In the displacement method of a lens,we obtain two images for different positions of the lens. One image is magnified,and the other is diminished. If $m$ is the magnification of the first image,then the focal length of the lens is given by:

$A$ 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 $= 4/3$)

$A$ convex lens is placed between an object and a screen which are at a fixed distance $D$ apart. For one position of the lens,the magnification of the image formed on the screen is $m_1$. When the lens is shifted by a distance $d$,the magnification of the image formed on the same screen is $m_2$. The focal length of the lens is:

$A$ student measures the focal length of a convex lens by obtaining an image of an object pin at a distance '$v$' from the lens,with the object placed at a distance '$u$' from the lens. What will the graph between '$u$' and '$v$' look like?

$A$ real object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between the object and its image will be

Parallel rays are incident on a thick plano-convex lens having radius of curvature $R$,refractive index $\mu$ and thickness $t$. When rays are incident on the plane surface,they converge at a distance $x$ from the plane surface. When rays are incident on the curved surface,they converge at a distance $y$ from the curved surface. Then: