Question

a object is placed at a distance of 10cm from a concave mirror of focal length 10 cm. Find the i) position of image ii) nature of image iii) size of image.​

Answer 1

Answer:

• position of image (v) = 10 cm

• virtual and erect

• same size as the object

Step-by-step explanation:

Object distance(u) = -10 cm

Focal length(f) = -10 cm

Image distance (v) = ?

size of the image = ?

Now by the mirror formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{- 10} = \frac{1}{v} + \frac{1}{ - 10}$

$\frac{1}{v} = \frac{1}{ - 10} - \frac{1}{ - 10}$

$\frac{1}{v} = \frac{ - 1 -( - 1) } {10}$

$\frac{1}{v } = \frac{ - 1 + 1}{10}$

$v = 10cm \\ that \: means \: image \: is \: formed \\ 10cm \: behind \: the \: mirror$

Now we will find magnification(m)

$m = \frac{ - v}{u}$

$m = \frac{ - 10}{ - 10 } \\ m = 1$

Here magnification is positive that means image

is virtual and erect.

If magnification = 1

that means( hi )=( ho ).

Hence size of image will be as same as the size of the object.