# An object is placed 15 cm in front of a convex lens of focal length 10cm. Find the nature and...

## Question:

An object is placed 15 cm in front of a convex lens of focal length 10cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20cm be placed so that the final image is formed at the position of the object itself?

## Convex Lens

A convex lens is a converging lens. A converging lens can produce images that are both real and virtual and any size. To locate the image and image size, you can use the following equations:

{eq}\displaystyle \frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}\\ \displaystyle \frac{h_i}{h_0}=\frac{-s_i}{s_o} {/eq}

1.Convex Lens

We can use the following equations to solve this problem:

{eq}\displaystyle \frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}\\ \displaystyle M = \frac{h_i}{h_0}=\frac{-s_i}{s_o} {/eq}

We know the following:

{eq}s_o = 15\,cm \\ f = 10\,cm {/eq}

The position of the image is calculated by

{eq}\displaystyle \frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f} \\ \displaystyle \frac{1}{15\,cm}+\frac{1}{s_i}=\frac{1}{10\,cm} \\ \displaystyle \frac{1}{s_i}=\frac{1}{10\,cm} - \frac{1}{15\,cm} \\ \displaystyle \frac{1}{s_i}=\frac{1}{10\,cm}\cdot \frac{3}{3} - \frac{1}{15\,cm} \cdot \frac{2}{2} \\ \displaystyle \frac{1}{s_i}=\frac{3}{30\,cm} - \frac{2}{30\,cm} \\ \displaystyle \frac{1}{s_i}=\frac{1}{30\,cm} \\ \displaystyle s_i = \frac{30\,cm}{1} \\ \displaystyle s_i = 30\,cm {/eq}

The image's magnification is

{eq}\displaystyle M =\frac{-s_i}{s_o} \\ \displaystyle M = \frac{-(30\,cm)}{(15\,cm)} \\ \displaystyle M = -2 {/eq}

Therefore, the image is located 30 cm behind the lens, since the image distance is positive meaning the image is located on the opposite side of the object, the image is magnified by 2x and is inverted since the magnification is negative.

2. Concave Mirror

The radius of curvature of a mirror is related to its focal length by

{eq}\displaystyle f = \frac{R}{2} {/eq}

So the focal length of the given concave mirror is

{eq}\displaystyle f = \frac{20\,cm}{2} \\ \displaystyle f = 10\,cm {/eq}

Using the mirror equation, we want the image to be at the object position, that is {eq}d_i = d_o {/eq}:

{eq}\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}\\ \displaystyle \frac{1}{d_o}+\frac{1}{(d_o)}=\frac{1}{10\,cm} \\ \displaystyle \frac{1}{d_o}+\frac{1}{d_o}=\frac{1}{10\,cm}\\ \displaystyle \frac{2}{d_o}=\frac{1}{10\,cm} \\ \displaystyle \frac{2}{d_o}(10\,cm)=1 \\ \displaystyle 2(10\,cm)=1 (d_o)\\ \boldsymbol{ d_o = 20\,cm} {/eq}