Two objects $P$ and $Q$ are placed at $10 \ cm$ and $30 \ cm$ in front of a convex lens of focal length $20 \ cm$. The correct option for the image of $P$ and $Q$ is:

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 \,cm$. If the radii of curvature are equal, the value of the radius of curvature (in $cm$) is

The focal length of a thin lens made from a material of refractive index $1.5$ is $15 \ cm$. When it is placed in a liquid of refractive index $\frac{4}{3}$,its focal length will be $..........\ cm$.

The formation of a real image using a biconvex lens is shown below:
If the whole setup is immersed in water without disturbing the object and the screen positions,what will one observe on the screen?

Assertion: $A$ double convex lens $(\mu = 1.5)$ has a focal length of $10 \, cm$. When the lens is immersed in water $(\mu = 4/3)$,its focal length becomes $40 \, cm$.
Reason: $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

The word $KVPY$ is written on a board and viewed through different lenses such that the board is at a distance beyond the focal length of the lens. Ignoring magnification effects,consider the following statements:
$(I)$ First image has been viewed from the planar side of a plano-convex lens and second image from the convex side of a plano-convex lens.
$(II)$ First image has been viewed from the concave side of a plano-concave lens and second image from the convex side of a plano-convex lens.
$(III)$ First image has been viewed from the concave side of a plano-concave lens and second image from the planar side of a plano-convex lens.
$(IV)$ First image has been viewed from the planar side of a plano-concave lens and second image from the convex side of a plano-convex lens.
Which of the above statements are correct?