$A$ 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......$cm^2$

There is an equiconvex glass lens with radius of each face as $R$,$_a{\mu _g} = 3/2$,and $_a{\mu _w} = 4/3$. If there is water in the object space and air in the image space,then the focal length is:

$A$ biconvex lens of refractive index $1.5$ has a focal length of $20 \,cm$ in air. Its focal length when immersed in a liquid of refractive index $1.6$ will be:

If the behavior of light rays passing through a convex lens is as shown in the adjoining figure,then:

$A$ photograph of a landscape is captured by a drone camera at a height of $18 \ \text{km}$. The size of the camera film is $2 \ \text{cm} \times 2 \ \text{cm}$ and the area of the landscape photographed is $400 \ \text{km}^2$. The focal length of the lens in the drone camera is: (in $\text{cm}$)

$A$ 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 $\frac{\Delta f}{f} > 0$.