$A$ bulb is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance $d$ ahead of the second wall. The required focal length will be:

Write the lens maker's equation for a thin lens.

An equiconvex lens of radius of curvature $14 \ cm$ is made up of two different materials. The left half and right half of the vertical portion are made up of materials with refractive indices $1.5$ and $1.2$ respectively,as shown in the figure. If a point object is placed at a distance of $40 \ cm$,calculate the image distance. (in $cm$)

$A$ thin biconvex lens is made of glass $(\mu = 1.50)$ and both surfaces have a radius of curvature of $20 \ cm$. $A$ beam of incident light is parallel to the principal axis of the lens. The lens converges it at a distance $L \ cm$. Then $L = \dots$

An equiconvex lens made of glass of refractive index $\frac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\frac{4}{3}$. The percentage change in the focal length 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?