The distance between an object and the screen is $100 \ cm$. $A$ lens produces an image on the screen when placed at either of the positions $40 \ cm$ apart. The power of the lens is

$A$ concave-convex lens of refractive index $1.5$ has radii of curvature of its surfaces as $30 \ cm$ and $20 \ cm$,respectively. The concave surface is facing upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid-glass combination will be

$A$ square card of side length $1\, mm$ is being seen through a magnifying lens of focal length $10\, cm$. The card is placed at a distance of $9\, cm$ from the lens. The axis is perpendicular to the plane of the card. The apparent area of the card through the lens is......$cm^2$

Derive the thin lens equation.

The figure shows a thin lens with centres of curvature $C_1$ and $C_2$. Find its focal length in $cm$. (Take refractive index $\mu = 1.5$)

$A$ thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$,as shown in the figure. The radii of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1 = n_2 = n$. The focal length is $f + \Delta f$ when $n_1 = n$ and $n_2 = n + \Delta n$. Assuming $\Delta n \ll (n - 1)$ and $1 < n < 2$,which of the following statement$(s)$ is/are correct?
$(1)$ The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
$(2)$ $\left|\frac{\Delta f}{f}\right| < \left|\frac{\Delta n}{n}\right|$
$(3)$ For $n = 1.5, \Delta n = 10^{-3}$ and $f = 20 \text{ cm}$,the value of $|\Delta f|$ will be $0.04 \text{ cm}$.
$(4)$ If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$.