q$, and $R$ with refractive indices $1$, $1.25$, and $1.5$, respectively. The medium $Q$ having a thickness of $5 \text{ cm}$ is placed between extended media $P$ and $R$. An object $O$ is placed at the centre of medium $Q$. If viewed from medium $P$ near the normal direction, the apparent depth of $O$ is $h_1$. For similar observation from medium $R$, the apparent depth is $h_2$. The value of $|h_1 - h_2|$, in $\text{cm}$, is:

$A$ convex lens is placed $10\, cm$ from a light source and it forms a sharp image on a screen,kept $10\, cm$ from the lens. Now,a glass block (refractive index $\mu = 1.5$) of $1.5\, cm$ thickness is placed in contact with the light source. To get the sharp image again,the screen is shifted by a distance $d$. Then $d$ is: