The radii of curvature of the faces of a double convex lens are $10\,cm$ and $15\,cm$. Its focal length is $12\,cm$. What is the refractive index of glass?

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

The height of the image formed by a converging lens on a screen is $8\,cm$. For the same position of the object and screen,another image of size $12.5\,cm$ is formed on the screen by shifting the lens. The height of the object is...

Monochromatic light rays parallel to $x$-axis strike a convex lens $AB$. If the lens oscillates such that $AB$ tilts up to a small angle $\theta$ (in radian) on either side of $y$-axis,then the amplitude of oscillation of the image will be ($f =$ focal length of the lens):

$A$ bi-concave symmetric lens made of glass has a refractive index of $1.5$. It has both surfaces with the same radius of curvature $R$. On immersion in a liquid of refractive index $1.75$,it will behave as a:

$A$ thin convex lens is made of a material of refractive index $1.6$. An object is kept at a distance of $u$ from the lens on the principal axis as shown in the figure. The radii of curvature of the surfaces are $10 \, cm$ and $5 \, cm$. Now,the lens is reversed such that the face having radius of curvature $5 \, cm$ lies close to the object. The difference in image position as obtained for both the cases is equal to ......$u$.