The far point of a nearsighted person is $6.0 \, m$ from her eyes,and she wears contact lenses that enable her to see distant objects clearly. $A$ tree is $18.0 \, m$ away and $2.0 \, m$ high. How high is the image formed by the contact lenses?

Radii of curvature of a converging lens are in the ratio $1 : 2$. Its focal length is $6 \,cm$ and refractive index is $1.5$. Then its radii of curvature are ......... respectively.

$A$ convex lens forming a real image of magnification $m_1$ on a screen is moved through a distance $x$. $A$ new image of magnification $m_2$ is again formed on the screen. The focal length of the lens is:

$A$ converging lens forms an image of an object on a screen. The image is real and twice the size of the object. If the positions of the screen and the object are interchanged,leaving the lens in the original position,the new image size on the screen is

$A$ turnip sits before a thin converging lens,outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center,you can change the curvature of its front and rear sides. Suppose that the image must be formed on a card which is at a certain distance behind the lens,while you move the turnip away from the lens,then you should

$A$ biconvex lens of radius of curvature $R$ is made up of a variable refractive index $\mu = 2[1 + \frac{|y|}{d}]$. Assume $2d << R$. $A$ point object is placed at a distance $R$ on the principal axis as shown in the figure. If the spread of the image lies over a span of $n$ meters,find the value of $n$.