You are given two lenses,a converging lens with focal length $+10\, cm$ and a diverging lens with focal length $-20\, cm$. Which of the following would produce a virtual image that is larger than the object?

$A$ plano-convex thin lens is used to obtain the image of a point object $O$ on the screen $S$ as shown in the figure. The thickness of the lens in the middle is $0.5 \ cm$ and the refractive index of the material of the lens is $1.5$. If the separation $D$ between the object and the screen is to be minimum,then the aperture diameter of the lens should be.....$cm$.

When a convex lens is immersed in a liquid of refractive index equal to $80 \%$ of the refractive index of the material of the lens,the focal length of the lens increases by $100 \%$. The refractive index of the liquid is

For an object placed at a distance $2.4\,m$ from a lens,a sharp focused image is observed on a screen placed at a distance $12\,cm$ from the lens. $A$ glass plate of refractive index $1.5$ and thickness $1\,cm$ is introduced between the lens and the screen such that the glass plate faces are parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen (in $,m$)?

$A$ thin lens made of glass of refractive index $\mu_g = 1.5$ has a focal length of $12 \ cm$ in air. If it is immersed in water of refractive index $\mu_w = 4/3$,what will be its new focal length in $cm$?

Explain how the image is formed by a thin convex lens and derive the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.