The focal length of a glass convex lens is $2 \, cm$. What will be its focal length in $cm$ when it is immersed in a liquid of refractive index $1.25$?

$A$ thin converging lens of focal length $25 \,cm$ forms a sharp image of an object on a screen placed at a distance of $75 \,cm$ from the lens. Later, the screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object is to be shifted so that its image on the screen is sharp again is

In the displacement method,there are two positions of a lens for which we get a clear image on a screen. If the first position of the lens is at $40 \, cm$ from the object and the second position is at $80 \, cm$ from the object,the focal length of the lens is ......... $cm$.

The focal lengths of a converging lens are $f_V$ and $f_R$ for violet and red light,respectively. Which of the following relations is correct?

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

$A$ lens is made of glass having an index of refraction $1.5$. One side of the lens is flat and the other side is convex with a radius $R$. If an object is placed $60 \ cm$ towards the convex side of the lens,the image is formed at $120 \ cm$ on the other side of the lens. The value of $R$ is (in $cm$)