$A$ convex lens is made of glass $(\mu_g = 1.5)$ and has a focal length of $4 \, cm$ in air. If it is immersed in water $(\mu_w = 1.33)$,what will be its new focal length in $cm$?

$A$ lens has a power of $-4.0 \text{ Diopter}$. This means . . . . . . .

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?

$A$ thin convex lens made from crown glass $\left( \mu = \frac{3}{2} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$,it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is:

$A$ convex lens made of glass (refractive index $= 1.5$) has a focal length of $24 \ cm$ in air. When it is totally immersed in water (refractive index $= 1.33$),its focal length changes to: (in $cm$)

$A$ boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length $10\, cm$. The diameter of the sun is $1.39 \times 10^9\, m$ and its mean distance from the earth is $1.5 \times 10^{11}\, m$. What is the diameter of the sun's image on the paper?