Question

Hello!

Here is the list of questions :

1. An object of 6 cm height is placed at a distance of 30 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror, will the image be formed? what are the characteristics of the image ?

Answer 1

HEYA ...
Answr...
Here, u=-6cm, f=+12cm , v=?

Using mirror formula

1/f = 1/v+1/u

1/v = 1/f - 1/u

1/v = 1/12 - 1/-6

1/v =1/4

V=4cm behind the mirror

So the nature of the image is virtual ,diminished.

Answer 2

HEYA!!!!

Here is your answer :
______________________________

Given,

focal length = 10 cm

object distance = -30cm

height = 6cm

After sign convention the data becomes,

f = -10cm

u = -30cm

ho = +6cm

v = ?

According to mirror formula,


$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ = > - \frac{1}{10} = \frac{1}{v} + \ \: \frac{1}{( - 30)} \\ \\ \\ = > - \frac{1}{10} = \frac{1}{v} - \frac{1}{30} \\ \\ = > \frac{1}{v} = \frac{1}{30} - \frac{1}{10} \\ \\ = > \frac{1 - 3}{30} = - \frac{2}{30} = - \frac{1}{15} \\ \\ = > v \: = - 15cm$

The image is formed at a distance of 15cm in front of the concave mirror and the image is real as it is formed on the same side of the object.

Magnification,


$m \: = \frac{h1}{h2} = - \frac{ v}{u} \\ \\ = > \frac{ - ( - 15)}{( - 30)} \\ \\ = > - \frac{15}{30} \\ \\ = > - \frac{1}{2} \\ \\ i.e \: \frac{h1}{h2} = - \frac{1}{2} \\ \\ = > \frac{h1}{6 } = - \frac{1}{2} \\ \\ = > h1 = \frac{ - 6}{2} = - 3cm$

As, h1 = - 3cm or m = -1/2 ( negative ) the image is inverted and diminished.

Hence, a real, inverted and diminished image is formed.


HOPE THIS HELPS U. . . .