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

$A$ rod of length $2 \text{ cm}$ makes an angle of $\frac{2 \pi}{3} \text{ rad}$ with the principal axis of a thin convex lens. The lens has a focal length of $10 \text{ cm}$ and is placed at a distance of $\frac{40}{3} \text{ cm}$ from the object as shown in the figure. The height of the image is $\frac{30 \sqrt{3}}{13} \text{ cm}$ and the angle made by it with respect to the principal axis is $\alpha \text{ rad}$. The value of $\alpha$ is $\frac{\pi}{n} \text{ rad}$,where $n$ is:

$A$ lens of focal length $f$ and diameter $d$ forms an image of intensity $I$. If the central part of the lens of diameter $\frac{d}{2}$ is covered with a paper,what will be the new focal length and the new intensity of the image?

$A$ convex lens:

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

In an optics experiment,with the position of the object fixed,a student varies the position of a convex lens and for each position,the screen is adjusted to get a clear image of the object. $A$ graph between the object distance $|u|$ and the image distance $|v|$,from the lens,is plotted using the same scale for the two axes. $A$ straight line passing through the origin and making an angle of $45^{\circ}$ with the $x$-axis meets the experimental curve at $P$. The coordinates of $P$ will be